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(A) Let the radius of the inner solid sphere be $r$ and the radius of the outer shell be $R$.According to Gauss's law,the electric field $E$ in the re

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(A) Let the radius of the inner solid sphere be $r$ and the radius of the outer shell be $R$.According to Gauss's law,the electric field $E$ in the region between the sphere and the shell $(r The charge enclosed is the charge on the inner sphere,which is $Q$.Thus,$E = \frac{Q}{4 \pi \epsilon_0 x^2}$.The potential difference $V$ between the surface of the inner sphere and the surface of the outer shell is given by $V = \int_{r}^{R} E \, dx = \int_{r}^{R} \frac{Q}{4 \pi \epsilon_0 x^2} \, dx$.Evaluating the integral,$V = \frac{Q}{4 \pi \epsilon_0} \left[ -\frac{1}{x} \right]_{r}^{R} = \frac{Q}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{R} \right)$.Since the potential difference $V$ depends only on the charge $Q$ of the inner sphere and the radii $r$ and $R$,changing the charge on the outer shell does not affect the potential difference between the two surfaces.Therefore,the new potential difference remains $V$.

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