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(B) Let $Q_1$ be the charge on the inner shell and $Q_2$ be the charge on the outer shell.The potential at the inner shell $(R_1)$ is given by $V_1 =

Vedclass · Accepted Answer

$\frac{V_1 R_1 (R_2 - x) + V_2 R_2 (x - R_1)}{(R_2 - R_1) x}$

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(B) Let $Q_1$ be the charge on the inner shell and $Q_2$ be the charge on the outer shell.The potential at the inner shell $(R_1)$ is given by $V_1 = \frac{k Q_1}{R_1} + \frac{k Q_2}{R_2}$.The potential at the outer shell $(R_2)$ is given by $V_2 = \frac{k Q_1}{R_2} + \frac{k Q_2}{R_2}$.Subtracting the two equations: $V_1 - V_2 = k Q_1 (\frac{1}{R_1} - \frac{1}{R_2}) = k Q_1 \frac{R_2 - R_1}{R_1 R_2}$.Thus,$k Q_1 = \frac{(V_1 - V_2) R_1 R_2}{R_2 - R_1}$.The potential at a distance $x$ $(R_1 Since $V_2 = \frac{k Q_1}{R_2} + \frac{k Q_2}{R_2}$,we have $\frac{k Q_2}{R_2} = V_2 - \frac{k Q_1}{R_2}$.Substituting this into the expression for $V(x)$: $V(x) = k Q_1 (\frac{1}{x} - \frac{1}{R_2}) + V_2 = k Q_1 \frac{R_2 - x}{x R_2} + V_2$.Substituting the value of $k Q_1$: $V(x) = \frac{(V_1 - V_2) R_1 R_2}{R_2 - R_1} \cdot \frac{R_2 - x}{x R_2} + V_2$.Simplifying: $V(x) = \frac{(V_1 - V_2) R_1 (R_2 - x) + V_2 x (R_2 - R_1)}{x (R_2 - R_1)} = \frac{V_1 R_1 (R_2 - x) + V_2 R_2 (x - R_1)}{x (R_2 - R_1)}$.

$A$ capacitor is formed by two concentric spherical shells. The potential of the shell with radius $R_1$ is $V_1$ and the potential of the shell with radius $R_2$ is $V_2$. What is the potential at a point at a distance $x$ from the center? $(R_2 > x > R_1)$

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