Question

A particle A of charge q is placed near a uniformly charged infinite plane sheet with surface charge density σ, then it experiences a force F₁ while when this particle is placed near a metal plate with surface charge density σ, then it experiences a force F₂. Then

$\frac{|F_1|}{|F_2|}$ is equal to

• A. 1
• B. 2
• C. $\frac{1}{2}$
• D. 3

Answer 1

Answer: (C)

$\frac{1}{2}$

Answer 2

For an infinite plane sheet carrying charge density $\sigma$, the electric field is

$E_1 = \frac{\sigma}{2\epsilon_0}$.

Thus, the force on a charge $q$ is

$F_1 = qE_1 = \frac{q\sigma}{2\epsilon_0}$.

For a metal plate (conductor), the free charges arrange themselves so that the electric field outside is

$E_2 = \frac{\sigma}{\epsilon_0}$.

Therefore, the force on the charge is

$F_2 = qE_2 = \frac{q\sigma}{\epsilon_0}$.

Taking the ratio of the magnitudes:

$\frac{|F_1|}{|F_2|} = \frac{\frac{q\sigma}{2\epsilon_0}}{\frac{q\sigma}{\epsilon_0}} = \frac{1}{2}$.