Two parallel plates of area 0.02 m² carry charges +Q and –Q respectively, with Q=4 nC. Find the magnitude of the electric field between the plates.

Step-by-step solution

1. Step 1: Determine the surface charge density: $\sigma =\frac{Q}{A}=\frac{4\times 10^{-9},\text{C}}{0.02,{\text{m}}^2}=2.0\times 10^{-7},{\text{C/m}}^2$.
2. Step 2: For two infinite plates with equal and opposite charges, the field between them is $E=\frac{sigma}{varepsilo{n}_0}$.
3. Step 3: Substitute the constants: ${\epsilon }_0=8.854\times 10^{-12},{\text{C}}^2{\text{/N m}}^2$.
4. Step 4: Compute: $E=\frac{2.0\times 10^{-7}}{8.854\times 10^{-12}}\approx 2.26\times 10^4,\text{N/C}$.
5. Step 5: Round to three significant figures: $2.3\times 10^4,\text{N/C}$.

Concept tested: Electric field between parallel plates

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked