Download Chapter 25: Electric Potential

Chapter 25: Electric Potential
As mentioned several times during the quarter Newton’s law of gravity and Coulomb’s law are
identical in their mathematical form. So, most things that are true for gravity are also true for
electrostatics!
Here we want to study the concepts of work and potential as they apply to the electric field.
In the study of mechanics we talk about work done by (or on) the gravitational field.
Example: The work (W) done by gravity when a 1 kg mass (m) falls a distance (d) of 1 meter is:
W=mgd=(1kg)(9.8m/s2)(1m)=9.8 Joules
Constant force : Work = F ⋅ d = Fd cos θ
We also talk about the potential energy of an object.
In general : Work = ∫ F ⋅ dx = ∫ Fdx cos θ
Example: A 1kg mass sitting 1m above the earths surface has a potential energy (U) of:
U=mgd=(1kg)(9.8m/s2)(1m)=9.8 Joules
The relationship between work done by gravity on an object and its change in potential energy is:
Wdone by gravity= -∆U= -(Ufinal-Uinital)
When our 1 kg object falls 1m its potential energy decreases:
∆U= Ufinal-Uinital= 0 – 9.8J =-9.8J
The work done by the earth’s gravitational field is:
Wdone by gravity= -∆U= -(-9.8J)= +9.8J
Remember:
When an object falls
due to gravity its potential
energy decreases.
Imagine instead of gravity the force on the object was electrostatic.
Example: A positive charge (q) of 1 C is in an electric field (E) of 9.8N/C that points down.
How much work is done by the E-field if the charge moves 1 m in the direction of E?
E +q
WE-field=qEd=(1C)(9.8N/C)(1m)=9.8Joules
How much does the charge’s potential energy change when it moves 1m in the direction of E?
∆U= -Wdone by E-field=-(9.8J)= -9.8J
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The Electric Potential Defined
It turns out to be very useful to define a quantity called the electric potential (V).
The electric field can be calculated from the electric potential and visa versa.
The electric potential is just the electric potential energy per unit charge:
V =
U
q
electric potential
Actually, it is the electric potential difference we want since in analogy with potential energy
it is only the change in potential energy that counts:
example: In our previous example we said that a 1 kg mass held 1 m above the earth’s surface
had a potential energy of 9.8 J. In doing this problem we (implicitly) assumed that the potential
energy at the earths surface was 0J. More correctly, we should say the potential energy
difference of a 1 kg mass 1m above the earth’s surface is 9.8J.
∆V = V f − Vi =
U f −Ui
q
=
∆U
q
electric potential difference
The electric potential difference is a scalar quantity. This is one of its virtues. It allows us to
calculate the electric field, a vector, from a scalar!
Finally, we can relate the electric potential difference to the work done by an electrostatic force:
∆V = V f − Vi = −
W
q
note the minus sign!
The unit of potential difference is the volt.
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1 volt=1 joule per coulomb
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The Electric Potential Defined continued
In electrostatics we usually set the reference at infinity (∞): U∞=0. With this reference V∞=0 too.
We can define the electric potential V at any point in an electric field in terms of the work (W∞)
it takes to move charge from infinity to a point f:
W
V = V f − V∞ = −
∞
q
The electric potential is a property of the electric field. It is defined independent of any charges
placed in the electric field.
E
f
i
+q
Example: A positive charge q=+1C moves 1m in an electric field of 9.8N/C as in the figure.
a) The work done by the E-field is: WE-field=qEdcosθ=qEdcos(180)=-(1C)(9.8N/C)(1m)=-9.8J
b) The charge’s potential energy difference is: ∆U= -Wdone by E-field=-(-9.8J)=+9.8J
c) The electric potential difference is: ∆V=∆U/q=+9.8 volts
An important property of the potential difference is that its value is independent of the path
taken to get from point i to point f.
a
b
y
i
d
E
f
Let’s calculate the potential difference in going from i to f by two different routes.
i) the direct route i to f: ∆V=∆U/q=(-WE-field)/q=-Edcosθ=-Edcos(0)=-Ed
ii) the indirect route i to f:
along path i) to a): ∆V=0 as no work is done since: WE-field=Eycosθ=Eycos(90)=0
along path a) to b): ∆V=∆U/q=(WE-field)/q=-Edcosθ=-Edcos(0)=-Ed
along path b) to f): ∆V=0 as no work is done since: WE-field=Eycosθ=Eycos(270)=0
Thus ∆V is independent of the path taken!
The electrostatic force is a conservative force and therefore the potential difference is
independent of path.
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Equipotential Surfaces
It is also useful to speak of equipotential surfaces or lines.
These are points in space at the same potential. Since along an equipotential surface we
have Vf-Vi=0 (duh!) no work is done moving along an equipotential path.
HRW Fig. 25-3
Equipotential lines and a point charge.
The figure on the right shows the electric field lines
(radially outward) and the equipotential lines (concentric circles)
