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Topic 3 of 9

Electric Potential & Energy

The electric potential $V$ is a scalar field whose negative gradient gives $\vec E$. Working with $V$ is usually easier than with $\vec E$ directly: it's a scalar sum rather than a vector sum. The energy stored in an electric field is $u=\varepsilon_0 E^2/2$ per unit volume.

Key Concepts

Electric Potential

V(\vec r)=-\int_{\mathcal{O}}^{\vec r}\vec E\cdot d\vec l

(reference point

\mathcal{O}

usually at infinity). For a point charge:

V=kq/r

\vec E=-\nabla V

Superposition

V

is a scalar, so the potential of multiple charges is the algebraic sum:

V=\sum_i kq_i/r_i

. Much simpler than the vector sum for

\vec E

Energy of a Charge Distribution

The energy stored in assembling a continuous charge distribution:

W=\frac{\varepsilon_0}{2}\int E^2\,d^3r

. Energy density:

u=\varepsilon_0 E^2/2

Method of Images

A point charge

+q

at height

d

above a grounded infinite conductor can be replaced by

+q

plus an image charge

-q

at depth

d

below the surface. The boundary condition

V=0

at the surface is automatically satisfied.

Key Equations

Potential of point charge

V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}

Scalar; superpose for multiple charges.

Energy density

u = \frac{\varepsilon_0}{2}E^2 = \frac{1}{2\mu_0}B^2

Energy stored per unit volume in EM fields.

Energy of point charges

W = \frac{1}{2}\sum_i q_i V(\vec{r}_i)

Total electrostatic energy; factor ½ avoids double-counting.

Worked Example

Potential of Two Charges

Problem

Charge $q_1=+3.0$ nC is at the origin and $q_2=-1.0$ nC is at $x=0.40$ m. Find the potential at $P=(0.40,0.30,0)$ m.

Solution

Distance from $q_1$ to $P$ : $r_1=\sqrt{0.16+0.09}=0.50$ m.

Distance from $q_2$ to $P$ : $r_2=\sqrt{(0.40-0.40)^2+0.09}=0.30$ m.

V = k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right) = 8.99\times10^9\left(\frac{3\times10^{-9}}{0.50}+\frac{-10^{-9}}{0.30}\right)

V = 8.99\times10^9(6\times10^{-9}-3.33\times10^{-9}) = 8.99\times10^9\times2.67\times10^{-9} = 24.0\text{ V}

Answer

V=24.0

Practice

Exercises

7 problems

1 of 7

Drag the slider along the V = kQ/r curve and watch the equipotential rings change. For Q = 5.0 nC, find V at r = 0.30 m (in V).

electric potential

Explore radius r V(0.30 m) = 149.8 V

V = kQ/r falls off as 1/r — halving the distance doubles the potential. Equipotential surfaces are spheres.

V at r=0.3 m

V = V

V = kQ/r, k=8.99×10⁹, Q=5 nC, r=0.3 m

2 of 7

Watch the test charge move from r₁ to r₂ along the potential curve. The shaded area is ΔV. Find the work W (in nJ) to move q = 2.0 nC from r₁ = 0.50 m to r₂ = 0.20 m near Q = 4.0 nC.

work & potential

Work done = charge × potential difference. Moving toward a like-sign source takes positive work.

W = q·ΔV

W = nJ

q=2 nC, Q=4 nC, r₁=0.5 m → r₂=0.2 m

3 of 7

Energy stored assembling two charges $q_1=q_2=1.0\times10^{-9}$ C separated by $d=0.10$ m (in J). $W=kq_1q_2/d$ .

Unlock Exercise 3