JEE PYQ: Electrostatics Question 22
Question 22 - 2020 (04 Sep Shift 1)
Two point charges $4q$ and $-q$ are fixed on the $x$-axis at $x = -\frac{d}{2}$ and $x = \frac{d}{2}$, respectively. If a third point charge ‘$q$’ is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will:
(1) increase by $\frac{3q^2}{4\pi\varepsilon_0 d}$
(2) increase by $\frac{2q^2}{3\pi\varepsilon_0 d}$
(3) decrease by $\frac{q^2}{4\pi\varepsilon_0 d}$
(4) decrease by $\frac{4q^2}{3\pi\varepsilon_0 d}$
Show Answer
Answer: (4)
Solution
$\Delta u = q(V_f - V_i)$. At origin: $V_i = \frac{k \cdot 4q}{d/2} + \frac{k(-q)}{d/2} = \frac{6kq}{d}$. At $x = d$: $V_f = \frac{k \cdot 4q}{3d/2} + \frac{k(-q)}{d/2} = \frac{8kq}{3d} - \frac{2kq}{d} = \frac{2kq}{3d}$. $\Delta u = q\left(\frac{2kq}{3d} - \frac{6kq}{d}\right) = -\frac{4q^2}{3\pi\varepsilon_0 d}$. Energy decreases.
BETA
