JEE PYQ: Electrostatics Question 49
Question 49 - 2019 (12 Apr Shift 2)
Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho(r) = kr$, where $r$ is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance $a$, from the centre. If $A$ and $B$ do not experience any force, then:
(1) $a = 8^{-1/4}R$
(2) $a = \frac{3R}{2^{1/4}}$
(3) $a = 2^{-1/4}R$
(4) $a = R/\sqrt{3}$
Show Answer
Answer: (1)
Solution
The force on charge A at distance $a$ from centre is zero when the electric field due to the sphere at $a$ equals the field due to charge B. $E_{\text{sphere}} = \frac{Qr^2}{2\pi\varepsilon_0 R^4}$ (for the given density). Setting up equilibrium: $\frac{kQa^2}{R^4} = \frac{kQ}{(2a)^2}$, giving $a^4 = \frac{R^4}{8}$, so $a = 8^{-1/4}R$.
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