Rankers Physics

Electric Field: Practice Problem & Solution

Two point charges A and B, having charges $+Q$ and $-Q$ respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes : (2019)
F
$\frac{9F}{16}$
$\frac{16F}{9}$
$\frac{4F}{3}$

Solution Explained:

To solve this problem, we apply the core principles of Electric Field. Understanding the underlying formula is key to arriving at the correct answer below:

Initial force $F = \frac{kQ^2}{r^2}$. When 25% ($Q/4$) of A is transferred to B, new charge on A is $Q - Q/4 = 3Q/4$ and on B is $-Q + Q/4 = -3Q/4$. The new force is $F' = \frac{k(3Q/4)(3Q/4)}{r^2} = \frac{9}{16}\frac{kQ^2}{r^2} = \frac{9F}{16}$.

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