for a positive point charge. Careful inspection of the geometry
shows that the lines of the electric field and the equipotential are
perpendicular to each other. This is true in all circumstances. If it
were not true then there would be component of the electric field along an
equipotential and therefore work would be done moving along an
equipotential. But this would violate the definition of an equipotential!
The figure on the right shows a constant electric field and its
lines of equipotential. As expected, the electric field lines and the
equipotentials are perpendicular to each other.
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Calculating the Electric Potential from the E Field
We can get an expression for the potential difference in terms of the E-field using work (W).
Here we move a positive test charge q0 a distance s in an electric field E.
dW = F ⋅ ds = q 0 E ⋅ ds
The total work done moving the charge q0 a distance s (from point i to f) in an electric field E is:
f
f
f
i
i
i
W = ∫ F ⋅ ds = ∫ q 0 E ⋅ ds = q 0 ∫ E cos θds θ is the angle between E and ds.
Using the definition of potential difference we find:
f
W
V f − Vi = −
=−
q0
q 0 ∫ E ⋅ ds
i
q0
f
f
= − ∫ E ⋅ ds
i
i
Let’s calculate the potential from a point charge (q) electric field.
Our path takes us from R to ∞, defining V∞=0:
∞
∞
∞
∞
q
dr
q
=
V∞ − V R = −V R = − ∫ E ⋅ ds = − ∫ E cos θdr = − ∫ Edr = −
4πε 0 ∫R r 2 4πε 0 r
R
R
R
VR =
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V = − ∫ E ⋅ ds
If we choose Vi=0 we get:
q
4πε 0 R
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∞
=−
R
q
4πε 0 R
We picked a path where E and dr were
parallel (θ=0).
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The Electric Potential (V) of a Group of Charges
We can calculate the V due to a group of point charges by extending our result for a point charge:
q
VR =
4πε 0 R
n
V=
qi
∑ 4πε
i =1
0 ri
potential at a distance R from a single point charge
potential at point (x,y,z) from n point charges
It is important to remember the following about V:
i) We are assuming that V=0 at infinity.
ii) This is the potential at some point in space (x, y, z): V=V(x, y, z)
iii) ri is distance that charge i (qi) is from the point (x,y,z). r is always positive.
iv) The charge qi can be positive of negative. The sign of V depends on the signs of the qis
v) The potential is a scalar quantity and we are using superposition to calculate its value here.
Example: Four charges are at the center of a square with side =L as shown in the figure.
1) What is the potential at the center of the square?
+2C The distance of each charge to the center of the square is L/√2
+2C
x
4
Vcenter =
qi
∑ 4πε
i −1
0 ri
=
+ 2C
4πε 0 L / 2
+
+ 2C
4πε 0 L / 2
+
− 2C
4πε 0 L / 2
+
− 2C
4πε 0 L / 2
=0
2) What is the potential at x, a point midway between the two positive charges?
The negative charges are r=(5/4)1/2L from x.
-2C
-2C
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Vx =
+2C
+2C
−2C
−2C
4C
+
+
+
=
[2 − 4 / 5 ]
4πε 0 L / 2 4πε 0 L / 2 4πε 0 L 5 / 4 4πε 0 L 5 / 4 4πε 0 L
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The Potential Due to a Continuous Charge Distribution
We can derive the expression for the potential due to a continuous charge distribution in a
fashion similar to the one used for the electric field from a charge distribution.
warning!
q
dq
V=
⇒ dV =
V is the potential
4πε 0 r
4πε 0 r
not Volume
Some things to note about dV:
i) r is the distance to dq. It is always positive.
ii) dq can be positive or negative depending on what type of charges we have.
iii) V is a scalar!! No dot product in the integral.
So, to find the potential (assuming it is zero at infinity) we must do the following integral:
V=
1
4πε 0
∫
dq
r
Example: Suppose a charge Q is uniformly distributed in a circle of radius R as shown in
R
in the figure. What is the potential at the center of the circle assuming V∞=0?
Here we have a linear charge density λ. Note: for this problem r is constant (=radius of circle).
Previously we found that dq=λds, and using s=arc length (s=Rθ) we get dq=λRdθ.
V=
1
4πε 0
∫
dq
1
=
4πε 0
r
∫
λds
R
=
1
4πε 0
2π
∫
λRdθ
0
R
λ
=
4πε 0
2π
∫
0
dθ =
λ (2π )
Q
λ
=
=
4πε 0
2ε 0 4πε 0 R
The last step used λ=Q/(2πR)
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The Potential Due to a Continuous Charge Distribution
continued
Let’s do a more challenging problem, one where the distance to the charge is not constant.
Example: A thin uniformly charged rod of length L with linear charge density λ.
The distance from dq to P is:
r = h2 + x2
P
For a linear charge density along a line we have: dq=λdx
h
r
V=
dq
x=L/2
x=0
L/2
L/2
∫
−L / 2
dx
h2 + x2
4πε 0
∫
dq
1
=
4πε 0
r
L/2
∫
−L / 2
λdx
h2 + x2
This integral is given in App. E, as #17:
∫
For our problem we get:
λ
V=
4πε 0
1
dx
h2 + x2
= ln( x + h 2 + x 2 )
 L / 2 + h 2 + ( L / 2) 2
λ
λ
2
2
2
2
=
[ln( L / 2 + h + ( L / 2) ) − ln(− L / 2 + h + (− L / 2) )] =
ln 
4πε 0
4πε 0  − L / 2 + h 2 + ( L / 2) 2





