JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 4)
Consider two charged metallic spheres S1 and
S2 of radii R1 and R2, respectively. The electric
fields E1 (on S1) and E2 (on S2) on their surfaces
are such that E1/E2 = R1/R2. Then the ratio
V1 (on S1) / V2 (on S2) of the electrostatic
potentials on each sphere is :
(R1/R2)2
(R2/R1)
(R1/R2)3
R1/R2
Explanation
We know,
E1 = $${{K{Q_1}} \over {R_1^2}}$$ and E2 = $${{K{Q_2}} \over {R_2^2}}$$
Given
$${{{E_1}} \over {{E_2}}} = {{{R_1}} \over {{R_2}}}$$
$$ \Rightarrow $$ $${{{{K{Q_1}} \over {R_1^2}}} \over {{{K{Q_2}} \over {R_2^2}}}} = {{{R_1}} \over {{R_2}}}$$
$$ \Rightarrow $$ $${{{Q_1}} \over {{Q_2}}} = {{R_1^3} \over {R_2^3}}$$
Now $${{{V_1}} \over {{V_2}}} = {{{{K{Q_1}} \over {{R_1}}}} \over {{{K{Q_2}} \over {{R_2}}}}}$$ = $${{R_1^2} \over {R_2^2}}$$
E1 = $${{K{Q_1}} \over {R_1^2}}$$ and E2 = $${{K{Q_2}} \over {R_2^2}}$$
Given
$${{{E_1}} \over {{E_2}}} = {{{R_1}} \over {{R_2}}}$$
$$ \Rightarrow $$ $${{{{K{Q_1}} \over {R_1^2}}} \over {{{K{Q_2}} \over {R_2^2}}}} = {{{R_1}} \over {{R_2}}}$$
$$ \Rightarrow $$ $${{{Q_1}} \over {{Q_2}}} = {{R_1^3} \over {R_2^3}}$$
Now $${{{V_1}} \over {{V_2}}} = {{{{K{Q_1}} \over {{R_1}}}} \over {{{K{Q_2}} \over {{R_2}}}}}$$ = $${{R_1^2} \over {R_2^2}}$$
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