Electrostatics

JEE Main level, single correct, medium

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Question 301 of 311

Let a total charge 2Q2Q be distributed in a sphere of radius R,R, with the charge density given by ρ(r)=kr,\rho (r)=kr, where rr is the distance from the centre. Two charges AA and BB , of −Q-Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If AA and BB do not experience any force, then:

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Answer: (D) a=8−14Ra={8}^{\dfrac{-1}{4}}R

As total charge on sphere is =2Q=2Q
2Q=∫0R(kr)(4πr2dr)⇒kπr4=2Q2Q=\int _{0}^{R}\left(kr\right)\left(4\pi {r}^{2}dr\right)\Rightarrow k{\pi r}^{4}=2Q
Charged enclosed in the sphere of radius aa
q=∫0akr×4πr2dr=kπa4q=\int _{0}^{a}kr\times 4\pi {r}^{2}dr=k\pi {a}^{4}
As force on QQ is zero
14πϵ0×Q2(2a)2=kπa44πϵ0×a2×Q\dfrac{1}{4\pi {\epsilon }_{0}}\times \dfrac{{Q}^{2}}{{\left(2a\right)}^{2}}=\dfrac{k\pi {a}^{4}}{4\pi {\epsilon }_{0}\times {a}^{2}}\times Q
⇒Q24=kπa4Q\Rightarrow \dfrac{{Q}^{2}}{4}=k\pi {a}^{4}Q
=a4(2QR4)Q={a}^{4}\left(\dfrac{2Q}{{R}^{4}}\right)Q
R4=8a4⇒R=814a{R}^{4}=8{a}^{4}\Rightarrow R={8}^{\dfrac{1}{4}}a or a=8−14Ra={8}^{-\dfrac{1}{4}}R

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