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Contents:
1. COULOMB’S LAW,
2. ELECTRIC FEILD, ELECTRIC INTENSITY AND CALCULATION.
3. ELECTRIC FLUX AND CALCULATION FLUX DUE TO CHARGE.
4. GAUSS’S LAW ANDTHREE APPLICATIONS.
5. ELECTRIC POTENTIAL ENERGYAND POTENTIAL DIFFERENCE.
6. ABSOLURE POTENTIAL AND ABSOLUTE ENERGY
7. KINETIC ENERGY AND LAW OF CONSERVATION OF ENERGY
8. ELECTRON VOLT , RELATION BETWEEN ELECTRIC FIELD AND POTENTIAL , AND EQUIPOTENTAL SURFACE
9. CAPACITOR
10. CAPACITANCE OF PARALLEL PLATES CAPACITOER
11. COMBINATION OF CAPACITORS
12. VARIABLE CAPACITORS
13. EQUATIONS.
14. DIMENSIONS
15. SHORT QUESTIONS AND ANSWERS
Electrostatics: It is the branch of physics explaining phenomena arising due to the existence of electric
charges, which do not move means, they are static. Electrostatic phenomena arise from the forces that
electric charges exert on each other. Electroscope is device used to detect static charge.
Electrostatic induction: It is a redistribution of electrical charge in an object, caused by the influence of
nearby charges.
Conductors: Materials which contain movable charges that can flow with minimal resistance.
Insulators: Materials with few or no movable charges, or with charges which flow with extremely high
resistance.
Semiconductors: Materials whose behavior ranges between that of a conductor and that of an insulator
under different conditions. Their conducting behavior may be heavily dependent on temperature. They
are useful because we are able to change their conducting behavior to be dependent on many other factors.
The Atom: An atom contains a positively charged nucleus and one or more negatively charged electrons.
The atom exists in three states: neutral, positively charged, and negatively charged. A neutral atom has the
same number of electrons and protons, a positively charged atom has more protons than electrons and a
negatively charged atom has more electrons than protons.
Charge : charge, property of matter that gives rise to all electrical phenomena. The basic unit of charge,
usually denoted by e, is that on the proton or the electron; that on the proton is designated as positive (+e)
and that on the electron is designated as negative (-e).
There are two kinds of charge, positive and negative
like charges repel, unlike charges attract. Charge is conserved
Positive charge comes from having more protons than electrons; negative charge comes from having
more electrons than protons
Charge is quantized, meaning that charge comes in integer multiples of the elementary
Charles Augustine de Coulomb did an experiment in the 1780's, in which he
measured the dependence of the electrostatic repulsion between two charged
spheres as a function of distance between the spheres using a sensitive
torsion balance. Name.- - - - - - - - - - - - - - - -- - S/O , D/O- - - - - - - - - - - - - - - belongs to - - - - - - - - - - - - College - - - - GROUP TIME - - - - - - -GROUP #----------------
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
The electrostatic force between two point charged particles is directly
proportional to the product of the charges on these particles and inversely proportional
to the square of the distance between them and force acts along a straight line joining
the centers of these particles.
q q 
F  12 2
 r 
Explanation:
Consider two point charges q1 and q2 create equal electrostatic force on each other which is
directly proportional to the magnitude of each charge.
F  q1 q 2 
And electrostatic force on each other which is inversely proportional to square of the separation between
their centers.
F
1
r2
Mathematical Derivation:
q1 q 2
r2
q q 
F = K 1 2 2 
 r 
In this equation, “K” is called “Coulomb’s Constant”, its value depend on the medium in which the
electric charges interact.
1
=9×109 N m 2 c-2
That, K =
4 π εo
According to the statement of Coulomb’s law,
F 
The “o” is called “Permittivity of free space”; its value is 8.854210-12 C² N-1 m-², when the charges are
placed in vacuum. Thus, the electrostatic coulomb’s force can be written as,
1  q1 q 2 
, For vacuum.
F=
4 π ε o  r 2 
Effect of medium: When a dielectric medium is completely filled between charges the force between the
same two charge decreases. A dielectric substance has relative permittivity “r”.
1
 q1 q 2 
For air, r = 1(least value).
, For any dielectric medium
F' =
4 π ε o ε r  r 2 
Vector form: As we know that electrostatic force is a vector quantity, then it can be written as,

F' =
1
 q1 q 2 
r
4 π ε o ε r  r 2 
Relative permittivity
F
ε r = It is defined as the ratio of the force in vacuum to the force in any other medium between the same
F'
pair of charges separated by the same distance r. The relative permittivity εr is greater than 1 for any
medium other than vacuum or air.
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 Coulomb:

The coulomb is defined as the quantity of charge that has passed through the
cross-section of an electrical conductor carrying one ampere within one second.
1 Coulomb = 6.3x1018 elementary charges
1 elementary charge = 1.6x10-19 Coulomb.
The electric field can be represented with electric field lines.
Their density is a measure of the strength of the electric field at that point.
1.
They start from a positive charge and end in a negative charge
2.
Field lines are smooth continuous curves without any break.
3.
The tangent to it at any point gives the direction of electric field at that point.
4.
Electric field lines never intersect each other.
5.
They always enter or emerge normal to the surface of a conductor.
6.
They tend to contract longitudinally (on account of attraction between
unlike charges)
7.
They tend to expand laterally (on account of repulsion between like
charges)
8.
Stronger fields have closer lines.
9.
Field strength and line density decreases as move away from the charges.
10. The negative flux just equals in magnitude the positive flux, so that the
net, or total, electric flux is zero.

Electric field of a charge is an area in which the
charge has electric force.
An electric field exists in a region if an electric charge at rest. Since an electric charge
experiences a force if it is in the vicinity of a charged body, there is an electric field
surrounding any charged body, the force due to the charge is detectable. The Electric
field is represented by electric lines of force.

The strength of electric field at a particular point is called as
Electric Field intensity. In other words, the force per unit positive charge
situated at that point is called Electric field intensity.

