$W = \int_A^B \vec{F} \cdot d\vec{l} = q\int_A^B \vec{E} \cdot d\vec{l}$

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)

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

(a) Potential at center (r = 0):

$V(0) = \frac{kQ}{2R^3}(3R^2 - 0) = \frac{3kQ}{2R}$

$V(0) = \frac{3(8.99 \times 10^9)(6.0 \times 10^{-8})}{2(0.10)}$

$V(0) = \frac{1618.2}{0.20} = 8091 \text{ V}$

(b) Potential at surface (r = R):

Using inside formula: $V(R) = \frac{kQ}{2R^3}(3R^2 - R^2) = \frac{kQ}{2R^3}(2R^2) = \frac{kQ}{R}$

Or using outside formula: $V(R) = \frac{kQ}{R}$

$V(R) = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.10}$

$V(R) = 5394 \text{ V}$

(c) Potential at r = 0.20 m > R:

$V(0.20) = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.20}$

$V(0.20) = 2697 \text{ V}$

Verification:

• V(0) = (3/2)V(R) = (3/2)(5394) = 8091 V ✓
• V(0.20) = (1/2)V(R) since r = 2R ✓

Potential ratio: $\frac{V(0)}{V(R)} = \frac{3/2 \cdot kQ/R}{kQ/R} = \frac{3}{2} = 1.5$

Answers: (a) V(0) = 8.09 kV (center) (b) V(R) = 5.39 kV (surface) (c) V(0.20 m) = 2.70 kV (outside)

(a) Potential at center:

Potential is scalar sum: $V = k\sum \frac{q_i}{r_i} = \frac{k}{r}(q_1 + q_2 + q_3)$

$V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9} + 4.0 \times 10^{-9} - 4.0 \times 10^{-9})$

$V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9})$

$V = (5.20 \times 10^{10})(4.0 \times 10^{-9}) = 208 \text{ V}$

(b) Electric field at center:

Set up coordinates: q₁ at top, q₂ at bottom-left, q₃ at bottom-right

Field from each charge: $E_i = \frac{kq_i}{r^2} = \frac{(8.99 \times 10^9)(4.0 \times 10^{-9})}{(0.173)^2}$

$E_i = \frac{35.96}{0.0299} = 1203 \text{ N/C}$

Vector addition:

By symmetry, q₁ and q₂ (both positive, same magnitude):

• Their fields point away from charges
• The horizontal components cancel
• Vertical components add

The field from q₃ (negative) points toward it (upward by our setup).

Due to symmetry of two positive charges and one negative of equal magnitude:

Net field components:

• E₁ points down from top: E₁ = 1203 N/C (down)
• E₂ points from bottom-left: at 60° from horizontal
• E₃ points toward bottom-right: at 60° from horizontal (opposite side)

E₂ and E₃ have components that partially cancel...

Actually, with two +4 nC and one -4 nC arranged symmetrically, the net field has magnitude:

$E_{net} = \frac{3k|q|}{r^2} = 3(1203) = 3609 \text{ N/C}$

Direction: toward the negative charge.

(c) Work to bring charge from infinity:

$W = qV = (2.0 \times 10^{-9})(208) = 4.16 \times 10^{-7} \text{ J}$

$W = 416 \text{ nJ}$

Answers: (a) V = 208 V (b) E ≈ 3.6 kN/C (toward q₃) (c) W = 416 nJ