# Two point charges 4q and -q are fixed on the x-axis at x=-d/2 and x= +d/2 respectively ….

Q: Two point charges 4q and -q are fixed on the x-axis at x=-d/2 and x= +d/2 respectively . If a third point charge q is taken from origin to x = d along the semicircle as shown in the figure , the energy of charge will

(a) increase by $\displaystyle \frac{3 q^2}{4\pi\epsilon_0 d}$

(b) decrease by $\displaystyle \frac{q^2}{4\pi\epsilon_0 d}$

(c) decrease by $\displaystyle \frac{4 q^2}{3\pi\epsilon_0 d}$

(d) increase by $\displaystyle \frac{2 q^2}{3\pi\epsilon_0 d}$

Click to See Solution :
Ans: (c)

Sol:

Electrostatic Potential Energy when charge q is at origin O

$\displaystyle U_1 = \frac{1}{4\pi\epsilon_0} [\frac{4q \times q}{d/2} + \frac{(-q) \times q}{d/2}]$

$\displaystyle U_1 = \frac{1}{4\pi\epsilon_0} \frac{6q^2}{d}$

Electrostatic Potential Energy when charge q moved to x = d

$\displaystyle U_2 = \frac{1}{4\pi\epsilon_0} [\frac{4q \times q}{3d/2} + \frac{(-q) \times q}{d/2}]$

$\displaystyle U_1 = \frac{1}{4\pi\epsilon_0} \frac{2q^2}{3d}$

The change in Poential Energy ,

$\displaystyle \Delta U = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d} (\frac{2}{3}-6)$

$\displaystyle \Delta U = – \frac{4 q^2}{3\pi\epsilon_0 d}$