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ELECTROSTATIC
1.
The study of electrostatic introduces us to the vast world by electricity, magnetism, current, working
of circuits and charges. Right from the attraction between a proton and neutron to the working of
large high voltage generator can be explained by the simple concepts of electrostatics, the most
basic of phenomenon and yet the most interesting of all.
Introduction:
Electrostatic is the study of charges that are at rest and their interactions.
Electric
Charge:
It is an intrinsic property of protons and electrons, proton and electron are smallest unit of positively
and negatively charge.
The S.I.unit of charge is Coulomb.
1)
2)
Like
Charge
Some
charges
is
quantized
properties
repel
multiple
of
of
while
charge
unlike
on
an
charges:
attract.
electron.
3) Net amount of charge in a system is conserved.
Discuss this topic
Coulomb’s
1)
Force
Where
And k =
Law:
between
r
=
where
the
two
separation
charges
between
=
q1
and
q 2.
= permittivity of the medium.
Vector
form-
Important
points:
* Principle of superposition: Electrostatic force between any two charges is independent of other
charges.
2) For a dielectric (insulator) medium
Where
k=
e = permittivity of the medium
dielectric
Note: Electric force is an action–reaction pair and conservative in nature.
constant.
Illustration:
Two charges are placed at a distance 1.0 apart what is the minimum possible magnitude of electric
force.
Question:
Q-1:
Why
solution: Since q1, q2 should be minimum and q1min = q2min = e =
C
Q-2: Five point charges of values +q are arranged on the vertices of a regular hexagon as shown in
the figure, and a –q charge is placed in the center. Find the force on –q charge.
Fig (1)
solution: Imagine two charges of
on the vacant vertex, the net force on center is now only
due to -q as all other forces cancel out (principle of superposition).
Fig (2)
Electric Field: It is the electrostatic force per unit charge.
The space around a charge where its electric force can be experienced by another charged
particle.
Important
1)
points:
Field due to point charge =
2)
Principle of superposition: The electric field due to multiple no of charges gets
vectorially.
added
Note: For continuous charge distribution
Where dq is charge due to infinitesimally small element.
Electrical
Lines
of
These are imaginary lines drawn to aid our visualization of properties of electric field.
forces:
Question:
Q-1: Why electric field lines never intersects?
Ans: Intersection of electric field lines implies that there are two direction of electric field at
point which is impossible, therefore they never intersects.
that
Q-2: Why are electric lines always normal to the surface of a conductor?
Ans: If electric lines are not normal then there is a component along the surface which means that
change of potential along the surface which is not acceptable.
Q-3: Why do electric field lines start from positive and end on negative charge?
Ans: Because the direction of electric field is from positive to negative i.e. the force on a unit
positive charge is away from positive and towards negative charge.
Field lines from +ve Pointcharge.
Field lines towards -ve Point charge.
Equal +ve charge.Two equal but oppositecharged particle
Illustration:
Q-1: A field of 103N/C acts at a point. What will be the force on a charge of 3´10-6C and -2´10-6C,
kept there?
solution:
For +3MC
For -2MC
Discussion:
The direction of force changes for different charges. For a positive field(due to +ve charge)
1)
The
force
is
in
direction
of
field
on
the
+vecharge
i.e.
repulsive.
2) The force is opposite to field direction onthe –ve charge i.e. attractive.
Similar inference can be drawn for negative charges.
Electric field due to:
a)Charged
ring
of
charge
Q
and
radius
R.
1) The Center=0.
Derivation:
Linear
Consider
charge
the
field
density
at
center
But the field due to point diametrically opposite =
(l)
due
on
to
any
ring
element
=
=
in opposite direction.
\Net field at center = 0 (By symmetry)
2)On the axis =
On axis ofring at distance x.
Derivation:
Fig (5)
As obviousfrom the diagram the field component along the line gets added due to
oppositeelement.
*)By
substituting
x=0
in
2ndresult,
we
can
get
the
firstresult.
*) Student should verify that the graph is of the followingmanner.
Fig (6)
b)Due to a straight charged rod of length 2L with charge per unit length ‘ ’ ata distance ‘a’ on
its
E
perpendicular
bisector.
=
Derivation:
The rod is divided into infinitely small elements andthe field due to symmetrically; opposite part add
up as shown in figure (7).
Net field at P.
E=
dE =
Fig(7)
Useful Tips:
1) If x>>a, E =
2) If L
like a point charge.
i.e. for infinite length
Question:
1) How did
Ans:
the
electric
field
cancel
inone
direction
and
add
in
another?
