Electric Field and Coulomb's Law

Coulomb's law, electric field from continuous charge distributions

Written and reviewed by the Study Mondo Education TeamLast updated February 16, 2026

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Electric Field and Coulomb's Law

Coulomb's Law

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

📚 Practice Problems

1Problem 1hard

❓ Question:

A thin rod of length L = 0.5 m has a non-uniform linear charge density λ(x) = λ₀x, where λ₀ = 4.0 × 10⁻⁶ C/m² and x is measured from one end. Find the electric field at a point P located a distance d = 0.2 m from the end where x = 0, along the axis of the rod.

💡 Show Solution

Given:

L = 0.5 m
λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
d = 0.2 m
k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: $dq = λ(x)dx = λ_0 x \, dx$

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.

2Problem 2medium

❓ Question:

A ring of radius R = 0.1 m carries a uniformly distributed charge Q = 2.0 × 10⁻⁸ C. (a) Find the electric field at a point on the axis at distance x = 0.05 m from the center. (b) At what distance x does the electric field reach its maximum value? (c) What is the maximum field strength?

💡 Show Solution

Given:

R = 0.1 m
Q = 2.0 × 10⁻⁸ C = 20 nC
k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: $E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}$

(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

3Problem 3medium

❓ Question:

An electric dipole consists of charges +q and -q separated by distance d = 2.0 mm. The dipole moment is p = 5.0 × 10⁻¹² C·m. Find (a) the magnitude of each charge, (b) the electric field at a point on the perpendicular bisector at distance r = 10 cm (where r >> d), and (c) the electric field at a point on the axis at distance r = 10 cm from the center.

💡 Show Solution

Given:

d = 2.0 × 10⁻³ m = 2.0 mm
p = 5.0 × 10⁻¹² C·m
r = 0.1 m = 10 cm
k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

$q = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}$

(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)**

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Electric Field and Coulomb's Law

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Electric Field and Coulomb's Law

Coulomb's Law

Electric Field

Continuous Charge Distributions

Linear Charge Density

Surface Charge Density

Volume Charge Density

Example: Infinite Line of Charge

Example: Ring of Charge

Example: Disk of Charge

Example: Spherical Shell

Dipole

📚 Practice Problems

1Problem 1hard

❓ Question:

2Problem 2medium

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3Problem 3medium

❓ Question:

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