Question

Two parallel large thin metal sheets have equal surface charge densities $\left(\sigma=26.4 \times 10^{-12} C / m ^{2}\right)$ of opposite signs. The electric field between these sheets is

• A. $1.5\,N/C$
• B. $1.5\times 10^{-10} \,N/C$
• C. $3\,N/C$
• D. $3\times 10^{-10}\,N/C$

Answer 1

Answer: (C)

The situation is shown in the figure. Plate $1$ has surface charge density $\sigma$ and plate $2$ has surface charge density $\sigma$. The electric field at point $P$ due to two charged plates add up, giving

$E =\frac{\sigma}{2 \varepsilon_{0}}+\frac{\sigma}{2 \varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}} $
Given, $ \sigma =2.64 \times 10^{-12} C / m ^{2}, $
$\varepsilon_{0} =8.85 \times 10^{-12} C ^{2} / N - m ^{2}$
Hence, $ E =\frac{26.4 \times 10^{-12}}{8.85 \times 10^{-12}} \approx 3\, N / C$