E =

F
qo
Explanation:
Let's suppose that an electric charge “q” creates an electric field; as the source
charge. The strength of the source charge electric field could be measured by
any
other charge placed in its surroundings as a test charge since it is used to test the field strength. The test
charge has a quantity of charge qo, when placed within the electric field. The magnitude of the electric
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field is simply defined, as the force per charge on the test charge. The electric field
strength is denoted by the symbol E, then the equation can be rewritten as,


F
qo
The unit of electric intensity is NC-1 or Vm-1.Electric field strength is a vector quantity; it
has both magnitude and direction.
F = E qo , is the magnitude of electrostatic force. The direction of
force is the direction of electric intensity.
E =
Explanation:
Let's suppose that two point charges, “q” and “qo” are placed at “r” distance apart as the source and
test charge respectively. The electrostatic force experienced by the charges on each other is given by
according electrostatics coulomb’s force.
Mathematical Derivation:

1  q qo  ^
As we know,
F=
r
4 π ε o  r 2 

F
1  q qo  ^ 1
=
r
qo
4 π ε o  r 2  q o
Divide both sides by q o we get,
1 q  ^
r
4 π ε o  r 2 
1 q 
, is the magnitude of electric intensity.
E=
4 π ε o  r 2 

We get,
E=
Conclusion:

The magnitude of electric intensity is directly proportional to the charge and inversely
proportional to the square of the distance, from the charge to that point.
The electric flux is property of an electric field that is the number of electric lines of
force passing through area. It is the measurement of flow, of electric lines of force through area.



The dot product between electric intensity E and area A is called “Electric Flux”.


 = E .A
Explanation:


Let's suppose that surface of area A be so small that is in the uniform electric field E . The electric flux
passing thorough the surface is as scalar product,


Or
 = E .A
Its magnitude is given by,  = E A Cos 
Electrical flux has units of volt metres (V m), or, N m2 C−1= kg m3 s−3 A−1.
Special cases:
i)
The surface is placed perpendicular on to the electric field, the direction of electric intensity and area
are parallel, say  = 0o
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 = E A Cos 0o
 = E A(1)
Or
 = E A It means, “Maximum” Electric flux passing through the surface.
ii) The surface is placed 45o on to the electric field, the direction of electric intensity and
area are making45o, Therefore,
 = E A Cos 45o
 = E A(0.707)
Or
 = 0.707 E A
iii ) The surface is placed parallel to the electric field the area is perpendicular onto the electric intensity,
say  = 90o
Therefore,
 = E A Cos 90o
=EA(0)
 = 0 It means “minimum” or zero Electric flux passing
through the surface.
iv) The surface is placed perpendicular on to the electric field, the direction
of electric intensity is opposite to the area, say  = 180o
Therefore,
 = E A Cos 180o
 = E A(-1)
Or
 = -E A
Therefore,
Explanation:
Let's suppose that a positive charge “+q” is placed at the center of sphere. It is surrounded by a

closed spherical surface of radius “r”. The uniform electric intensity E produced due to a charge with in

1 q ^
spherical region.
E=
r
4 π ε o  r 2 
The electric lines of force are passing perpendicular through a surface of sphere. The magnitude of electric
intensity is same at “r” distance on the spherical surface area.
Mathematical Derivation:
The electric flux “” through such smaller vector area is given by,


φ = E . A
Δφ = E ΔA Cosθ
Δφ = E ΔA Cos0o
Δφ = E ΔA 1
Δφ = E ΔA
1 q
φ 
A
4 π εo r 2
N
 Δφ=
i =1
Because,
N
 ΔA =A = 4 r2
i
1 q N
 ΔA
4 π ε o r 2 i=1
is the total are of sphere.
i=1
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1 q
4 π r2 
2 
4 π εo r
q
We get, electric flux due to charge, φ 
εo
Conclusion:

The electric flux due to charge placed at the center of sphere is independent of size.
When, q = 0 (no charge), then,  = 0 (no electric flux).
Thus,
φ

Total electric flux through a closed surface enclosing a charge is
1
times
εo
the magnitude of the charge enclosed
e = Q
1
εo
Proof:
Let us consider charge is uniformly distributed on the surface, where q = q1+ q2 + - - - + qn . The
1
1
1
electric flux due to the individual charge is given by, 1 = q1
, 2 = q2
, - - -, n = qn
εo
εo
εo
We know that, electric flux is scalar quantity; the total electric flux e is given by,
 e = 1 + 2 + 3 +- - + n
q
q
q
q
φe = 1 + 2 + 3 + - - - + n
εo
εo εo
εo
1
φe =  q1 + q 2 + q 3 + - - - + q n 
or
εo
1
Hence,
e = ( q )
εo
1
or
e = Q
This is called Gauss’ law.
εo
FIRST
APPLICATION :
Let us consider a spherical shell of radius “r” uniformly charged by charge
“Q”. The intensity of electric field at a point due to this shell depends onto the
position of the point with respect to this shell. The intensity of electric field at
different positions of point is calculated as follows:
a) Electric field of a uniform spherical surface charge at a distance “r” from
the center:
Consider a uniform spherical distribution of charge on a conductor
would be free to move and would end up on the surface. This charge density
is uniform throughout the sphere. Charge Q is uniformly distributed
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throughout a sphere of radius “a”. First consider r > a; the electric field is at a radius r.
That is, find the electric field at a point outside the sphere. Just as for the point charge with
Gauss's Law.
we find  =E A
Mathematical Derivation:
 =E (4  r2)
Q
and we know =
εo
Q
=E(4  r2)
εo
1 Q
E=
4πε o r 2

That is, the electric field outside the sphere is exactly the same as if there were only a point charge Q.
b).When a point is on to the surface of the spherical shell (outside the sphere):
Suppose a point “P” is lying on the spherical shell .The electric intensity, of the electric field, is to
be calculated at the distance “a” from the center at the given shell. Where, a > r, at every point of this inner
sphere, the electric field will be equal and normal on to the surface.
According to the gauss’s law,
Mathematical Derivation:
q
φ=
εo
Hence,
q
εo
q
E ( 4πa² ) =
εo
 E.ΔA
=
(4πa²) σ
εo
q
From the definition of charge density, σ=
A
Therefore,
q =  ( 4 a² )
σ
E=
εo
 This shows that, the electric field intensity at a
point outside the sphere will be same as if the
whole charge is concentrated at the center of
sphere.
c). When a point lies inside the charged hallow sphere:
Consider a hollow conducting sphere with
radius R. Let q be the charge on this sphere. To find the
field at a point P, draw a Gaussian surface (dotted circle)
of radius r. Since, this surface does not enclose any charge, we have.
Mathematical Derivation:
We get,
E ( 4πa² ) =
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According to the Gauss’s law,
 =E A
Where, q = 0, this result is independent of the radius, provided that it is less than that
charged sphere.
Therefore,
E = 0,
And in terms of charge density,  = 0
Conclusion:

