Two parallel plates of area 0.01 m² are separated by 4 mm. They carry charges +6 nC and –6 nC respectively. Find the magnitude of the electric field between the plates.

Step-by-step solution

1. The electric field between two parallel plates is $E=sigma/{\epsilon }_0$, where σ is the surface charge density.
2. Compute σ: $\sigma =Q/A=6\times 10^{-9}C/0.01m^2=6\times 10^{-7}C/m^2$.
3. Insert into $E=\sigma /{\epsilon }_0$: $E=6\times 10^{-7}/(8.85\times 10^{-12})\approx 1.50\times 10^{5}N/C$.
4. The field is directed from the positively charged plate to the negatively charged plate.
5. Thus the magnitude of the field between the plates is $1.50\times 10^{5}N/C$.

Concept tested: Electric field between parallel plates

Published 13 September 2026 · How this solution was prepared and checked