Let a total charge 2Q be distributed in a sphere of radius R, with… | JEE Main 2019 (12 Apr, Shift 2)

Q105JEE Main 2019 · 12 APR · SHIFT 2Difficult

Electric Flux & Gauss's Law

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:

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Answer: (a)

Total charge: $\int_0^R kr \cdot 4\pi r^2 dr = k\pi R^4 = 2Q \Rightarrow k = \frac{2Q}{\pi R^4}$ . Charge within $a$ : $\frac{2Qa^4}{R^4}$ .
Field at A due to the sphere: $E = \frac{2Qa^2}{4\pi\varepsilon_0 R^4}$ ; force on $-Q$ is inward, $\frac{2Q^2a^2}{4\pi\varepsilon_0 R^4}$ .
Repulsion from B (at $2a$ ) is outward: $\frac{Q^2}{4\pi\varepsilon_0 (2a)^2}$ .
Equate: $\frac{2a^2}{R^4} = \frac{1}{4a^2} \Rightarrow a^4 = \frac{R^4}{8} \Rightarrow a = 8^{-1/4}R$
So the answer is (a).

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Electric Flux & Gauss's Law question from JEE Main 2019 (12 Apr, Shift 2)

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