First condition:
Both conducting spheres are shown.
Vin−Vout=(r1KQ)−(r2KQ)
⇒V=KQ(r11−r21)
Second condition:
Shell is now given charge −4Q .
Vin−Vout=(r1KQ−r24KQ)−(r2KQ−r24KQ)
=r1KQ−r2KQ
=KQ(r11−r21)=V
JEE Main 2019 — Physics Electromagnetism
A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –4Q, the new potential difference between the same two surfaces is:
Held on 8 Apr 2019 · Verified 6 Jul 2026.
–2V
2V
V
4V
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