Observe the direction of electric field of two points which are symmetricallyopposite.
Along
the
Along the perpendicular,
axis,
net
field
=
Net field =
Similarly all fields along the perpendicular to axis cancel out.
Flux and Gauss Law:
Flux-It literally means the no of field lines passing through it. Itis defined as.
Gauss law- The electric flux
enclosed by thesurface.
through any closed surface is equal to
times the net charge
Illustration:
A square frame of edge 5cm is placed such that a uniformelectric field of 10 V/m makes 30 0with
positive normal.
Application of gauss law:
1)Uniformly charged sphere (solid) with charge Q and radius R.
Fig
(9)
Case-1:
(r>R)
Fig(10)
Charge
enclosed
=
Q
Case-2: (r<R)
Question:
Q-1
How?
Ans: The field lines should be radially outward and thereforecollinear with outer normal.
Cos
2)Infinitely long linear charge distribution:
Derivation:
Consider the Gaussian surface shown:
Fig (11)
Qenclosed= ll
For surface (1) and (3)
=1
As electric field perpendicular to area vector for curved surface (2)
By Gauss’s law
E (2pRl) =
3)Field due to infinite charged plane sheet with charge density ( )=
=
Derivation:
Consider a cylindrical Gaussian surface as shown.
Flux through curved surface = 0
Fig (12)
Conductors:
The electric field inside conductorsis zero in electrostatic equilibrium.
Cavity
in
a
conductor:
Consider
the
metallic
shell
with
a
postcharge
in
the
center.
Since
E
=
0
inside
a
conductor
qenclosedbyG.S
=
0
Þ
-q
charge
is
induced
on
inner
surface,
as
shown
in
figure(13).
Since shell is neutral, +q is induced on outer surface as shown infigure (14).
Fig (13) Fig(14)
Question:
Explain why electric field inside a conductoris zero, with the help of induction.
Solution:A conductorhas large no of free electrons. When it is kept in an electric field,electrostatic
force acts on these electrons which draw most of the electron toone side causing negative charge
on that side and positive charge on oppositeside due to deficiency of electrons. The field due to
these induced chargesbalances the external electric field and thus there is no electric field in
themetal.
Fig (15)
Enet=E+Ein= 0
On the surface of charged Conductor:
Electric field means a Conductor =
Force per unit area or electrostatic pressure experienced by a chargedconductor =
(The derivation of above formula is not very important for IIT JEEpreparation)
Energy density of electric field =
Electric
It is the work needed to bring a unitpositive charge from infinity to the given point.
(Because, since the field is conservative.)
Q-1: The value of integral depends only on end points A and B why?
Ans:
(As it is opposite to electrostatic force)
Note:
1)It is a scalar quantity and has units Joules per coulomb (J/C) called Volt.
2)Potential difference between two points A and B is V B-VA=
Where W ABis work done in moving from A to B.
Electric potential due to a point charge =
2.
Principle of superposition:
Potential:
Net potential due to
InCartesian Co-ordinate-
a
system
of
charge
is,
V
=V 1+V2+V3+V4+----------------
Illustration:
1) Three charges q1, q2,q3are placed at (r, 0, 0), (0, 3r, 0), (0, 0, 2r) respectively. Findpotential at
origin.
Ans:
Net electric potential = Vq1+Vq2+Vq3
=
=
Equipotential surface:
If a surface is drawn in such a way that electric potential is same at allpoints of the surface it is
called equipotential surface.

Workdone in moving a charge on equipotential surface = 0.
Electric potential on –
Derivation:
Divide the ring into infinite elements and potential due to eachelement.
Fig (16)
If r = 0, V at center =
as shown in figure (17)
Fig (17)
2) axis of uniformly charged disc =
Derivation: Consider disc having uniform surface charge6(c/m 2)
Fig (18)
Potential due to ring of radius r =
Question:
Q-1: How dq =
Ans: dq = 6 x Surface area
=
2 r.dr= area of the ring.
3)
Due
Derivation:Shell of radius R has charge Q
Case-1: (r>R)
Case-2: (r<R)
Einside= 0
to
a
Shell:
Electric Potential Energy:
1) Potential energy between two charges q 1and q2separated by distance r =
.
2) P.E due to System of charge =
Change should be replaced with sign.
Question:
Q-1:
Why
½
in
the
above
expression?
