Question 212: Electrostats & Capacitance

Question

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

Accepted Answer

Let E be electric field in air
T sin$$	heta$$ = qE
T cos$$	heta$$ = mg
tan$$	heta$$ = $${{qE} \over {mg}}$$
$$Q = \left[ {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} ight][{V_1} + {V_2}]$$
$$E = {Q \over {A{ \in _0}}} = \left[ {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} ight]{{[{V_1} + {V_2}]} \over {A{ \in _0}}}$$
$${C_1} = {{{ \in _0}A} \over {d - t}} \Rightarrow E = {{{C_2}[{V_1} + {V_2}]} \over {({C_1} + {C_2})(d - t)}}$$
Now, $$	heta  = {	an ^{ - 1}}\left[ {{{q.E} \over {mg}}} ight]$$
$$	heta  = {	an ^{ - 1}}\left[ {{q \over {mg}} 	imes {{{C_2}({V_1} + {V_2})} \over {({C_1} + {C_2})(d - t)}}} ight]$$

Answer

$${	an ^{ - 1}}\left[ {{q \over {mg}} 	imes {{{C_1}({V_2} - {V_1})} \over {({C_1} + {C_2})(d - t)}}} ight]$$

Answer

$${	an ^{ - 1}}\left[ {{q \over {mg}} 	imes {{{C_2}({V_2} - {V_1})} \over {({C_1} + {C_2})(d - t)}}} ight]$$

Answer

$${	an ^{ - 1}}\left[ {{q \over {mg}} 	imes {{{C_2}({V_1} + {V_2})} \over {({C_1} + {C_2})(d - t)}}} ight]$$

Answer

$${	an ^{ - 1}}\left[ {{q \over {mg}} 	imes {{{C_1}({V_1} + {V_2})} \over {({C_1} + {C_2})(d - t)}}} ight]$$

Electrostats & Capacitance

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

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