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

Answer 1

Answer: (A)

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

$E =E_{1}+E_{2}=\frac{\sigma}{2 \varepsilon_{0}}+\frac{\sigma}{2 \varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}$
Given, $\sigma =26.4 \times 10^{-12} C / m ^{2}$
$\varepsilon_{0} =8.85 \times 10^{-12} C ^{2} / N - m ^{2}$
Hence, $E =\frac{26.4 \times 10^{-12}}{8.85 \times 10^{-12}} \approx 3\, N / C$