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(D) The potential at a point $P$ at a distance $x$ from the center,where $r_2 < x < r_1$,is the sum of the potentials due to both spheres.$1$. The pot

Vedclass · Accepted Answer

$\frac{q_1}{4\pi \varepsilon_0 r_1} + \frac{q_2}{4\pi \varepsilon_0 x}$

Vedclass · Answer

(D) The potential at a point $P$ at a distance $x$ from the center,where $r_2 $1$. The potential due to the outer sphere (radius $r_1$,charge $q_1$) at any point inside it is constant and equal to the potential at its surface: $V_1 = \frac{q_1}{4\pi \varepsilon_0 r_1}$.$2$. The potential due to the inner sphere (radius $r_2$,charge $q_2$) at a point outside it at distance $x$ is: $V_2 = \frac{q_2}{4\pi \varepsilon_0 x}$.$3$. The net potential at distance $x$ is $V = V_1 + V_2 = \frac{q_1}{4\pi \varepsilon_0 r_1} + \frac{q_2}{4\pi \varepsilon_0 x}$.

Two concentric hollow metallic spheres of radii $r_1$ and $r_2$ $(r_1 > r_2)$ contain charges $q_1$ and $q_2$ respectively. The potential at a distance $x$ between $r_1$ and $r_2$ will be

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