JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 10)
Concentric metallic hollow spheres of radii R and 4R hold charges Q1
and Q2
respectively. Given that
surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is :
$${{3{Q_2}} \over {4\pi {\varepsilon _0}R}}$$
$${{3{Q_1}} \over {4\pi {\varepsilon _0}R}}$$
$${{3{Q_1}} \over {16\pi {\varepsilon _0}R}}$$
$${{{Q_2}} \over {4\pi {\varepsilon _0}R}}$$
Explanation
_3rd_September_Evening_Slot_en_10_1.png)
$$\sigma = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi 16{R^2}}}$$
$$ \Rightarrow $$ $$16{Q_1} = {Q_2}$$
$$ \therefore $$ $${V_R} - {V_{4R}} = {{K{Q_1}} \over R} + {{K{Q_2}} \over {4R}} - {{K{Q_1}} \over {4R}} - {{K{Q_2}} \over {4R}}$$
$$ = {{3K{Q_1}} \over {4R}} = {{3{Q_1}} \over {16\pi {\varepsilon _0}R}}$$ [As K = $${1 \over {4\pi {\varepsilon _0}}}$$ ]
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