Intensity of electric field at every point inside the uniformly charged hollow empty
sphere is zero.
SECOND APPLICATION :
Explanation:
Let's suppose an infinitely large and thin
plane sheet which carries a positive charge a
per unit area. The electric field is perpendicular to the plane.
q
σ=
- - - - - - -► eq. # 1
A
Let  be the “charge density” of an infinite plane sheet.
For Gaussian surface we can choose a point P, is right circular
cylinder with faces parallel to the plane of charge. The field lines
are parallel to the sides of cross sectional area of the cylinder. It
has two faces P and P’ will contribute to the flux as the electric
field lines parallel to them.
Mathematical Derivation:
q  Total charge 
By putting “q “ from
φe  Total electric flux  =
εo
equation # 1 we get, According to the Gauss’s law,
σA
φ e  Total electric flux  =
εo
The flux through the P surface is, p = E A Cos 
p = E A Cos 0o
Or
p = E A( 1 )
p = E A
and similarly the flux through the P’ surface is,
p’ = E A
The total electric flux, “e” through the surface will be,  e = P + P’
or
e = E A + E A
e = 2 E A
σA
By putting the value of “e” in equation # 2, we get, 2 E A =
εo
σ
E=
or
2 εo
We know that, electric intensity is a vector quantity, its direction given by unit vector r ^.
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
Hence,
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E =
σ ^
U
2 εo
Conclusion:

The electric intensity is independent of distance. That is because the field lines
everywhere are straight, parallel, and uniformly spaced. This is because, the sheet is
infinitely large.
THIRD APPLICATION :
Explanation:
Consider two parallel conducting plates are equal size it is assumed that each plate has equal and opposite
charge. One plate has positive charge with charge density “+”, while the other plate negatively charged
with charge density “-”. Imagine a point P is between the two parallel oppositely charged plates. The
electric field due to the charged plate is uniform in the space between them. The electric intensity due to
positively charged plate is given by,

σ ^
E1 = +
U
2 εo

σ ^
U
And due to the negatively charged plate is, E 2 = +
2 εo
The total electric intensity due to the two oppositely charged plates



will be, E = E1  E 2

 σ ^
 σ ^
E = +
U
+


U
 2 εo 
 2 εo 

 σ ^
E = +2 
U
 2 εo 

 σ ^
E= 
U
 εo 
or
The electric intensity due to two oppositely charged plates is double the electric intensity
due to single charged plate.
Conclusion:

The amount of workdone, in displacing the charge
from one point to another against the electric field, is called “Electric
Potential energy”.
Explanation:
Let us consider a charge “+q”, produced an electric field of intensity

“ E ” another test charge “+qo” is placed in its field at point “P”. A test
charge is moved to wards the source charge through smaller

displacement  r , in the opposite direction of electric field. The work done
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between two points is the change of potential energy U.
Mathematical Derivation:


Work = F . d


F
qo
The magnitude of electrical potential energy is given by, U = E qo r Cos
U = E qo r Cos180o
U = -E qo r
UO - UP = E qo .  r
Because, E =

The work done per unit charge in moving a test charge from
point to another is the electrostatic potential difference between the
two points.
Explanation:

Let us consider a charge “+q” produced an electrostatic field E . The charge is to be moved from P to O,
the work be to be done in the opposite direction of electric field.
Mathematical Derivation:


U = E q o .  r
We know that,


E qo .  r
U
=
qo
qo


U
=E .  r
qo


UO
U
- P= E . r
qo
qo
The change in potential energy per unit charge between two fixed points is called “potential difference”.


VO - VP = E .  r


V = E .  r
Thus Potential difference is, “The dot product between electric intensity and displacement”. It is scalar
quantity, denoted by “V” having magnitude,
V = E r Cos
The unit of potential difference is “Volt”, denoted by “V”.
Volt: 

When one-Joule work is to be done in moving one coulomb charge against the
electric intensity, then potential difference between two points is one volt.
work
Potential difference =
charge
1 Joule
1 Volt =
1 Coulomb

The potential at a point in a field is equal
to the work done per coulomb in moving a
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positively charged particle from infinity to that point in the field.
Explanation:

Let us suppose a positive charge “+q”, produced an electric field of intensity E .
It is required to calculate the potential at an isolated point “B” in the electric field at a
distance “r”.
Let us calculate work done in carrying the charge “qo” from infinity point “B” to
that point A. The displacement is to be divided into very smaller equal displacement

elements“  r ”.
Mathematical Derivation:
Where,
r = rB - rA


Work done = F .  r

Or

Work = E q o .  r
Or
Work = E qo r Cosθ
Work = E q o  r Cos 180o
Hence,
Work = - E qo  r
We know that, E =
1
4 π εo
q
 r 2 
1 q
q o  r - - - - - - - -►eq. # (1)
4 π ε o  r 2 
BA
r +r
The average distance between the these two points is given by, rave = A B
2
2
By simplifying and squaring we get, r = rA rB , equation # 1 can be written as,
Work
=-
Or
Work
B A
Or
Work
BA
We get,
Work
BA
1  q 


4 π ε o  rBrA 
1
=q qo
4 π εo
=-
qo  rB - rA 
 rB

 rA rB
1
1
=q qo  4 π εo
 rA
 UA  +  UB  = -
rA 

rA rB 
1

rB 
1 1
1
q qo   
4 π εo
 rA rB 
1
UA
U
1
1
+ B =q  - 
qo
qo
4 π εo
 rA rB 
The electric field at point B is zero, thus electric potential energy at that point also zero, Say, U B = 0 and rB
= .
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Therefore,
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1
UA
0
1
1
=q  - 
qo
qo
4 π ε o  rA  

1 q
VA + 0 =   0
4 π ε o  rA

VA = -
1 q
 
4 π ε o  rA 
If, VA = V then, rA = r
Thus, absolute potential t any isolated point is given by,
V= -
1
4 π εo
q 
 r 
Conclusion:
The potential is directly proportional to the charge and inversely proportional to the distance at that point.

