1. Relation Between Electric Potential V & Electric Field E
Figure To find Vba in a nonuniform electric field E, we integrate E·dl from point a to point b.

2. F = qE & U = qV So, the relationship is: For the Electric Force:
The General Relationship between a Conservative
Force & its Potential Energy has the form:
For the Electric Force:
F = qE & U = qV
So, the relationship is:
Figure To find Vba in a nonuniform electric field E, we integrate E·dl from point a to point b.

3. The simplest case is a Uniform E Field

4. E field obtained from voltage
Example
E field obtained from voltage
Two parallel plates are charged to
produce a potential difference of
50 V. The plate separation is
d = m.
Calculate the magnitude of the electric
field E in the space between the plates.
Solution: E = V/d = 1000 V/m.

5. E field obtained from voltage
Example
E field obtained from voltage
Two parallel plates are charged to
produce a potential difference of
50 V. The plate separation is
d = m.
Calculate the magnitude of the electric
field E in the space between the plates.
E = (Vba)/d = (50)/(0.05)
E = 1,000 V/m
Solution: E = V/d = 1000 V/m.

6. Charged Conducting Sphere (a) r > r0 , (b) r = r0 , (c) r < r0
Example
Charged Conducting Sphere
A conducting sphere is uniformly charged with total charge Q. Calculate the potential at a distance r from the sphere’s center for
(a) r > r0 , (b) r = r0 , (c) r < r0
V is plotted here, and compared with the electric field E. Use the relation:
Solution: The electric field outside a conducting sphere is Q/(4πε0r2). Integrating to find the potential, and choosing V = 0 at r = ∞:
a. V = Q/4πε0r.
b. V = Q/4πε0r0.
c. V = Q/4πε0r0 (the potential is constant, as there is no field inside the sphere).

7. Example Charged Conducting Sphere
Calculate the potential at a distance r from the sphere’s center for
(a) r > r0 , (b) r = r0 , (c) r < r0
Use the relation:
Vba = - [Q/(4πε0)] ∫[dr/(r2)]
Limits ra = to rb
Solution: The electric field outside a conducting sphere is Q/(4πε0r2). Integrating to find the potential, and choosing V = 0 at r = ∞:
a. V = Q/4πε0r.
b. V = Q/4πε0r0.
c. V = Q/4πε0r0 (the potential is constant, as there is no field inside the sphere).

8. Setting the potential V to zero at r = ∞ gives the General Form of the Potential Due to a Point Charge:
Figure Potential V as a function of distance r from a single point charge Q when the charge is positive.
Figure Potential V as a function of distance r from a single point charge Q when the charge is negative.

9. Example: Work required to bring two positive charges close together:
Calculate the minimum work that must be done
by an external force to bring a charge q = 3 μC
from a great distance away (take r = ∞) to a
point m from a charge Q = 20 µC.
Solution: The work is equal to the change in potential energy; W = 1.08 J. Note that the field, and therefore the force, is not constant.

10. W = qV = q(Vb –Va) = qke[(Q/rb) – keQ(Q/ra)] 1.08 J
Example: Work required to bring two positive charges close together:
Calculate the minimum work that must be done
by an external force to bring a charge q = 3 μC
from a great distance away (take r = ∞) to a
point m from a charge Q = 20 µC.
W = qV = q(Vb –Va)
= qke[(Q/rb) – keQ(Q/ra)]
1.08 J
Solution: The work is equal to the change in potential energy; W = 1.08 J. Note that the field, and therefore the force, is not constant.

11. Example: Potential above two charges
(a) Calculate the electric potential at point A in the figure due to the two charges shown. (b) Repeat the calculation for point B. (Solution on white board).
Solution: The total potential is the sum of the potential due to each charge; potential is a scalar, so there is no direction involved, but we do have to keep track of the signs.
a. V = 7.5 x 105 V
b. V = 0 (true at any point along the perpendicular bisector)