Q65 / 90 · JEE Main 2019 (12 Apr Shift 2) · Physics

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:

  • A
    a=3R214
  • B
    a=R3
  • C
    a=2-14R
  • Correct answer
    D
    a=8-14R

Solution

As total charge on sphere is =2Q
2Q=∫0Rkr4πr2dr⇒kπr4=2Q
Charged enclosed in the sphere of radius  a 
q=∫0akr×4πr2dr=kπa4
As force on  Q  is zero
14πε0×Q22a2=kπa44πε0×a2×Q
⇒Q24=kπa4Q
=a42QR4Q
R4=8a4⇒R=814 a or  a=8-14R

Hence, the correct answer is option: 'D'.

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