“The amount of work done in bringing the
positive charge from infinity to that point, in the
electric field against the electric intensity, is called “absolute potential energy”.
Mathematical derivation:
1 q 
We know that , V =
4 π ε o  r 
work
The definition of electric potential energy is,
Potential =
charge
Work = Potential at distance ( r )  Charge ( qo )
1 q 
Or
Electric absolute potential energy =
qo
4 π ε o  r 
1  q qo 
Electrical potential energy U =
4 π ε o  r 
Explanation:
Suppose a charge “q” of mass “m” is moving under the
influence of electric field of intensity “E” from one plate to another with the velocity “v”. Hence, a charge
acquires kinetic energy, from the state of rest. It is given by,
Electrical kinetic energy = ½ m v2
According to the law of conservation of energy.
Loss of electric potential energy = Gain of electric kinetic energy
1
V q = m v2
2
1
V e = m v2
or
2
The modern unit of the electrical energy is “electron volt”.

An electron volt is a measure
of energy. An electron volt is the kinetic energy gained by an electron passing through a potential
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difference of one volt “An electron volt (eV) is the energy that an electron gains when it
travels through a potential of one volt”.
Conversion:
Work
We know that, V =
q
Energy (Work) = Vq
1 electron volt (eV) = potential difference  charge of
an electron.
1eV = V (1.6  10-19 C)
J
1eV= (1.610-19 C )
C
1eV = 1.6  10-19 J
An equipotential surface refers to a
surface where the potential is constant. The intersection of an equipotential
surface and a plane results into a path called an equipotential line. No work
is done in moving a charge from one point to the other along an
equipotential line or surface i.e. VA – VB = 0 the electric flux lines and the
equipotential surface and normal to each other.

A potential gradient is the rate of change of the potential
with respect to displacement.
 V 
Potential Gradient = 
 r 
Units are volts per meter (V/m). The electric field is the same as the potential
gradient but with opposite sign.
Relation between Electric Potential Gradient and Electric Intensity:
The relation between potential difference and electric intensity is given by an
equation,
 V 
 r  = - E
The change in potential per unit distance in an electric field is
called “Potential gradient”.
 V 
Potential Gradient = 
 r 
Potential gradient = - E
This is the relation between potential gradient and electric intensity.
The electric field is directed from higher potential to lower potential point while the potential gradient
is directed from lower to higher potential.

A capacitor is a device that stores electric charge on to
parallel plates.
Construction:
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A capacitor get's charged when a battery is connected across the two parallel
plates. The plate A attached to the positive terminal of the battery get's positively charged
+Q and the B plate joined to the negative terminal get's negatively charged -Q. Once
capacitor get's fully charged, flow of charge carriers stops in the circuit and in this condition
potential difference across the plates of capacitor is same as the potential difference across
the terminals of battery .
Explanation:
Suppose total charge Q is stored on the parallel plates and the potential V
of the battery produced across the plates of capacitor. The charge increases to the
potential difference increases applied across the plates or vice versa.
Mathematical Derivation:
Q V
Q=CV
Thus, Capacitance “C” is a measure of the amount of electric charge stored for a given electric
potential V. The capacitance of parallel plate’s capacitor is,
Q
C 
V
The amount of charge stored between the two plates for a potential difference existing across the
plates. It is called “Capacitance of parallel plates capacitor”.
The basic unit of capacitance is a farad.

The electric charge one coulomb stored on the parallel plates of a capacitor per one
Q
volt electric potential, the capacitance is called “One farad”. C =
V
1 coul.
1Farad =
1Volt
Explanation:
A capacitor consists of two parallel conducting plates, each of area” A”,
separated by a distance “d”. When the capacitor is charged, the plates carry equal
amounts of charge, one plate carries positive charge + Q, and the other carries
negative charge – Q on the surface. The electric field lines for a parallel-plate
capacitor that the field is uniform in the central region between the plates.
Mathematical Derivation:
The electric intensity due to two oppositely charged pates is given
by the Gauss’ law,
σ
E=
εo
Q
σ=
We know that,
A
Q
E=
Therefore,
- - - - - - - - -► eq. 1
A εo
The potential difference “V” is applied between the plates, then V = E d
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V
- - - - - - - - -►eq. 2
d
Q
V
=
Compare eq.1 and eq.2 we get,
A εo
d
We know that, Q = C V
CV
V
=
Therefore,
A εo
d
A εo
C=
We get,
d
Conclusion:

