Question

A charged particle having a charge of $-2\times {10}^{-6}C$ is placed close to a non conducting plate having a surface charge density as $4\times {10}^{-6}C{m}^{-2}$. What will be the force of attraction between them :

• A. $0.25N$
• B. $.0.75N$
• C. $0.5N$
• D. $0.45N$

Answer 1

Answer: (D)

The electric field due to charge plate with charge density $\sigma$ is given by, $E=\frac{\sigma }{2{ϵ}_0}$

Now the force between charge plate and charge $q$ is $F=qE=q\frac{\sigma }{2{ϵ}_0}$

or $F=(-2\times {10}^{-6})\times \frac{4\times {10}^{-6}}{2\times 8.85\times {10}^{-12}}=-0.45N$

(Negative sign indicates the force between them is attractive)