recall and use E = Q / (4πε0 r 2) for the electric field strength due to a point charge in free space

Electric Field of a Point Charge

Learning Objective

Recall and use the expression

\$E = \frac{Q}{4\pi\varepsilon_0 r^{2}}\$

to calculate the electric field strength due to a point charge in free space.

Key Concepts

• Electric field (E): a vector quantity representing the force per unit positive test charge.
• Point charge (Q): an idealised charge concentrated at a single location.
• Distance (r): the straight‑line separation between the point charge and the point where the field is being evaluated.
• Permittivity of free space (\$\varepsilon0\$): a constant, \$\varepsilon0 = 8.854 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}\$.
• The field radiates outward from a positive charge and inward toward a negative charge.

Derivation Overview

The expression for the electric field of a point charge follows from Coulomb’s law for the force between two point charges:

\$F = \frac{1}{4\pi\varepsilon_0}\frac{|Qq|}{r^{2}}\$

Dividing the force by a small positive test charge \$q\$ gives the field:

\$E = \frac{F}{q} = \frac{Q}{4\pi\varepsilon_0 r^{2}}\$

Direction is given by the unit vector \$\hat{r}\$ pointing away from a positive \$Q\$ (or toward a negative \$Q\$):

\$\vec{E} = \frac{Q}{4\pi\varepsilon_0 r^{2}}\,\hat{r}\$

Using the Formula

1. Identify the magnitude of the point charge \$Q\$ (in coulombs).
2. Determine the distance \$r\$ from the charge to the point of interest (in metres).
3. Substitute \$Q\$, \$r\$, and \$\varepsilon_0 = 8.854\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}\$ into the formula.
4. Calculate \$E\$; the result is in newtons per coulomb (N C⁻¹) which is equivalent to volts per metre (V m⁻¹).
5. Assign the correct direction: radially outward for \$Q>0\$, radially inward for \$Q<0\$.

Worked Example

Problem: A point charge of \$+5.0\ \mu\text{C}\$ is located at the origin. Find the magnitude of the electric field at a point \$0.20\ \text{m}\$ from the charge.

Solution:

1. Convert the charge: \$Q = +5.0\ \mu\text{C} = 5.0\times10^{-6}\ \text{C}\$.
2. Distance: \$r = 0.20\ \text{m}\$.
3. Insert into the formula:

   \$E = \frac{5.0\times10^{-6}}{4\pi(8.854\times10^{-12})(0.20)^{2}}\$

4. Calculate:

   \$\$E \approx \frac{5.0\times10^{-6}}{4\pi(8.854\times10^{-12})(0.04)}

   \approx 2.24\times10^{5}\ \text{N C}^{-1}\$\$

5. Direction: radially outward from the origin because the charge is positive.

Common Mistakes to Avoid

• Forgetting to square the distance \$r\$.
• Mixing units (e.g., using cm instead of m).
• Omitting the \$4\pi\$ factor, which leads to a result that is too large by roughly \$12.57\$ times.
• Confusing the sign of \$Q\$ with the direction of the field vector.

Quick Reference Table

Suggested Diagram

Summary

The electric field produced by a point charge in free space follows the inverse‑square law:

\$\boxed{E = \dfrac{Q}{4\pi\varepsilon_0 r^{2}}}\$

Remember to keep track of units, square the distance, and assign the correct radial direction based on the sign of \$Q\$. Mastery of this formula underpins many later topics, such as electric potential, field superposition, and electrostatic forces between multiple charges.