The capacitance of a parallel-plate capacitor is directly
proportional to the area of its plates and inversely proportional to the plate
separation between the plates.
Effect of Dielectric:
If the dielectric completely fills the space between the plates,
the capacitance increases by r, which is called the dielectric constant known as relative permittivity. The
capacitance of the capacitor will be,
A εoε r
C' =
d
Thus, we see that a dielectric provides the following advantages:
• Increase in capacitance
• Increase in maximum operating voltage
Or,
E=
There are two methods to find out the equivalent capacitance
using methods described.1.parllel and 2. Series combination
Description:
Two or more
capacitors connected as, the left plates of all the capacitors are connected
by a wire to the positive terminal of the battery and right plates to the
negative terminal of the voltage source. Thus, the potential difference
across each capacitor is same and equal to the potential applied. Hence,
such network connection is called “parallel combination”.
Explanation:
Suppose two capacitors C1 and C2 are connected as shown in
the figure. One plate of both capacitors is connected at junction A, while
the other plate at the B, the junctions is connected with the positive and
negative terminals of source respectively. The voltage V across the capacitors is equal to that across the
battery terminals. The charges on each capacitor is Q1 and Q2
while.
However, different charge on each capacitor is given by,
Mathematical Derivation:
The total charge Q of the combination capacitor, is Q = Q1 + Q2
Q1 = V C1, and Q2 = V C2
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Or
Q = V C1 + V C2
Q =V (C1 + C2)
Q
=  C1 + C 2 
V
These two capacitors are replaced by one equivalent capacitor having a capacitance Ce.
Q
Ce 
Therefore,
V
Thus,
Ce= (C1 + C2)
Conclusion:
The equivalent capacitance of a parallel combination of capacitors is equal to the sum of individual
capacitances of capacitors.
Description:
Two or more capacitors connected as the left
plate of first capacitor and the right plate of last capacitor is
connected to the terminals of a battery. The remaining plates are
connected to each other. The charges on each capacitor connected are
same. Such type of network is called “Series combination of
capacitors”.
Explanation:
Let’s suppose two capacitors C1, and C2 are connected as
shown. If +Q charge are given to left plates of C1, then by induction Q charge appears on the right plate. +Q appears on the left plate of
capacitor C2 and so on. Thus, same charge Q appears on each capacitor. The potential difference across
each capacitor however is different. Therefore,
Mathematical Derivation:
The total potential V of the combination capacitor, is
We know that, Q = C1V1 and Q = C2 V2
Q
Q
V1 = , and V2 =
C2
C1
Q Q
V= 
C1 C2
1
1
V=Q   
 C1 C2 
V 1
1 
=   
Q  C1 C2 
These two capacitors are replaced by one equivalent capacitor having a capacitance Ce.
V
1
Therefore,
=
Q
Ce
1
1
1
=
+
Thus,
Ce
C1 C2
Conclusion:
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This implies that when the capacitors are connected in series, the reciprocal of the
equivalent capacitance equals to the reciprocals of the individual capacitance.
1) Variable capacitor:
Dielectric Capacitors are usually of the variable
type such as used for tuning transmitters, receivers and transistor radios. Variable
capacitors are used for adjustment etc. of frequency. In mechanically controlled variable
capacitors, the distance between the plates, or the amount of plate surface area which
overlaps, can be changed.
The most common form arranges a group of semicircular metal plates on a rotary
axis (“rotor”) that are positioned in the gaps between a set of stationary plates (“stator”)
so that the area of overlap can be changed by rotating the axis. Air or plastic foils can be
used as dielectric material. By choosing the shape of the rotary plates, various functions
of capacitance vs. angle can be created, e.g. to obtain a linear frequency scale. Various
forms of reduction gear mechanisms are often used to achieve finer tuning control, i.e. to spread the
variation of capacity over a larger angle, often several turns.
2)Trimmer capacitors
Trimmer capacitors (trimmers) are miniature variable capacitors. They are designed to be
mounted directly onto the circuit board and adjusted only when the circuit is built. A small
screwdriver or similar tool is required to adjust trimmers. The process of adjusting them requires
patience because the presence of your hand and the tool will slightly change the capacitance of the
circuit in the region of the trimmer. Trimmer capacitors are only available with very small
capacitances, normally less than 100pF. It is impossible to reduce their capacitance to
zero, so they are usually specified by their minimum and maximum values, for example
2-10pF
3) Multiple capacitors:
This type of capacitor consists a mica dielectric. The capacitance is “n” times the
capacitance between two successive plates, where “n” is the number of dielectrics
between all the plates .The whole arrangement is sealed into plastic case.
Very often, multiple stator/rotor sections are arranged behind one another on the same
axis, allowing for several tuned circuits to be adjusted using the same control, e.g. a preselector, an input
filter and the corresponding oscillator in a receiver circuit. Capacitors with multiple sections often include
trimmer capacitors in parallel to the variable sections, used to adjust all tuned circuits to the same
frequency.
4) Paper capacitor:
The film and foil types of capacitors are made from long thin strips of thin metal foil with the dielectric
material sandwiched together which are wound into a tight roll and then
sealed in paper or metal tubes. These film types require a much thicker
dielectric film to reduce the risk of tears or punctures in the film, and is
therefore more suited to lower capacitance values and larger case sizes.
Metallized foil capacitors have the conductive film metallized sprayed
directly onto each side of the dielectric which gives the capacitor selfhealing properties and can therefore use much thinner dielectric films. This allows for
higher capacitance values and smaller case sizes for a given capacitance. Film and foil
capacitors are generally used for higher power and more precise applications.
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5) Electrolytic Capacitor:
Aluminum is used for the electrodes by using a thin oxidization membrane. Large
values of capacitance can be obtained in comparison with the size of the capacitor,
because the dielectric used is very thin. The most important characteristic of electrolytic
capacitors is that they have polarity. They have a positive and a negative
electrode.[Polarised] This means that it is very important which way round they are
connected. If the capacitor is subjected to voltage exceeding its working voltage, or if it is
nnected with incorrect polarity, it may burst. It is extremely dangerous, because it can quite
literally explode. Make absolutely no mistakes.
.
F
1  q1 q 2 
1 q
q q 
1. F=K  1 2 2  =
2. E =
3. F= E q o 4. E =
5. φ = E . A 6. φ = E A Cos θ
2 

qo
4 π ε o  r 2 
 r  4πε o  r 
q  Total charge 
1 
7. φ = q  
8. φMax. Flux = E A
9. φMin.Flux = 0 10.  Charge density  σ =
A
o 
work  energy 
σ
1σ
11. E =  
12.  Potential  V =
13. E =   14. V = E . r 15. ΔV = E Δr Cosθ
q  charge 
2  εo 
 εo 
1
C'
1 q
21. Ve = m v 2
= ε r 19.  Potential energy  U = E q . r 20. V =


2
C
4πε o  r 
A εo εr
1
1
1
1
=
+

23. C'for dielectric =
24. Ce  C1  C2     .
25.
d
Ce
C1
C 2 C3
16. U = V e 17 Q = C V 18.
22. C =
A εo
d
PHYSICAL QUANTITY & SYMBOL
Electric Charge, (Q = I t)
Electric Intensity, E =
F
qo
Electric Potential, (V =E d )
Electric Flux, (  = E A )
DIMENSION
 A T
UNIT
C = amp.sec
 M L T -3 A -1 
N C-1 = kg.m.sec-3.amp-1.
 M L2 T -3 A -1 
 M L3 T -3 A -1 
J-coul -1 = V= kg.m2.sec-3.amp-1.
N - m2 C-1 = Kg. m3. sec-3. amp-1.
A εo
=[Charge]/[Potential]  M -1 L-2 T 4 A 2  Farad = kg-1.m-2sec4.amp2 =A2 C.V-1
d
 M L2 T -2 
Electric potential energy ,U =Ve
Joule or electron volt = kg.m2. sec-2
Capacitance, C =
 A T L
Electric dipole moment
coul.meter = amp.sec. m.
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Dielectric constant, k =
ε
εo
dimensionless
Charge Density, σ =
q
A
Energy, U = V e
 A T
Permittivity of Free Space, ε o =
q1 q 2
4 π F r2
F
qo
Electric Flux, (  = E A )
A εo
d
Electric potential energy ,U =Ve
Electric force, F= E q o
 A 2 T 4 M -1 L-3 
C N -1 m-2 = amp2.sec4.kg-1.m-3
N C-1 = kg.m.sec-3.amp-1.
 M L2 T -3 A -1 
 M L3 T -3 A -1 
N - m2 C-1 = Kg. m3. sec-3. amp-1.
 M -1 L-2 T 4 A 2 
Farad = kg-1. m-2 sec4.amp2.
 M L2 T -2 
 A T L
Electric dipole moment
ε
εo
C = amp.sec
[M L T-3A-1]
Electric Potential, (V =E d )
Capacitance, C =
Coul. / m2
q1 q 2
 A 2 T 4 M -1 L-3  C N -1 m-2 = amp2.sec4.kg-1.m-3
2 
4πFr
 M L2T -2 A -1 T -1  A T = M L2 T -2
kg.m2.sec2. = J
Electric Charge, (Q = I t)
Dielectric constant, k =
N
 A T L-2 
Permittivity of Free Space, ε o =
Electric Intensity, E =
no unit
 MLT -2 C-1   C =  M LT -2 
Electric force, F= E q o
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J-coul -1 = V= kg.m2.sec-3.amp-1.
Joule or electron volt = kg.m2. sec-2
coul.meter = amp.sec. m.
dimensionless
 MLT -2 C-1   C =  M LT -2 
q
 A T L-2 
A
Charge, molar
[Charge]/[Quantity] [T.A.mol-1 ]
Energy flux [Energy]/[Time].
[ M L2 T 3 ]
Charge Density, σ =
no unit
N
m-3.s.A = Coul. / m2
C.mol-1
J.s-1= W Same as power.
Q.No.1 What happens to the force between two charges, if the distance between them is halved?
q q 
Answer
We know that, F   1 2 
 r² 
q q
r
If, r =
then, we know that, F  1 2 2
2
r
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Therefore,
F' 
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q1 q 2
2
r
 
