Electric Potential/Energy
Phys102
General Physics II
Electric Potential
Electric potential energy
ƒ The electric force is mathematically the same as
gravity so it too must be a conservative force. We
will find it useful to define a potential energy as is
the case for gravity. Recall that the change in the
potential energy in moving from one point i to
point f is the negative of the work done by the
electric force.
ƒ ∆U = Uf -Ui = - Work done by the electric force
f
G G
∆U = − ∫ F ⋅ ds
i
ƒ Since F=q
F=qE,
f
G
•Topics
•Electric potential energy and electric potential
•Equipotential Surface
•Calculation of potential from field
•Potential from a point charge
•Potential due to a group of point charges, electric dipole
•Potential due to continuous charged distributions
•Calculating the field from the potential
•Electric potential energy from a system of point charge
•Potential of a charged isolated conductor
Electric potential energy
ƒ ∆U = Uf -Ui = - Work done by the electric
force
G G
W = F .d
G
G
F = qE
G G
W = qE.d = qEd cosθ
G
∆ U = − q ∫ E ⋅ ds
i
1
Electric potential
ƒ
Equipotential Surfaces
Electric Potential = Potential energy / unit charge
V= = U/q
ƒ
Electric Potential difference = Potential energy change/ unit charge
charge
∆V = Vf -Vi = (U
(Uf – Ui)/q=
)/q= ∆ U/q
∆V = Vf -Vi =- W/q
ƒ The potential difference is the negative of the work done
per unit charge by an electric field on a positive unit charge
when it moves from point i to point f.
ƒ
First draw the field lines,
then find surface perpendicular
to these lines.
The SI units of V are Joules/ Coulomb or Volt.
E is N/C or V/m.
Equipotential Surfaces
Equipotential Surfaces
Uniform E field
E = Ex , Ey = 0 , Ez = 0
V = Ex d
V = constant in y and z
directions
Point charge
(concentric shells)
Two point charges
(ellipsoidal concentric shells)
2
f
∫
G
G
∆U = Ub -Ua = - Work done by the electric force = − F ⋅ ds
i
f
Electric force is a conservative force ∆V = V − V = − EG .dsG
f
i
Therefore independent of path
i
∫
∆ V= = ∆ U/q
f
G G
∆V = V f − Vi = − ∫ E.ds
i
f
G G
We are free to choose V to be 0 at any location.
V = − E.ds
Normally V is chosen to be 0 at infinity for a
i
point charge.
∫
Examples
Examples
ƒ
What is the electric potential difference for a positive charge moving in
an uniform electric field?
ƒ
E
∆V =
E
d
a
q
x direction
b
∆U
= Vb − Va = 0 − 9
q
∆U = −9q
= -9V x 10-6 C
b
b
a
a
∆V = − ∫ E ⋅ ds = − E ∫ dx = − E ( xb − xa )
d
∆V = -Ed
In a 9 volt battery, typically used in IC circuits, the positive terminal has
a potential 9 v higher than the negative terminal. If one micromicro-Coulomb
of positive charge flows through an external circuit from the positive
positive to
negative terminal, how much has its potential energy been changed?
changed?
∆U = q ∆V
∆U = -qEd
∆U = -9V x 10-6 Joules
∆U = -9 microJoules
Potential energy is lower by 9
microJoules
3
Electric potential due to a point charge
Examples
G G
V f − Vi = − ∫ E ⋅ ds
f
A proton is placed in an electric field of
going 10 cm, what is its speed?
Use conservation of energy.
a
E=105
E = 105 V/m
d = 10 cm
b
V/m and released. After
∞
∆V = Vb – Va = -Ed
v=
∆U = q ∆V
∆U + ∆K = 0
∆K = (1/2)mv2
∆K = -∆U
V f − Vi = − ∫ E ⋅ dr = −kq ∫
2qEd
v=
m
+
R
1.67 × 10
v = 1.4 ×106
kg
0 − Vi = −k
Electric potential for a positive point charge
r = x2 + y2
q
R
V=
kq
R
k=
m
s
V is a scalar, not a vector
V is positive for positive
charges, negative for
negative charges.
(1/2)mv2 = -q ∆V = +qEd
V (r) = kq /r
∞
1
1
dr = kq
r2
rR
We choose V=0 at r = ∞.
2 ⋅1.6 × 10 −19 C ⋅105 Vm ⋅1m
− 23
kq
r2
E=
i
V=
1
4πε 0
1 q
4πε 0 r
Electric potential due to a positive point charge
Hydrogen atom.
•What is the electric potential at a distance of 0.529 A from
the proton?
-
+
r
r = 0.529 A
• V = kq/r = 8.99*1010 N m2//C2 *1.6*10-19 C/0.529*10-10m
• V = 27. 2 J/C = 27. 2 Volts
4
Electric potential due to point charges
V=
1 q
4πε 0 r
Electric potential due to point charges
ƒ For many point charges, the potential at a
point in space is the sum V= Σ kqi/ri.
