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(A) The electric potential $V$ at a point due to a spherical shell is given by:$1$. Inside the shell $(r < 	ext{radius})$: $V = \frac{KQ}{	ext{radiu

Vedclass · Accepted Answer

$\frac{KQ}{6R}$

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(A) The electric potential $V$ at a point due to a spherical shell is given by:$1$. Inside the shell $(r $2$. Outside the shell $(r > 	ext{radius})$: $V = \frac{KQ}{r}$For a point at $r = \frac{3R}{2}$:- This point is outside the inner shell (radius $R$), so the potential due to the inner shell is $V_1 = \frac{KQ}{3R/2} = \frac{2KQ}{3R}$.- This point is inside the outer shell (radius $2R$), so the potential due to the outer shell is $V_2 = \frac{K(-Q)}{2R} = -\frac{KQ}{2R}$.The total potential $V$ is the sum of the potentials due to both shells:$V = V_1 + V_2 = \frac{2KQ}{3R} - \frac{KQ}{2R}$$V = \frac{4KQ - 3KQ}{6R} = \frac{KQ}{6R}$

Two concentric hollow spherical conductors of radii $R$ and $2R$ have charges $Q$ and $-Q$ respectively. Find the electric potential at a distance of $\frac{3R}{2}$ from the centre. $\left[K=\frac{1}{4\pi\varepsilon_0}\right]$

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