If the electric potential of the inner metal | vedclass

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(A) Let $q_1$ be the charge on the inner sphere of radius $a$ and $q_2$ be the charge on the outer shell of radius $b$.The potential at the surface of

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$10$

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(A) Let $q_1$ be the charge on the inner sphere of radius $a$ and $q_2$ be the charge on the outer shell of radius $b$.The potential at the surface of the inner sphere is given by the sum of potentials due to both spheres: $V_a = \frac{kq_1}{a} + \frac{kq_2}{b} = 10 	ext{ V}$.The potential at the surface of the outer shell is given by: $V_b = \frac{kq_1}{b} + \frac{kq_2}{b} = \frac{k(q_1 + q_2)}{b} = 5 	ext{ V}$.The potential at the centre of the spheres is the sum of the potentials due to both spheres at that point. Since the potential inside a spherical shell is constant and equal to the potential at its surface,the potential at the centre is: $V_c = \frac{kq_1}{a} + \frac{kq_2}{b}$.Comparing this with the expression for $V_a$,we see that $V_c = V_a = 10 	ext{ V}$.

If the electric potential of the inner metal sphere of radius $a$ is $10 \text{ V}$ and that of the outer spherical shell of radius $b$ is $5 \text{ V}$,then the potential at the centre will be ...... $\text{V}$.

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