JEE Advance - Physics (2018 - Paper 1 Offline - No. 9)
Three identical capacitors $${C_1},{C_2}$$ and $${C_3}$$ have a capacitance of $$1.0\,\mu F$$ each and they are unchanged initially. They are connected in a circuit as shown in the figure and $${C_1}$$ is then filled completely with a dielectric material of relative permittivity $${\varepsilon _r}.$$ The cell electromotive force $$\left( {emf} \right)\,\,{V_0} = 8V.$$ First the switch $${S_1}$$ is closed while the switch $${S_2}$$ is kept open. When the capacitor $${C_3}$$ is fully charged, $${S_1}$$ is opened and $${S_2}$$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $${C_3}$$ is found to be $$5\,\mu C.$$ The value of $${\varepsilon _r} = $$ _________________.
Answer
1.50
Explanation
Given C1 = C2 = C3 = 1.0 $$\mu$$F and V0 = 8 V
When S1 is close and S2 is open. Charge on C3 is 8 $$\mu$$C.
The circuit is
When S1 is open and S2 is close. Charge on C3 is 5 $$\mu$$C. Therefore, charge on C1 and C2 will be 3 $$\mu$$C.
The potential difference across C3 is equal to the potential difference across C1 and C2 together i.e.,
$${{{q_1}} \over {C{'_1}}} + {{{q_2}} \over {{C_2}}} = {{{q_3}} \over {{C_3}}}$$,
$${{3\mu C} \over {{ \in _r}(1\mu F)}} + {{3\mu C} \over {1\mu F}} = {{5\mu C} \over {1\mu F}}$$.
Solve to get $${ \in _r} = 1.5$$.
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