Electric Field of a Point Charge

A single point charge produces a radial electric field — every field line points directly away from the charge (for positive charges) or directly toward it (for negative charges). The magnitude of the field depends only on the distance:

$E(r) = \dfrac{k\,|q|}{r^{2}}$

where $k = 9 \times 10^{9}\;\text{N}\cdot\text{m}^{2}/\text{C}^{2}$ (Coulomb's constant). The field strength drops off as the inverse square of distance — at twice the distance, the field is one-quarter as strong.

Step-by-Step Solution

Given: A point charge $q = +2\;\text{μC} = 2 \times 10^{-6}\;\text{C}$.

Find: The electric field magnitude at distances of 0.5 m, 1 m, and 2 m, and the direction of the field.

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Step 1 — Apply Coulomb's law for the field at $r = 0.5\;\text{m}$.

$E = \dfrac{k\,q}{r^{2}} = \dfrac{(9 \times 10^{9})(2 \times 10^{-6})}{(0.5)^{2}}$

$= \dfrac{1.8 \times 10^{4}}{0.25}$

$= 7.2 \times 10^{4}\;\text{V/m} = 72{,}000\;\text{V/m}$

Step 2 — At $r = 1\;\text{m}$.

$E = \dfrac{(9 \times 10^{9})(2 \times 10^{-6})}{1^{2}} = 1.8 \times 10^{4}\;\text{V/m} = 18{,}000\;\text{V/m}$

That's exactly 1/4 of the field at 0.5 m, since the distance doubled and the field goes as $1/r^{2}$.

Step 3 — At $r = 2\;\text{m}$.

$E = \dfrac{(9 \times 10^{9})(2 \times 10^{-6})}{4} = 4500\;\text{V/m}$

That's 1/16 of the field at 0.5 m. Quadruple the distance → field drops by 16×.

Step 4 — Direction at any point.

The electric field points radially outward from a positive charge. At a point with position vector $\vec r$ measured from the charge:

$\vec E(\vec r) = \dfrac{kq}{r^{2}}\,\hat r$

where $\hat r = \vec r/r$ is the unit vector from the charge to the field point. For a negative charge, just flip the direction — the field points inward.

Step 5 — Force on a test charge.

The force on a test charge $q_0$ placed in this field is $\vec F = q_0\vec E$. So a $+1\;\text{nC}$ charge at $r = 1\;\text{m}$ would feel:

$F = (10^{-9})(18{,}000) = 1.8 \times 10^{-5}\;\text{N}$

Tiny but measurable.

Step 6 — Field lines as a visualization tool.

We draw field lines that:

• Start on positive charges and end on negative charges (or extend to infinity).
• Never cross.
• Are denser where the field is stronger.

For a single positive point charge, the lines are an infinite radial spray — symmetric in all directions.

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Answer: A point charge $q = 2\;\text{μC}$ produces a radially outward field with magnitudes:

• At $0.5\;\text{m}$: $\boxed{72{,}000\;\text{V/m}}$
• At $1\;\text{m}$: $18{,}000\;\text{V/m}$
• At $2\;\text{m}$: $4500\;\text{V/m}$

The field strength drops as $1/r^{2}$ — doubling the distance reduces the field by a factor of 4.

Try It

• Adjust the charge $q$ — the field at any radius scales linearly.
• Toggle the sign of the charge — field lines reverse direction.
• The faint probe vectors show the local field direction and magnitude.