Question

A charged particle having a charge of $-2\times {10}^{-6}\text{C}$ is placed close to a non-conducting plate having a surface charge density $4\times {10}^{-6}{\text{C/m}}^2$. Find the force of attraction between the particle and the plate.

• A. $0.22\text{N}$
• B. $0.33\text{N}$
• C. $0.44\text{N}$
• D. $0.55\text{N}$

Answer 1

The correct option is C $0.44\text{N}$

Figure for given problem:


Electric field near the charged non-conducting plate,
$E=\frac{\sigma }{2{ϵ}_0}$

Substituting the values,
$E=\frac{4\times {10}^{-6}}{2\times 9\times {10}^{-12}}$

$\therefore E=2.2\times {10}^5\text{N/C}$
Now, force of attraction between the particle and the plate,
$F=qE$
$\Rightarrow F=2\times {10}^{-6}\times 2.2\times {10}^5$
$\Rightarrow F=0.44\text{N}$
Hence, option (c) is the correct answer.