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Topic 4 of 15

Electric Potential

The electric field tells us the force a charge feels; the electric potential tells us the potential energy per unit charge at every point in space. Because potential is a scalar (just a number, not a vector), it is far easier to add contributions from multiple sources. Once we have V everywhere, we can recover the field from it — making potential one of the most powerful concepts in electrostatics.

Key Concepts

Electric Potential Energy

The work done by an external agent bringing charge

q

from infinity to a point in a field, stored as potential energy. For two point charges:

U = kq_1q_2/r

. Negative

U

means the configuration is bound (attractive); positive means repulsive.

Electric Potential

The electric potential

V

at a point is the potential energy per unit positive charge:

V = U/q_0

. Units: volts (V = J/C). For a point charge:

V = kq/r

. Potential is a scalar — contributions from multiple charges add algebraically.

Potential Difference

The potential difference

\Delta V = V_B - V_A

between two points equals the negative of the work done per unit charge by the electric field moving from A to B:

\Delta V = -W_{AB}/q

. Only differences in

V

are physically meaningful.

Equipotential Surfaces

Surfaces on which

V

is constant. No work is done moving a charge along an equipotential. Equipotentials are always perpendicular to electric field lines. For a point charge they are concentric spheres.

Relationship Between E and V

The electric field is the negative gradient of potential:

E_x = -\partial V/\partial x

. In 1D:

E = -\Delta V / \Delta x

. The field points from high potential to low potential — "downhill" in the potential landscape.

Key Equations

Potential of a point charge

V = \frac{kq}{r}

Electric potential at distance r from a point charge q. Sign matters: positive for +q, negative for −q.

Potential energy of two charges

U = \frac{kq_1 q_2}{r}

Potential energy stored in a pair of point charges separated by r. Positive = repulsive system; negative = attractive.

Work done by field

W = q(V_A - V_B) = -q\,\Delta V

Work done by the electric force on charge q moving from A to B. Positive work when positive charge moves from high V to low V.

Field from potential (1D)

E = -\frac{\Delta V}{\Delta x}

Magnitude of electric field equals the rate of decrease of potential with distance. Field points in the direction of decreasing V.

Worked Example

Potential and Work for Two Point Charges

Problem

Charges $q_1 = +3\,\mu$ C (at origin) and $q_2 = -5\,\mu$ C (at $x = 0.40$ m) are fixed. (a) Find the electric potential at point P at $x = 0.20$ m. (b) How much work is required to bring $q_3 = +2\,\mu$ C from infinity to P?

Solution

(a) Potential is a scalar sum of contributions from each charge. Distance from $q_1$ to P is $r_1 = 0.20$ m; distance from $q_2$ to P is $r_2 = 0.20$ m:

V_1 = \frac{kq_1}{r_1} = \frac{(8.99\times10^9)(3\times10^{-6})}{0.20} = +134{,}850 \text{ V}

V_2 = \frac{kq_2}{r_2} = \frac{(8.99\times10^9)(-5\times10^{-6})}{0.20} = -224{,}750 \text{ V}

V_P = V_1 + V_2 = 134{,}850 - 224{,}750 = -89{,}900 \text{ V} \approx -90.0 \text{ kV}

(b) Work done by external agent equals the change in potential energy:

W = q_3 V_P = (2\times10^{-6})(-89{,}900) = -0.180 \text{ J}

The negative sign means the field actually does positive work bringing $q_3$ in — the external agent must restrain it.

Answer (a) V_P ≈ −90.0 kV; (b) W = −0.180 J (field does +0.180 J of work).

Practice

Exercises

7 problems

1 of 7

What is the electric potential (in kV) at a distance $r = 0.30$ m from a point charge $q = +5\,\mu$ C?

Your answer

2 of 7

What is the electric potential energy (in J) of two charges $q_1 = +2\,\mu$ C and $q_2 = +3\,\mu$ C separated by $r = 0.25$ m?

Your answer

3 of 7

Charges $q_1 = +4$ nC (at $x = 0$ ) and $q_2 = -4$ nC (at $x = 0.60$ m) form a dipole. What is the electric potential (in V) at the midpoint $x = 0.30$ m?

Unlock Exercise 3