2
q q
F'  1 2 2
r 
4
 
4 q1 q 2
F' 
Or
r2
F'  4 F
This shows that, if the distance between two charges is halved then force increases four times the initial
value.
Q. No.2
What happens to the force between two point charges if each charge is tripled and the
distance between them is made one third?
qq
Answer
We know that, F = K 1 2 2
r
3q 3q
F' = K 1 22
1 
 3 r 
9q q
F' = K 1 2
or
1 2
r
9
81q1q 2
F' = K
r2
F' = 81F . This dhows that force increases 81 times when each charge is tripled and the distance between
them is made one third.
Q.No.3
Two identical metal spheres carry positive charge and negative charge; they touched
together and again separated. What are their charges?
Answer
The identical charged two spheres carrying positive and negative charge are touched
together. They cancel their charge. Hence, they have no any charge after separation.
Q. No.4
Explain why it is so much easier to remove an electron from an atom of large atomic weight
than it is to remove proton?
Answer
The outer electrons of large atomic weight have less force of attraction with protons. Thus an atom
contains number of orbits. So that, it is easier to remove electron from outer most shell an atom of large
atomic weight than proton.
Q.No.5
What will be the flux through a surface placed at right angle to the electric field?
Answer
We know that,  = E A Cos 
 = E A Cos 0o
 = E A, The flux will be maximum through the surface.
Q. No.6
Show that, N C-1 = V m-1
Answer
N C-1 = [N.m.C]. m-1
N C-1 = [J.C]. m-1
N C-1 = V.m-1
Hence shown
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Q. No.7 What will be flux through a closed surface, if it is enclosed with no charge?
q
Answer
We know that, φ =
εo
If, q = 0, then, =o
Hence, the flux will be zero through a closed surface.
Q. No.8
Convert 1 kWh in to joules.
Answer
We know that, Work = 1kilowatthour
1kwh = 1000 w x3600 sec
1kwh = 36x 10 5J.sec / sec
1kwh = 3.6 x10 6J
Q. No.9
Repulsion is a sure test for electrification. Explain?
Answer
Attraction is possible even when one of the bodies is charged. But repulsion is possible
only when BOTH the bodies are electrically charged. The force exerted by one charge over another, to
keep away from one another. Hence, repulsion is the sure test of electrification.
Q. No.10
What are equi potential surfaces?
Answer
The surfaces on which electric lines of force are perpendicular and the value of potentials
are same known as “equi potential surfaces”.
Q. No.11
The work done in moving the one coulomb charge between two points in an electric
field is 10 joules. What is the potential difference between them?
Answer
From the definition of potential,
work
Potential =
charge
10J
Potential =
. hence , Potential = 10 volt
1coul
Q. No.12
What kind of energy a charged capacitor has got?
Answer
A charged capacitor has got electrostatic potential energy.
Q. No.13
Convert 1eV in to joules.
Answer
1eV = 1.6 x 10-19C J / C
1eV = 1.6 x 10-19 J
Q. No.14
Define electron volt?
Answer
It is the electrical energy supplied by the unit potential on to one electronic charge in electric
field.
Q.No.15 How three capacitors, each of 3 micro farad are arranged to obtain minimum capacitance?
Answer
The capacitors are arranged in series to obtained minimum valued capacitance.
1
1
1
1
=
+

