JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 9)
A simple pendulum of mass 'm', length 'l' and charge '+ q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be
_27th_July_Evening_Shift_en_9_1.png)
$${\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_1}({V_2} - {V_1})} \over {({C_1} + {C_2})(d - t)}}} \right]$$
$${\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_2}({V_2} - {V_1})} \over {({C_1} + {C_2})(d - t)}}} \right]$$
$${\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_2}({V_1} + {V_2})} \over {({C_1} + {C_2})(d - t)}}} \right]$$
$${\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_1}({V_1} + {V_2})} \over {({C_1} + {C_2})(d - t)}}} \right]$$
Explanation
_27th_July_Evening_Shift_en_9_2.png)
Let E be electric field in air
T sin$$\theta$$ = qE
T cos$$\theta$$ = mg
tan$$\theta$$ = $${{qE} \over {mg}}$$
_27th_July_Evening_Shift_en_9_3.png)
$$Q = \left[ {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right][{V_1} + {V_2}]$$
$$E = {Q \over {A{ \in _0}}} = \left[ {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right]{{[{V_1} + {V_2}]} \over {A{ \in _0}}}$$
$${C_1} = {{{ \in _0}A} \over {d - t}} \Rightarrow E = {{{C_2}[{V_1} + {V_2}]} \over {({C_1} + {C_2})(d - t)}}$$
Now, $$\theta = {\tan ^{ - 1}}\left[ {{{q.E} \over {mg}}} \right]$$
$$\theta = {\tan ^{ - 1}}\left[ {{q \over {mg}} \times {{{C_2}({V_1} + {V_2})} \over {({C_1} + {C_2})(d - t)}}} \right]$$
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