Electrostatics

JEE Main level, single correct, medium

187 of 311

Question 187 of 311

Two uniformly charged spherical conductors AA and BB of radii 5mm5mm and 10mm10mm are separated by a distance of 2cm2cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere AA and BB will be

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Answer: (B) 2:12:1

When two conductors are connected by a conducting wire, their potentials become equal. Therefore, VA=VB{V}_{A}={V}_{B}
The potential of a spherical conductor is given by, V=kQRV=\dfrac{kQ}{R}
⇒kQARA=kQBRB\Rightarrow \dfrac{k{Q}_{A}}{{R}_{A}}=\dfrac{k{Q}_{B}}{{R}_{B}}
⇒Q∝R\Rightarrow Q\propto R
Now the electric field near the surface of the conductor will be, E=kQR2E=\dfrac{kQ}{{R}^{2}}
⇒EAEB=kQARA2kQBRB2=RBRA\Rightarrow \dfrac{{E}_{A}}{{E}_{B}}=\dfrac{\dfrac{k{Q}_{A}}{{R}_{A}^{2}}}{\dfrac{k{Q}_{B}}{{R}_{B}^{2}}}=\dfrac{{R}_{B}}{{R}_{A}}
⇒EAEB=105=21\Rightarrow \dfrac{{E}_{A}}{{E}_{B}}=\dfrac{10}{5}=\dfrac{2}{1}

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