Ans: If ½ were not produced each pairwould be counted twice. For example- For q1and q2
3)
Illustration:
Two point charges are located on x-axis q1=3microcoulomb at x=0 and q2= -1 microcoulomb at
x=2m.
a) Findwork to be done to bring q 3= 1microcoulomb from infinity to x=1m.
b) Find total P.E of the system.
solution:
=Uf-Ui
Work
done
=
difference
in
P.E.
Earthing:Potential of earth is zero. If a conductor is earthedits potential becomes zero.
Tips
for
Concentric
conducting
sphericalshells:
1) Net charge inside a closed Gaussian surface drawn in anyshell is zero.
2)
Potential
of
the
earthed
conductor
is
zero.
3)Two
connected
conductors
are
at
same
potential.
4) Charge remainsconstant in all conductors except those which are earthed.
5) Charge oninner surface of innermost shell = 0 (if no charge is keptinside)
Charge
on
outer
surface
of
outermost
6) Equal and opposite charge appear on oppositefaces.
shell
=
totalcharge
enclosed.
Illustration:
Figure (19) shows two conducting thin concentric shells ofradii r and 3r. The outer shell carries
charge q. Inner shell is neutral. Findthe charge that will flow from inner shell to earth after the
switch S isclosed.
Fig (19)
solution: Let q be the charge on inner shell when it isearthed V inner= 0.
i.e.
charge will flow from inner shell to earth.
Electric dipole:An arrangement of two equal and oppositecharges separated by small distance is
known as electric dipole.
Fig
1)
2)
3)
(20)
It
The
It
is
line
is
a
directed
from
–ve
joining
the
charges
a)Electric field on the axis (l<<r) =
Derivation:
Fig (21)
b) On perpendicular bisector =
Derivation:
is
vector
charge
to+ve
the
axis
of
quantity.
charge.
dipole.
Fig(22)
The
component
Alonghorizontal.
Enet=
of
E
cancels
E+qCosq
out
in
vertical
+
direction.
E-qCosq
(-ve indicates direction opposite todipole moment)
If x>>1
c) Electric field at any point A(r,0)
Fig (23)
Dipole
1)
2)
is
an
external
Force
uniform
electric
field
ondipole=0.
3) Potential energy = -PECos q = -P.E
Fig (24)
Q=
0,
U
will
be
Q=1800, U will bemaximum, unstable equilibrium.
minimum,
stable
equilibrium.
Illustration:
1) A dipole of dipole moment P lies in uniform electricfield with its dipole moment along E. If dipole
is
rotated
through900,
Find
work
done?
Ui=
-P.E
=
-P.E
Cosq
=-P.E.
Uf=
-PECos900=
0
Work done = -DU =Ui-Uf=-P.E.
2) A dipole with dipole moment
Cm is aligned at 800with the direction ofa electric field of
magnitude 104N/C Calculate magnitude oftorque.
|Z| = PESinq
Examples:
1) A charge q = 2´10-6C is placed at (1m, 1m, 1m) and electricfield at P (0m,-2m, 1m).
solution:
2) Two free particles with charges q and 4q are a distance L apart. Athird charge is placed so that
system is in equilibrium. Find location,magnitude and size of third charge.
solution: Let new chargebe +Q, It will be along the line joining the two charges to balance the
forces.Let it be located as shown.
Fig (25)
Since
For q:
For 4q:
they
are
in
equilibrium
(-ve x means charge will be in middle of the two charges.)
3) A charged particle of mass m and charge q is released from rest in an electric field of constant
magnitude E then the KE of the particle after time t is.
solution:
Force = qE
Acceleration =
.
4) P and Q are two concentric metallic spheres. P is positively charged and Qis earthed then
a)
Charge
density
on
Q
is
b)
Electric
field
between
P
c)
Electric
potential
inside
d) Electric fieldoutside Q is zero.
same
and
P
as
Qis
is
on
P.
uniform.
zero.
Fig(26)
solution:
VQ= 0
Let
charge
on
P
be
q1
and
on
Q
be
q2
i.e. (a) is false.
(b) Is false as there is potential difference between the two shells and theelectric field lines
separate outwards.
(c) is false
.
(d) is true as q enclosed by Q surface = q2+q1=0.
5) A certain charge Q is divided into two parts q and Q-q. For the maximumCoulomb force between
them the ratio (q/Q) is:
1) 1/6, 2) 1/8, 3) 1/4, 4) 1/2
solution:
Ans = 4).
6) Three large conducting plate sheets are kept as shown in figure (27). Findelectric field a)
between (1) and (2), b) between (2) and (3).