Ce
C1
C 2 C3
3
1
1 1 1
=   =
3
Ce
3 3 3
1
3
=
Ce 3
Ce = 1 micro farad
Q.No.16
Name some examples in which capacitors are used?
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Answer
The capacitors are used to store charge and energy; they are used in i)
flashguns ii) stabilizer iii) fans iv) motors v) computers vi) electronic devices, etc.
Q. No.17 How two or more capacitors of equal values, should be connected to obtain
maximum capacitance?
Answer: To obtain maximum capacitance at constant potential, the capacitors are
connected in parallel.
Ce = C 1 + C 2
Q. No.18 A capacitor is connected across battery. Why does each plate receive a charge of same
magnitude? Will it be true even if the plates are of different sizes?
Answer
When a capacitor is connected across the battery, pates of a capacitor receive a charge of same
magnitude due to induction. This is also true even when, the plates are of different sizes.
Q.No.19 If the absolute potential is zero at a point, must the electric field be zero there as well?
Answer
If the absolute potential is zero at a point, the electric field, intensity must be zero there as well.
Q.No.20 Will solid metal sphere hold a large electric charge than a hollow sphere of the same
diameter? Where does the charge reside in each case?
Answer
No, solid metal sphere will not hold a large electric charge than a hollow sphere of same
diameter. The charge reside in each case the outer surface.
Q. No.21 A parallel plates capacitor has a fixed charge on its plates .The plates are now pulled further
apart. Some work must be done during this pull. Why? Does the potential difference change during the
process?
Answer
Yes, work must be done in pulling the charge.
And, we know that, V = E d
σ
The electric intensity due two parallel oppositely charged plates is, E = ,
εo
σd
Therefore, V =
εo
Q
If, σ =
, as charge density,
A
Qd
Then, V =
.
A εo
This shows that, if Q and A are kept constant then, V  d
Q. No.22 Can an electric potential exist point in region where electric field is zero. Can the potential
be zero at a place where the field intensity is not zero?
Answer
No, an electric potential cannot exist at a point in a region where the electric field is zero.
No, the potential doesn’t be zero at a place, where the electric field intensity is not zero.
Q.No.23
How will the capacitor of parallel plates capacitor be affected, if the separation between
its plates is doubled and size of the plates is reduced to half?
A εo εr
Answer
We know that, C =
d
1
C 
This shows that,
and
C  A
d
It means, if the separation between plates is doubled then capacitance decreases to one half the initial
value. Similarly, if the size is reduced to half then capacitance also decreases to half the original value.
Q. No.24 Define absolute potential energy?
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Answer
The Potential on a positive test charge placed in the electric field, at that point is called
“absolute potential energy”.
1 q
U=
qo
4 π ε o  r 2 
U = V qo
Q. No.25 Define electric intensity and give its unit in SI system.
Answer
The force per unit charge is called electric intensity.
F
E = . Its unit is N / coul. in SI System.
q
Q. No.26
Define coulomb’s law.
Answer
The magnitude of electrostatic force is directly proportional to the product of charges and
inversely proportional to the square of distance between them, is called “Coulomb’s law”.
1  q1 q 2 
F=
4 π ε o  r 2 
Q. No.27 Define volt.
Answer
One -joule work is to be done on to unit coulomb charge the potential is called “one volt.”
1J
work
1volt =
or
Potential difference =
1coul
charge
Q. No.28 Define potential difference.
Answer Potential Difference between two points is the difference in Potential between the two points and
is equal to the work done per unit charge ion carrying a positive test charge from one point to another
against the electric field and without any acceleration”
Work (U)
Potential Difference =
Charge(q)
V=E .d
Q. No.29
Define flux .
Answer
The dot product between electric intensity and vector area is called “electric flux”. φ = E . A
Q.No.30
What happens to the capacitance of parallel plate capacitor if,
a) the area of each plate increased and b) the distance between the plates is increased
Answer
The capacitance of parallel plate’s capacitor,
a) “Increased”, if the area of plates increased and b) “Decreases”, if the distance between the plates
increased.
Q.No.31 How the force will change between two points charges if each charge is double and the
distance between them also double.
Answer:
There will be no change in electrostatic force.
1  2q1 2q 2 
F=
Because,


4 π ε o   2r 2 
1  4 q1 q 2 
4 π ε o  4 r 2 
1  q1 q 2 
Or, F =
4 π ε o  r 2 
F=
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Q. No.32 Give the advantages of dielectric medium used in capacitor.
Answer If dielectric medium used in capacitor then capacitance increases. We know that,
A εoεr
C =
d
Q. No.33 How many electrons contain one coulomb charge?
Answer
We know that, Q = n e
Q
n=
e
1coul.
n=
.Thus, n = 6. 25x1018, electrons contain by one coulomb charge.
1.6 10-19 coul
Q.No.34 What will happen to the capacitance of a charged capacitor, if a dielectric completely fills the
space between the plates? Explain.
Answer
If a dielectric constant completely filled the space between the plates of a capacitor then
capacitance of a capacitor increases.
Q. No.35 Why, it is logical to say that potential of an earth connected object is zero?
Answer
The potential of earth is zero. So that, it is correct to say logical that, potential of earth connected object is
zero.
Q. No.36 If the capacitor is charged to a certain potential difference then it is immersed in oil. What
happens to its capacitance?
Answer
If the capacitor is charged to a certain potential difference then it is immersed in oil.
The capacitance of a capacitor increases.
Q. No.37 What is Gaussain surface:
Answer Gauss's law is valid for symmetrical charge distribution. Gauss's law is very helpful in
calculating electric field in those cases where electric field is symmetrical around the source producing it.
Electric field can be calculated very easily by the clever choice of a closed surface that encloses the source
charges. Such a surface is called "Gaussian surface". This surface should pass through the point where
electric field is to be calculated and must have a shape according to the symmetry of source.
Q. No.38
Define flux density .
φ
Answer Total electric flux passing through a unit area is called electric flux. E =
, its unit is web.m-2.
A
Q. No.39 Can two equi potential surfaces intersect each other?
Answer
No, two equi potential surfaces can’t intersect each other, because, these surfaces are always
perpendicular to electric field. The value of potential is same at every position.
Q. No.40
Will the energy stored by three capacitors be greater if they were connected in series or in
parallel?
1
Answer
We know that, Energy = C V 2
2
This shows that, when capacitors are connected in parallel then energy will be greater.
Q. No.41 A capacitor of 6µf is connected in parallel with the combination of 4µF capacitor and a 3µF
capacitor that are in series. Find (a) the net capacitance of the entire combination. and (b) The Potential
Difference across the 3µF capacitor when 20V is maintained across the 6µF capacitor.
AnswerCapacitors 3 µF and 4 µF are in series
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let Cs = capacitance of a single capacitor which replaces the above two capacitors in series.
Then, 1/Cs = 1/4 + 1/3 =(4+3)/12=7/12
therefore Cs = 12/7 (see Fig a )
let Cp = capacitance of the capacitor which replaces the capacitors in parallel on Fig b
Cp = (12/7+6) µF = 54/7 µF = net capacitance
20 V potential difference is across 6 µF as well as 12/7 µF as they are in parallel
Charge through 12/7 µF
=12/7 µF x 20 = 240/7 µC
Charge on 4 µF and 3 µF capacitors will be the same as they are in
series.Potential difference across 3 µF = (240 µC) / (7 x 3 µF) =
80/7 V = 11.4 V
Q: No: 42 Why is static electricity more apparent in winter?
Answer: You notice static electricity much more in winter (with clothes in a dryer, or taking a sweater off,
or getting a shock when you touch something after walking on carpet) than in summer because the air is
much drier in winter than summer. Dry air is a relatively good electrical insulator, so if something is
charged the charge tends to stay. In more humid conditions, such as you find on a typical summer day,
water molecules, which are polarized, can quickly remove charge from a charged object.
Q: No: 43 why there is no net charge on gaussian surface?
Answer: There is no reason why there should be zero charge on a Gaussian surface. For example, imagine
a conducting sphere on which I have placed a certain amount of charge. All the charge resides on the
surface. If I say that Gaussian surfaces are concentric spheres, then the charged surface is one of them.
Incidentally, any surface you choose to think about is a Gaussian surface as long as it is closed, that is
encloses some volume; it just so happens that the most useful Gaussian surfaces are equipotentials.
Q No: 44 Electric field lines never intersect each other. Why?
Answer:
If they intersect, two tangents can be drawn at the point of intersection indicating two
different directions of electric field at the same position which is impossible.
Q No: 45 Electric field lines are always normal to the surface of a conductor. Why?
Answer: If the electric field line is not normal, it can be resolved into two components, a normal
component and a tangential component. The tangential component will make the free charges of the
conductor to move along the surface of the conductor. But no such movements are observed indicating
that there is no tangential component which implies that the electric field lines are always normal to the
surface of a conductor.
Q No: 46 What is capacitor?
Answer: Which stores charge is called capacitor. Capacitor stores capacity of charge
Q No: 47 What is the Field between two Oppositely Charged Parallel Conducting parallel Plates?
Answer: The field between the plates is basically uniformly spaced but there is a little fringing, at the
edges. We are not concerned with the edges so we can assume it is uniform. These parallel plates are the
basis of capacitors.
These two plates can be treated as two sheets of charge of opposite charge and we can
use Gauss's Law. Outside the two sheets, the components of the electric fields are of the
same magnitude but oppositely directed so their resultant is zero. At any point between
the plates the components are in the same direction so their resultant is the sum.
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Basically giving you the same result as that of the charged conducting plate: E=

