As total charge on sphere is =2Q
2Q=∫0R(kr)(4πr2dr)⇒kπr4=2Q
Charged enclosed in the sphere of radius a
q=∫0akr×4πr2dr=kπa4
As force on Q is zero
4πϵ01×(2a)2Q2=4πϵ0×a2kπa4×Q
⇒4Q2=kπa4Q
=a4(R42Q)Q
R4=8a4⇒R=841a or a=8−41R
JEE Main 2019 — Physics Electromagnetism
Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ(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:
Held on 12 Apr 2019 · Verified 6 Jul 2026.
a=2413R
a=3R
a=24−1R
a=84−1R
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