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

The electric field contribution at P (located at distance d from x = 0): $dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}$

Integration:

$E = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx$

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: $E = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du$

$E = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}$

$E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]$

$E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]$

Numerical calculation: $E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]$

$E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]$

$E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)$

$E = 1.94 \times 10^4 \text{ N/C}$

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.