JEE MAIN - Physics (2003 - No. 50)
A thin spherical conducting shell of radius $$R$$ has a charge $$q.$$ Another charge $$Q$$ is placed at the center of the shell. The electrostatic potential at a point $$P$$ a distance $${R \over 2}$$ from the center of the shell is
$${{2Q} \over {4\pi {\varepsilon _0}R}}$$
$${{2Q} \over {4\pi {\varepsilon _0}R}} - {{2q} \over {4\pi {\varepsilon _0}R}}$$
$${{2Q} \over {4\pi {\varepsilon _0}R}} + {q \over {4\pi {\varepsilon _0}R}}$$
$${{\left( {q + Q} \right)2} \over {4\pi {\varepsilon _0}R}}$$
Explanation
Electric potential due to charge $$Q$$ placed at the center of spherical shell at point $$P$$ is
$${V_1} = {1 \over {4\pi {\varepsilon _0}}}{Q \over {R/2}} = {1 \over {4\pi {\varepsilon _0}}}{{2Q} \over R}$$
Electric potential due to charge $$q$$ on the surface of the spherical shell at any point inside the shell is
$${V_2} = {1 \over {4\pi {\varepsilon _0}}}{q \over R}$$
$${V_1} = {1 \over {4\pi {\varepsilon _0}}}{Q \over {R/2}} = {1 \over {4\pi {\varepsilon _0}}}{{2Q} \over R}$$
Electric potential due to charge $$q$$ on the surface of the spherical shell at any point inside the shell is
$${V_2} = {1 \over {4\pi {\varepsilon _0}}}{q \over R}$$
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