Question

Comprehension II

Concept of image charge makes it possible to find electric interaction in some cases where large conducting plates are kept infront of charge. For this we consider potential of conducting plate as zero and find another (image) charge which along with given charge can make potential of conducting plate zero. In this way we can find electrostatic force between charge and conducting plate by replacing conducting plate with image charge.

Charge q is placed is infront of a conducting plate as shown in figure. Force on q by plate is

• A. 0.0
• B. $\frac{k{q}^2}{a^2}$
• C. $\frac{k{q}^2}{8a^2}(2\sqrt{2}-1)$
• D. $\frac{k{q}^2}{4a^2}(\sqrt{2}-1)$

Answer 1

The correct option is C $\frac{k{q}^2}{8a^2}(2\sqrt{2}-1)$

$F^{\prime }=\frac{k{q}^2}{8a^2},F=\frac{k{q}^2}{4a^2}$
$F_{net}$ = Fcos45 - F'
$F_{net}=\frac{k{q}^2}{8a^2}(2\sqrt{2}-1)$

$F=\frac{k{q}^2}{(2x{)}^2}\Rightarrow$ dW = F.dx
$W={\int }_{\infty }^l\frac{k{q}^2}{4x^2}dx\Rightarrow W=\frac{k{q}^2}{4l}$