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

What is an Electric Field?

Think of an electric field as a force map that surrounds a charged object.

If you place a tiny test charge (like a small ball) in this map, the field tells the ball how to move.

It’s invisible, but its effect is real—just like the wind pushes a kite even though you can’t see the air molecules.

The Key Formula

The electric field strength produced by a point charge in free space is:

\$E = \frac{Q}{4\pi\epsilon_0 r^2}\$

• \$Q\$ = charge of the point source (Coulombs)
• \$r\$ = distance from the charge (metres)
• \$\epsilon_0\$ = permittivity of free space ≈ \$8.85\times10^{-12}\$ F/m

Quick Example

A point charge of \$+2.0\,\mu\text{C}\$ is placed at the origin.

What is the electric field 0.10 m away?

1. Convert charge to Coulombs: \$2.0\,\mu\text{C} = 2.0\times10^{-6}\,\text{C}\$.
2. Plug into the formula:

   \$E = \frac{2.0\times10^{-6}}{4\pi(8.85\times10^{-12})(0.10)^2}\$

3. Calculate:

   \$E \approx \frac{2.0\times10^{-6}}{1.112\times10^{-12}} \approx 1.80\times10^{6}\,\text{N/C}\$

4. Result: \$E \approx 1.8\times10^{6}\,\text{N/C}\$ pointing away from the charge (since it’s positive).

Important Constants

Exam Tip 🚀

• Always check the units—they must cancel to give N/C.

• Remember that \$E\$ points away from a positive charge and toward a negative one.

• For multiple charges, use the principle of superposition: add the vectors of each field.

• If the problem asks for the field at a point, identify the distance \$r\$ correctly; it’s not the distance between charges unless specified.

• Practice converting microcoulombs to coulombs and meters to centimeters to avoid mistakes.