ƒ For many point charges, the potential at a
point in space is the sum V= Σ kqi/ri.
 q1 q 2 q 3 
V = k + + 
 r1 r 2 r 3 
y
1
r1
p
2
r2
r3
3
x
Electric potential due to an electric dipole
V=
1 q
4πε 0 r
Ways of Finding V
ƒ Direct integration. Since V is a scalar, it is easier to evaluate
evaluate V than E.
ƒ Find V on the axis of a ring of total charge Q. Use the formula for a
point charge, but replace q with elemental charge dq and integrate.
Point charge V = kq/r.
For an element of charge dq, dV = kdq/r.
r is a constant as we integrate.
V = ∫ kdq/r
= ∫ kdq/(z2+R2)1/2
= k/(z2+R2)1/2 ∫ dq
= k/(z2+R2)1/2 Q
This is simpler than finding E because V is
not a vector.
V = kq/r
5
Line of Charge
More Ways of finding V and E
•Use Gauss’ Law to find E, then use V= - ∫ E.ds to get V
G G
V = − ∫ E.ds
f
i
Suppose Ey = 1000 V/m. What is V? V = -1000 y
What is the electric potential of a
uniformly charged circular disk?
Calculating the Field from Potential
•More generally, If we know V, how do we find E? dV= - E. ds
–ds = i dx +j dy+k dz
and dV = - Exdx - Eydy - Ezdz
Ex = - dV/dx,
Ey = - dV/dy,
Ez = - dV/dz.
–So the x component of E is the derivative of V with respect to x, etc.
–If Ex = 0, then V = constant in that direction. Then lines or surfaces
–on which V remains constant are called equipotential lines or surfaces.
(Uniform field)
V = Ed
6
Electric Potential Energy of a system of point
charges
Potential of a charged isolated conductor
•Start out with a uniform electric field
with no excess charge on conductor.
Electrons on surface of conductor adjust
so that:
1. E=0 inside conductor
2. Electric field lines are perpendicular
to the surface. Suppose they weren’t?
3. Is the surface an equipotential?
G G
∆V = V f − Vi = − ∫ E.ds
f
i
How does a conductor shield the interior from
an exterior electric field?
•Start out with a uniform electric field
with no excess charge on conductor.
Electrons on surface of conductor adjust
so that:
1. E=0 inside conductor
2. Electric field lines are perpendicular
to the surface. Suppose they weren’t?
3. Does E = σ/ε0 just outside the conductor
4. Is σ uniform over the surface?
5. Is the surface an equipotential?
Dielectric Breakdown: Application of Gauss’s
Law
ƒ If the electric field in a gas exceeds a certain value, the gas
breaks down and you get a spark or lightning bolt if the gas is air.
In dry air at STP, you get a spark when E = 3*106 V/m. To
examine this we model the shape of a conductor with two
different spheres at each end:
V = constant on
surface of conductor
+
+
+
+
+
+
+
+
+
+
1
2
Radius R
6. If the surface had an excess charge, how would your answers change?
7
•The surface is at the same potential everywhere, but charge density
density and
electric fields are different. For a sphere, V= q/(4π
4πr2σ,
q/(4π ε0 r) and q = 4π
then V = (σ
(σ/ ε0 )*r.
Since E = σ/ ε0 near the surface of the conductor, we get V=E*r.
V=E*r.
Since V is a constant, E must vary as 1/r and σ as 1/r.
Hence, for surfaces where the radius is smaller, the electric field
field and
charge will be larger. This explains why:
V = constant on
surface of conductor
+
+
+
+
+
+
+
A metal slab is put in a uniform electric field of 106 N/C
with the field perpendicular to both surfaces
.
– Draw the appropriate model for the problem.
– Show how the charges are distributed on the
conductor.
– Draw the appropriate pill boxes.
– What is the charge density on each face of
the slab?
– Apply Gauss’s Law. ∫ E.dA = qin/ε0
+
+
+
1
Radius R
2
Sharp points on conductors have the highest electric fields and cause corona
discharge or sparks.
Pick up the most charge with charge tester from the pointy regions
regions of the nonnonspherical conductor.
Slab of metal In a uniform Electric field
Slab of metal
In a uniform Electric
field
-
E = 106 N/C
+
+
+
+
Gaussian
Pill Box
Evaluate E.dA = qin/ε0
• Left side of slab - E*A + 0*A = Aσ/ε0,
E = - σ /ε0, σ = - 106 N/C *10-11 C2/Nm2 = -10-5 N/m2
.Right side of slab
0*A + E*A = Aσ/ε0,
E = σ /ε0, σ = 106 N/C *10-11 C2/Nm2 = +10-5 N/m2
(note that charges arranges themselves so that field
inside is 0)
8