εo
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Electric Fields
Also similar to gravity, the electrostatic force is a non-contact force. Charged objects do
not have to be in contact with each other to exert a force on each other. Somehow, a
charged object feels the effect of another charged object through space. The property of
space that allows a charged object to feel a force is a concept called electric field. Although
we cannot see an electric field, we can detect its presence by placing a positive test charge at various points
in space and measuring the force the test charge feels.
While looking at gravity, the gravitational field strength was the amount of force obsered by a mass per
unit mass. The electric field strength is the amount of electrostatic force observed by a charge per unit
charge. Therefore, the electric field strength, E, is the electrostatic force observed at a given point in space
divided by the test charge itself. Electric field strength is measured in Newtons per Coulomb (N/C).
Electric Field Lines
Since we can't actually see the electric field, we can draw electric field lines to help us visualize the force a
charge would feel if placed at a specific position in space. To help us visualize the electric field, we can
draw electric field lines in space. These lines show the direction a positive charged particle would feel a
force if it were placed at that point in space. The more dense the lines are, the stronger the force a charged
particle would feel, therefore the stronger the electric field. As the lines get further apart, the strength of
the electric force a charged particle would feel is smaller, therefore the electric field is smaller.
By convention, we draw electric field lines showing the direction of force on a positive charge. Therefore,
to draw electric field lines for a system of charges, follow these basic rules:
1.
2.
3.
4.
5.
Electric field lines point away from positive charges, and toward negative charges.
Electric field lines never cross.
Electric field lines always intersect conductors at right angles to the surface.
Stronger fields have closer lines.
Field strength and line density decreases as you move away from the charges.
Let's take a look at a few examples of electric field lines, starting with isolated positive (left) and negative
(right) charges. Notice that for each charge, the lines radiate outward or inward spherically. The lines
point away from the positive charge, since a positive test charge placed in the field (near the fixed charge)
would feel a repelling force. The lines point in toward the negative fixed charge, since a positive test
charge would feel an attractive force.
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If you have both positive and negative charges in close proximity, you follow the same basic procedure:
Of course, electric field lines actually lie in three dimensions, as demonstrated in this video animation.
Comparing Electrostatics & Gravity
Because gravity and electrostatics have so many similarities, let's take a minute to do a quick comparison
of electrostatics and gravity.
Electrostatics
Force:
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Field Strength:
Field Strength:
Electrostatic Constant:
Charge Units: Coulombs
Gravity
Force:
Field Strength:
Field Strength:
Gravitational Constant:
Mass Units: kilograms
The big difference between electrostatics and gravity? The gravitational force can only attract, while the
electrostatic force can both attract and repel. Notice again that both the electric field strength and the
gravitational field strength follow the inverse-square law relationship. Field strength is inversely related to
the square of the distance.
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Electric Potential Difference
When we lifted an object against the force of gravity by applying a force over a distance,
we did work to give that object gravitational potential energy. The same concept applies
to electric fields as well. If you move a charge against an electric field, you must apply a
force for some distance, therefore you do work and give it electrical potential energy. The
work done per unit charge in moving a charge between two points in an electric field is
known as the electric potential difference, (V). The units of electric potential are volts, where a volt is
equal to 1 Joule per Coulomb. Therefore, if you did 1 Joule of work in moving a charge of 1 Coulomb in an
electric field, the electric potential difference between those points would be 1 volt. This is given to you in
the reference table as:
V in this formula is potential difference (in volts), W is work or electrical energy (in Joules), and q is your
charge (in Coulombs). Let's take a look at a sample problem.
Parallel Plates
If you know the potential difference between two parallel plates, you can easily calculate the electric field
strength between the plates. As long as you're not near the edge of the plates, the electric field is constant
between the plates, and its strength is given by:
You'll note that with the potential difference V in volts, and the distance between the plates in meters,
units for the electric field strength are volts per meter [V/m]. Previously, we stated that the units for
electric field strength were newtons per Coulomb [N/C]. It is easy to show that these units are equivalent:
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Question: The diagram represents two electrons, e1 and e2, located between two oppositely charged
parallel plates. Compare the magnitude of the force exerted by the electric field on e1 to the magnitude of
the force exerted by the electric field on e2.
Answer: The force is the same because the electric field is the same for both charges, as the electric field is
constant between two parallel plates.
Equipotential Lines
Much like looking at a topographic map which shows you lines of equal altitude, or equal gravitational potential
energy, we can make a map of the electric field and connect points of equal electrical potential. These lines, known as
equipotential lines, always cross electrical field lines at right angles, and show positions in space with constant
electrical potential. If you move a charged particle in space, and it always stays on an equipotential line, no work will
be done.
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