Force between two point charges:

$\vec{F} = k\frac{q_1q_2}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\hat{r}$

where:

• $k = 8.99 \times 10^9$ N·m²/C²
• $\epsilon_0 = 8.85 \times 10^{-12}$ C²/(N·m²) (permittivity of free space)

Vector form: $\vec{F}_{12} = k\frac{q_1q_2}{r_{12}^2}\hat{r}_{12}$

Electric Field

Electric field due to point charge:

$\vec{E} = k\frac{q}{r^2}\hat{r}$

Definition: $\vec{E} = \frac{\vec{F}}{q_0}$

where $q_0$ is test charge.

Superposition principle: $\vec{E}_{total} = \sum_i \vec{E}_i$

Continuous Charge Distributions

For continuous distribution with charge density $\rho$:

$\vec{E} = k\int \frac{dq}{r^2}\hat{r}$

Linear Charge Density

Charge per unit length: $\lambda = dq/dl$

$d\vec{E} = k\frac{\lambda \, dl}{r^2}\hat{r}$

Surface Charge Density

Charge per unit area: $\sigma = dq/dA$

$d\vec{E} = k\frac{\sigma \, dA}{r^2}\hat{r}$

Volume Charge Density

Charge per unit volume: $\rho = dq/dV$

$d\vec{E} = k\frac{\rho \, dV}{r^2}\hat{r}$

Example: Infinite Line of Charge

Uniform line charge density $\lambda$, find field at distance $r$:

By symmetry, field is radial. Consider element at distance $z$:

$dE_x = \frac{k\lambda \, dz}{(r^2 + z^2)}\frac{r}{\sqrt{r^2 + z^2}}$

$E = \int_{-\infty}^{\infty} \frac{k\lambda r \, dz}{(r^2 + z^2)^{3/2}}$

Using $z = r\tan\theta$:

$E = \frac{2k\lambda}{r} = \frac{\lambda}{2\pi\epsilon_0 r}$

Example: Ring of Charge

Ring of radius $R$, total charge $Q$, find field on axis at distance $x$:

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

At center ($x = 0$): $E = 0$ (by symmetry)

Far from ring ($x \gg R$): $E \approx kQ/x^2$ (point charge)

Example: Disk of Charge

Uniform surface charge density $\sigma$, radius $R$, field on axis:

$E_x = \frac{\sigma}{2\epsilon_0}\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)$

At surface ($x = 0$): $E = \sigma/(2\epsilon_0)$

Far from disk ($x \gg R$): $E \approx k\pi R^2\sigma/x^2 = kQ/x^2$

Infinite sheet ($R \to \infty$): $E = \sigma/(2\epsilon_0)$ (uniform!)

Example: Spherical Shell

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

Outside ($r > R$): $E = kQ/r^2$ (like point charge)

Inside ($r < R$): $E = 0$

(From Gauss's law, derived below)

Dipole

Two charges $+q$ and $-q$ separated by distance $d$:

Dipole moment: $\vec{p} = q\vec{d}$

(points from $-q$ to $+q$)

Field on axis (far field, $r \gg d$): $E_{axis} = \frac{2kp}{r^3}$

Field on perpendicular bisector: $E_{perp} = \frac{kp}{r^3}$

(a) Field at x = 0.05 m:

$E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}$

$E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}$

$E = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}$

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: $\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0$

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

$= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0$

$R^2 - 2x^2 = 0$

$x = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}$

(c) Maximum field strength:

$E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}$

$= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}$

$= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}$

$= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}$

$= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}$

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

(b) Field on perpendicular bisector (far field, r >> d):

$E_{perp} = \frac{kp}{r^3}$

$E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}$

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

$E_{axis} = \frac{2kp}{r^3}$

$E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}$

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: $E_{axis} = 2E_{perp}$

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**