Assertion : Two concentric charged shells are | vedclass

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(A) Let the inner shell $B$ have radius $R_1$ and charge $Q_1$,and the outer shell $A$ have radius $R_2$ and charge $Q_2$.The potential at the surface

Vedclass · Accepted Answer

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Vedclass · Answer

(A) Let the inner shell $B$ have radius $R_1$ and charge $Q_1$,and the outer shell $A$ have radius $R_2$ and charge $Q_2$.The potential at the surface of the outer shell $A$ is $V_A = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1 + Q_2}{R_2}$.The potential at the surface of the inner shell $B$ is $V_B = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q_1}{R_1} + \frac{Q_2}{R_2} \right)$.The potential difference between the shells is $V_B - V_A = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q_1}{R_1} + \frac{Q_2}{R_2} - \frac{Q_1 + Q_2}{R_2} \right) = \frac{1}{4 \pi \varepsilon_0} Q_1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.This shows that the potential difference depends only on the charge of the inner shell $Q_1$. Thus,the Assertion is correct.The potential due to the charge of the outer shell $Q_2$ at any point inside it (including the surface of the inner shell) is constant and equal to $\frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{R_2}$. Thus,the Reason is also correct and explains why the $Q_2$ term cancels out in the potential difference calculation.

Assertion : Two concentric charged shells are given. The potential difference between the shells depends on the charge of the inner shell.
Reason : Potential due to the charge of the outer shell remains the same at every point inside the sphere.

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Assertion : Two concentric charged shells are given. The potential difference between the shells depends on the charge of the inner shell.Reason : Potential due to the charge of the outer shell remains the same at every point inside the sphere.