JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 7)
The two thin coaxial rings, each of radius 'a' and having charges +Q and $$-$$Q respectively are separated by a distance of 's'. The potential difference between the centres of the two rings is :
$${Q \over {2\pi {\varepsilon _0}}}\left[ {{1 \over a} + {1 \over {\sqrt {{s^2} + {a^2}} }}} \right]$$
$${Q \over {4\pi {\varepsilon _0}}}\left[ {{1 \over a} + {1 \over {\sqrt {{s^2} + {a^2}} }}} \right]$$
$${Q \over {4\pi {\varepsilon _0}}}\left[ {{1 \over a} - {1 \over {\sqrt {{s^2} + {a^2}} }}} \right]$$
$${Q \over {2\pi {\varepsilon _0}}}\left[ {{1 \over a} - {1 \over {\sqrt {{s^2} + {a^2}} }}} \right]$$
Explanation
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$${V_A} = {{KQ} \over a} - {{KQ} \over {\sqrt {{a^2} + {s^2}} }}$$
$${V_B} = {{ - KQ} \over a} + {{KQ} \over {\sqrt {{a^2} + {s^2}} }}$$
$${V_A} - {V_B} = {{2KQ} \over a} - {{2KQ} \over {\sqrt {{a^2} + {s^2}} }}$$
$$ = {Q \over {2\pi {\varepsilon _0}}}\left( {{1 \over a} - {1 \over {\sqrt {{s^2} + {a^2}} }}} \right)$$
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