Let a total charge 2 Q be distributed in a sphere of radius R, with the charge density given by r(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 :

Engineering

Physics

Electric Flux and Gauss Law and its Applications

Question

Let a total charge 2 Q be distributed in a sphere of radius R, with the charge density given by r(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 :

A

$a = \frac{3 R}{2^{1 / 4}}$

B

a = 8^–1/4 R

C

a = 2^–1/4 R

D

$a = R / \sqrt{3}$

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Total charge = 2 Q

Charging density ρ = kr

Radius = R

F_BA = F_sphere

_{Force on charge B due to element}

_{$dF = \frac{k (dq) Q}{a^{4}} = \frac{kq (K 4 {πr}^{3}) dr}{a^{2}}$}

_{$F = \frac{kQK 4 π}{a^{4}} \int_{0}^{a} r^{3} dr = \frac{kQK 4 {πa}^{2}}{4}$}

F = kQKπa²

F_BA = F_sphere

_{$\Rightarrow \frac{{kQ}^{2}}{{(2 a)}^{2}} = kQ 4 {Ka}^{2} \Rightarrow By replace value of K from (1)$}

_{$\frac{Q^{2}}{4 a^{2}} = \frac{2 Q^{2}}{{πR}^{4}} {πa}^{2}$}

_{$\Rightarrow a^{4} = \frac{R^{2}}{8}$}

⇒ a = R 8^–1/4