We can “simplify” the above with a bit of algebra to get:
 L / 2 + h 2 + ( L / 2) 2 
λ

ln 
V=


2πε 0
h

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
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OK, so this wasn’t so easy.
At least we didn’t have to use vectors..
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Calculating the Electric Field from the Potential
One of the great things about the potential is that it is related to the electric field by a derivative!
Previously, we defined the potential to be:
f
W
∆V = V f − V i = −
= − E ⋅ ds
q0
∫
i
For an infinitesimal change in the potential: ∆V →dV and we can write:
dV = − E ⋅ ds
Let’s consider the case where the electric field is in the x direction (E=Ex) and ds is also
in the x direction (ds=dx).
dV = − E ⋅ ds = − E x dx
We can rearrange this to read:
dV
= −E x
dx
The x component of the electric field is just the negative of the potential with respect to x!
This result can be generalized for cases where the electric field has components along x, y, or z:
Ex = −
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∂V
∂x
Ey = −
∂V
∂y
Ez = −
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∂V
∂z
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Calculating the Electric Field from the Potential continued
Let’s try a few examples to see how this works….
Example: Let’s calculate the electric field of a point charge from the potential of a point charge.
y-axis
We want to calculate E at (x,y) if a charge +Q is at (0,0).
The potential at (x,y) due to a point charge +Q at (0,0) is:
(x,
y)
r
Q
Q
V ( x, y ) =
=
θ
4πε 0 r 4πε x 2 + y 2
x-axis
+Q
0
According to our prescription we can calculate Ex and Ey using derivatives:
Ex = −
Ey = −
Q
Q
Q 1 x
Q cos θ
∂V ( x, y )
−x
∂
=
=
=
−
=−
4πε 0 ( x 2 + y 2 ) 3 / 2 4πε 0 r 2 r 4πε 0 r 2
∂x
∂x 4πε 0 ( x 2 + y 2 )1 / 2
Q
Q
−y
Q 1 y
Q sin θ
∂V ( x, y )
∂
=
=
=
−
=−
4πε 0 ( x 2 + y 2 ) 3 / 2 4πε 0 r 2 r 4πε 0 r 2
∂y
∂y 4πε 0 ( x 2 + y 2 )1 / 2
To check if we got the correct answer let’s work the problem by calculating the electric field.
We know that the magnitude of the electric field at a distance r away from a point charge is:
Q
| E |=
4πε 0 r 2
Same as what we got
The components of the electric field Ex and Ey are:
from taking derivative
Q cos θ
Q sin θ
E x =| E | cos θ =
Ey =| E | sin θ =
of potential!
2
4πε r
4πε r 2
0
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Calculating the Electric Field from the Potential continued
We can use the relationship between the potential and electric field to show that the potential
in a conductor must be zero.
A potential: V(x,y)=c, with c= a constant has electric field:
Ex = −
∂V ( x, y )
∂c
=−
=0
∂x
∂x
Ey = −
∂V ( x, y )
∂c
=−
=0
∂y
∂y
Thus a conductor is an equipotential inside and on its surface!
Sometimes we don’t have an equation that describes the potential but instead we have a bunch
of measurements of the potential at different points in space. We can estimate the electric field
strength using Ex=-∆V/∆x.
Example: Suppose we have the following measurements:
x(m)
V(volts)
0.000
10
0.001
20
1.000
-5
1.001
0
The electric field at x~0m is: -(20-10)/(0.001-0)= -104V/m
The electric field at x~1m is: -(0- (-5))/(1.001-1)= -5x103V/m
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E is pointing in the –x direction
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