Two parallel large thin metal sheets have equal charge densities of $5.31 × 10^{-11}\, C m^{-2}$ of opposite signs. If the permittivity of free space is equal to $8.85 × 10^{-12}\, C^2\, N^{-1}m^{-2}$, the electric field between these sheets will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $6\, N C^{-1}$
Given:
Surface charge density, $\sigma = 5.31 \times 10^{-11} \ \text{C/m}^2$
Permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \ \text{C}^2/\text{N·m}^2$
Electric field due to a single sheet: $E_\text{sheet} = \frac{\sigma}{2 \epsilon_0}$
For two parallel sheets with opposite charges, the fields add up between the sheets:
$E = 2 \cdot \frac{\sigma}{2 \epsilon_0} = \frac{\sigma}{\epsilon_0}$
Substitute values:
$E = \frac{5.31 \times 10^{-11}}{8.85 \times 10^{-12}} \approx 6.0 \ \text{N/C}$
Electric field between the sheets: $E \approx 6.0 \ \text{N/C}$