Fig (27)
solution:
a) Between (1) and (2)
Field due to a plate =
.
b) Between (2) and (3)
7) A charged particle of mass m=2kg and q=1microcoulomb is thrown from groundat q=30 0with
speed 30 m/s. There is a horizontal electric fieldE=10 7V/m exists. Find the range on horizontal
ground
of
theprojectile.
solution:
8) A charge q is placed at l/2 distance above a square of length l as shownin figure (28). Find flux
through the square.
solution: Consider the charge to be kept in a cube of sidel.
Net
flux
=
Fig (28)
By symmetry flux through each face =
.
9) An alpha particle of KE = 6MeV is heading towards a stationary nucleus ofatomic no 30 calculate
distance of closet approach.
By conservation of energy:
½ mV2= Uf-Ui
Dump
Question:
Ans: Since charges are atinfinity
Why
Vi =
0?
Medium:
Two parallel conducting plates contain charge Q1and Q2 as shown. Find distribution of charges on
four surfaces.
Fig (29)
Consider
the
Gaussian
surfaceshown.
solution:Two faces lie in theconductor where field is zero, while other two are perpendicular to
fielddirection.
Net
charge
enclosed
=
0
So, the facing inner surface have equaland opposite charges.
Fig (30)
Let C be a point in the conductor –
Question:
Why net flux through Gaussian Surface = 0?
Fig (31)
There
Field
\E.dA = 0
is
at
no
2
So, net Flux = 0
Note: Short cut for parallel plates
field
and
4
at
are
1
and
perpendicular
3
toE.
1)Charge on facing faces of two plates are equal and opposite
2)Charge on outermost face =
.
Q-2: A curved line charge of density l forming a semi-curve as shown. Findelectric field at center.
Fig (32)
Consider
The
the
two
opposite
field
elements
as
cancels
shown.
alongx-direction.
Consider an infinitely small element substending angle d
Question:
1)What
would
change
if
the
curve
were
negatively
charged?
Ans: The net fieldwill be upwards.
2)If
half
of
the
curve
were
negatively
charged?
Ans: Virtual components cancelwhile horizontal remains. (Observe yourself)
Q-3: A uniform line charge exist from x=-a to x= a. Find electric field at point P a distance r along
the perpendicularbisector.
solution:
Fig (33)
Considering the field due to two opposite parts, the field cancels along thehorizontal.
Along Vertical:
Q-4: A charge Q is uniformly distributed over a spherical volume of radius R.Obtain an expression
for the energy of the system.
1)Inside the sphere(r<R)
2)Outside the sphere (r>R)
=
Question:
solution:Consider
a
G
sphere
Q-5: Find the potential on the edge of a disc P with surface charge density6.
Ans:
Fig (34)
To calculate the potential at P, the disc is divided as rings with P ascenter as shown in the figure.
Potential due to element between r andr+dr.
Now
r
=
dr =-2RSinqdq.
2RCosq
(as
diameter
subtends
900at
the
edge)
(Applying Integration by parts)
Question:
1)
Why
were
integration
limits
from
p/2
to
0?
Ans: Integration limits werefrom p/2 to 0 because at p/2, the area around the point would be
covered andwhen angle is zero, the other end of diameter gets covered. Therefore the wholedisc is
covered.
Q-6: The intensity of an electric field depends onco-ordinates x and y as follows,
Where P is a constant, find the charge within asphere of radius R with the center at the origin.
solution: At any point A(x, y, z) on the sphere, a unitvector is perpendicular to the sphere radially
outward.
Fig(35)
Now
can begot easily by integrating, for it is independent of x, y, and z.
Note: The question although easy may seem very difficult during exam becauseof the use of
Cartesian Co-ordinates. So student should go through the aboveexample carefully.
Q-7: A non-Conducting disc of radius R and Surface charge density isplaced on the ground with its
axis vertical. A particle of mass m and positivecharge q is dropped, along the axis of disc from
height
H
with
zero
initialvelocity.
The
particle
has
q/m=
4
e 0g/
a)
Find
value
of
H
if
the
particle
just
reaches
the
disc.
b) Sketchthe potential energy of the particle as a function of its height and find itsequilibrium
position.
solution:
a)
As
proved
earlier
Vi= potential on axis of a disc
Vf= potential at center
By
conservation
Decrease in gravitational potential energy
KEi=
KEf=0
(As
particle
is
mgH = q (Vf-Vi)
of
energy:
=Increase in electrostatic potential energy
at
rest
in
both
position)
b) Net potential energy = electrostatic P.E+ gravitational P.E
Fig (36)
Note:
1) Potential energy includes both electrostatic potential energy andgravitational potential energy.
