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MHT CET 2021 (23 Sep, Shift 1) · Electric Potential & Equipotentials

Electric Potential & Equipotentials question from MHT CET 2021 (23 Sep, Shift 1)

Q85MHT CET 2021 · 23 SEP · SHIFT 1Medium

Electric Potential & Equipotentials

A charged spherical conductor has radius 'rr'. The potential difference between its surface and 3 point at a distance '3r3r' from the centre is 'vv' The electric intensity at a distance '3r3r' from the centre of the conductor is

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Answer: (d)

  1. v=V(r)−V(3r)=kQ(1r−13r)=2kQ3r⇒kQ=3vr2v = V(r) - V(3r) = kQ\left(\frac{1}{r} - \frac{1}{3r}\right) = \frac{2kQ}{3r} \Rightarrow kQ = \frac{3vr}{2}
  2. E(3r)=kQ(3r)2=3vr2⋅9r2=v6rE(3r) = \frac{kQ}{(3r)^2} = \frac{3vr}{2\cdot 9r^2} = \frac{v}{6r}
  3. So the answer is (d) v6r\frac{v}{6r}.
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