2. Electric Potential and Capacitance
easy

Two hollow conducting spheres of radii $R_{1}$ and $R_{2}$ $\left(R_{1}>>R_{2}\right)$ have equal charges. The potential would be:

A

more on smaller sphere

B

equal on both the spheres

C

dependent on the material property of the sphere

D

more on bigger sphere

(NEET-2022)

Solution

$V =\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{ Q }{ R }$

$\frac{1}{4 \pi \epsilon_{0}}=\text { constant }$

$Q =\text { same (Given) }$

$\therefore V \propto \frac{1}{ R }$

$\therefore$ Potential is more on smaller sphere.

Standard 12
Physics

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