Potential energy change: $\Delta U = -W = -q\int_A^B \vec{E} \cdot d\vec{l}$

Electric Potential (Voltage)

$V = \frac{U}{q}$

Potential difference: $\Delta V = V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}$

(Path-independent for electrostatic fields)

Units: 1 volt = 1 joule/coulomb

Electric Field from Potential

$\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$

In one dimension: $E_x = -\frac{dV}{dx}$

Field points from high to low potential (down the potential gradient).

Potential of Point Charge

Choose $V = 0$ at $r = \infty$:

$V(r) = -\int_\infty^r \vec{E} \cdot d\vec{l} = -\int_\infty^r k\frac{q}{r^2} \, dr$

$V(r) = k\frac{q}{r}$

For multiple charges: $V = \sum_i k\frac{q_i}{r_i}$

(Scalar sum, easier than vector sum for $\vec{E}$!)

Potential of Continuous Distributions

$V = k\int \frac{dq}{r}$

Infinite Line Charge

$V = -\frac{\lambda}{2\pi\epsilon_0}\ln\frac{r}{r_0}$

(Must choose reference point $r_0$ since $V \to \infty$ at infinity)

Ring of Charge

On axis at distance $x$:

$V = \frac{kQ}{\sqrt{x^2 + R^2}}$

At center: $V = kQ/R$

Disk of Charge

On axis:

$V = \frac{\sigma}{2\epsilon_0}\left(\sqrt{x^2 + R^2} - x\right)$

Spherical Shell

Total charge $Q$, radius $R$:

Outside ($r > R$): $V = kQ/r$

On surface ($r = R$): $V = kQ/R$

Inside ($r < R$): $V = kQ/R$ (constant!)

Solid Sphere

Uniform charge density, total $Q$, radius $R$:

Outside ($r > R$): $V = kQ/r$

Inside ($r < R$): $V = \frac{kQ}{2R^3}(3R^2 - r^2)$

At center: $V = 3kQ/(2R)$

Equipotential Surfaces

Surfaces where $V$ = constant.

• Electric field perpendicular to equipotentials
• No work to move charge along equipotential
• Conductors are equipotentials

Electric Dipole Potential

Far field ($r \gg d$):

$V = k\frac{p\cos\theta}{r^2}$

where $\theta$ is angle from dipole axis and $p = qd$.

Electric field: $E_r = -\frac{\partial V}{\partial r} = \frac{2kp\cos\theta}{r^3}$

$E_\theta = -\frac{1}{r}\frac{\partial V}{\partial \theta} = \frac{kp\sin\theta}{r^3}$

Energy of Charge Distributions

Work to assemble charges:

$U = \frac{1}{2}\sum_{i} q_i V_i$

where $V_i$ is potential at location of $q_i$ due to all other charges.

For continuous distribution: $U = \frac{\epsilon_0}{2}\int E^2 \, dV$

(Energy stored in electric field)

$r = \sqrt{0.04 + 0.0225} = \sqrt{0.0625} = 0.25 \text{ m}$

Potential (scalar sum): $V = \frac{kQ}{r} = \frac{kQ}{\sqrt{x^2 + R^2}}$

$V = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})}{0.25}$

$V = \frac{71.92}{0.25} = 287.68 \text{ V}$

(b) Electric field from potential:

$E_x = -\frac{dV}{dx} = -\frac{d}{dx}\left[\frac{kQ}{\sqrt{x^2 + R^2}}\right]$

$= -kQ \cdot \left(-\frac{1}{2}\right)(x^2 + R^2)^{-3/2}(2x)$

$E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}$

$E_x = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})(0.20)}{(0.25)^3}$

$E_x = \frac{14.384}{0.015625} = 920.6 \text{ N/C}$

(c) Work done:

Work to bring charge from infinity: $W = qΔV = q(V_f - V_i) = q(V - 0)$

$W = (2.0 \times 10^{-9})(287.68)$

$W = 5.75 \times 10^{-7} \text{ J} = 575 \text{ nJ}$

Note: Work is positive because we're bringing like charges together (both positive).

Answers: (a) V = 288 V (b) E = 921 N/C (along axis, away from ring) (c) W = 575 nJ