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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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