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1. Electric Charges and Fields
normal
Two conducting spheres of radii $r_1$ and $r_2$ have same electric fields near their surfaces. The ratio of their electric potentials is
A
$\left( {r_1^2/r_2^2} \right)$
B
$\left( {r_2^2/r_1^2} \right)$
C
$(r_1/r_2)$
D
$(r_2/r_1)$
Solution
Electric field on the surface of conducting sphere $=\frac{k Q}{R^{2}}$
according to given condition
$\frac{k q_{1}}{r_{1}^{2}}=\frac{k q_{2}}{r_{2}^{2}}$
$\Longrightarrow \frac{q_{1}}{q_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}-(1)$
ratio of potentials at centre
$=\frac{\frac{k q_{1}}{r_{1}}}{\frac{k q_{2}}{r_{2}}}=\frac{q_{1}}{q_{2}} \cdot \frac{r_{2}}{r_{1}}$
substituting $\frac{q_{1}}{q_{2}}$ from $( 1)$
we get
ratio $=\frac{r_{1}}{r_{2}}$ Ans.
Standard 12
Physics
