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(A) The electric potential at the centre of a charged spherical shell of radius $R$ with charge $q$ is $V = \frac{q}{4 \pi \varepsilon_{0} R}$.For the

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$\frac{\sigma}{\varepsilon_{0}}(R+r)$

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(A) The electric potential at the centre of a charged spherical shell of radius $R$ with charge $q$ is $V = \frac{q}{4 \pi \varepsilon_{0} R}$.For the two concentric spheres,the total potential $V$ at the common centre is the sum of potentials due to each sphere:$V = V_{1} + V_{2} = \frac{q_{1}}{4 \pi \varepsilon_{0} R} + \frac{q_{2}}{4 \pi \varepsilon_{0} r}$.Given that the surface charge densities are equal,$\sigma = \frac{q_{1}}{4 \pi R^{2}} = \frac{q_{2}}{4 \pi r^{2}}$.This implies $q_{1} = 4 \pi R^{2} \sigma$ and $q_{2} = 4 \pi r^{2} \sigma$.Substituting these values into the potential equation:$V = \frac{4 \pi R^{2} \sigma}{4 \pi \varepsilon_{0} R} + \frac{4 \pi r^{2} \sigma}{4 \pi \varepsilon_{0} r}$.$V = \frac{R \sigma}{\varepsilon_{0}} + \frac{r \sigma}{\varepsilon_{0}} = \frac{\sigma}{\varepsilon_{0}}(R + r)$.

The concentric spheres of radii $R$ and $r$ have positive charges $q_{1}$ and $q_{2}$ with equal surface charge densities. What is the electric potential at their common centre?

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