Question:
1)
How
is
K.E=0
in
both
position?
Ans:Initially it has zero initial velocity K.Ei=0, finally it is justable to reach the disc.
Final velocity = 0 Þ K.Ef=0.
Hard
Q-1: A non conducting sphere of radius R has a positive charge which isdistributed over its volume
with density
per unit volume, where x is distance from the center.If dielectric
constant material the sphere is K=1, Calculate energy stored insurrounding space and total self
energy of the sphere.
solution: If strength of electric field at a point is E,energy stored per unit volume of dielectric at that
point is
hence to calculate density of electrostatic energy insurrounding space, first
electric field strength is to be calculated. But forits calculation, total charge on sphere must be
known.
Hence, first considera thin spherical shell inside the sphere as shown in figure.
Fig (37)
Let
radius
of
the
Totalcharge
shell
on
be
x
and
the
let
radical
thickness
shell
be
dx.
is
Now consider a concentric spherical shell (in surrounding space) of radius r(>R) and radical
thickness
dr.
Electric field at surface of thisshell.
Energy stored in the shell,
Energy stored in surrounding space,
Self energy of the sphere is total electrostatic energy stored in surroundingspace and inside the
sphere.
To calculate energy stored inside the sphere,consider a concentric spherical shell of radius x and
radical thickness dx.Electric field at surface of this shell is due to charge on inner concentricsolid
sphere
of
radius
x.
Charge on this concentric sphere is
Electric field at surface of this shell,
Energy stored in the shell,
Total energy inside the sphere is
Self energy of the sphere = Us+Ui
Q-2: Suppose in an insulating medium having dielectric constant K=1, andvolume density of
positive charge varies with coordinates according to law
. A particle of mass m having
positive charge q is placed in themedium at point A (0, y0) and projected with
Fig (38)
Neglecting gravity and frictional resistance of the medium and assumingelectric field strength to be
zero at y=0, Calculate slope of trajectory of theparticle as a function of y.
solution: Since, charge density varies with y coordinatesonly, therefore in medium an electric field
exists which is directed along yaxis. To calculate this electric field E(y). Consider a thin layer of
mediumhaving thickness dy at a distance y from x-axis as shown infigure.
Fig (39)
Let strength of electric field at positions y and (y+dy) be E and (E+dE)respectively.
Then
for
area
A
of
the
layer,
According
to
Gauss
law
Integrating
At
above
equation
y=0,E=0
and
with
y=y,
at
limits,
E=E
Since frictional resistance of the medium isnegligible, therefore there is no force on particle along xaxis
or
velocityV0of
particle
along
x-axis
remains
constant.
Due to Electricfield E, particle accelerates along positive y, this acceleration =
Let
y-component
of
velocity
of
particle
bev
Hence,
But
at
point
A(y=y0)
y
component
ofvelocity
is
v=0
Question:
1)
Why
there
is
no
force
inx-direction?
Ans: The applied electric field is in the y-direction only andit has been mentioned in the question
that friction, gravity have to beneglected. Thus there is no force in x-direction.
Q-3: Three identically charged, small sphere each of mass m are suspendedfrom a common point
by insulated light strings each of length l the spheres arealways on vertices of an equilateral
triangle of length of the sides x(<<l). Calculate the rate
with which charge on each sphere
increases if rate ofincrease of length of the sides of the equilateral triangle varies as
solution: Since
x.
is given as
therefore to calculate
.
first q is to be calculated in terms of
Since side of equilateral triangle is x, therefore electrostatic forcebetween two particle is
Positions of particle A, B, C with respect to each other are as shown infigure (40)-A.
Fig(40)-A Fig (40)-B
Resultant electrostatic force on particle A is
F = 2F1Cos300
In figure (A) G is centroid of triangle.
Distance of centroid G from every particle is
Since G is centroid therefore it always lies vertically below the point ofsuspension.
Fig (B) shows a vertical plane passing through particle A and centroid G. Allof the force acting on
particle A is shown in figure (B)
Considering horizontal forces on the particle A,
TSin
= F---------- (1)
For
vertical
Dividing
forces
eq
TCosq
(1)
=mg
by
Since x<<l therefore inclination q of each thread with the vertical isvery small.
From equation (3)
----------(2)
eq
(2)