JEE MAIN - Physics (2016 (Offline) - No. 14)
A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge $$Q$$ (having a charge equal to the sum of the charges on the $$4$$ $$\mu \,F$$ and $$9$$ $$\mu \,F$$ capacitors), at a point distance $$30$$ $$m$$ from it, would equal :
_en_14_1.png)
$$420N/C$$
$$480N/C$$
$$240N/C$$
$$360N/C$$
Explanation
_en_14_2.png)
Charge on $${C_1}$$ is $${q_1} = \left[ {\left( {{{12} \over {4 + 12}}} \right) \times 8} \right] \times 4 = 24\mu c$$
The voltage across $${C_P}$$ is $${V_P} = {4 \over {4 + 12}} \times 8 = 2v$$
$$\therefore$$ Voltage across $$9\mu F$$ is also $$2V$$
$$\therefore$$ Charge on $$9\mu F$$ capacitor $$ = 9 \times 2 = 18\mu C$$
$$\therefore$$ Total charge on $$4\mu F$$ and $$9\mu F$$ $$ = 42\mu c$$
$$\therefore$$ $$E = {{KQ} \over {{r^2}}} = 9 \times {10^9} \times {{42 \times {{10}^{ - 6}}} \over {30 \times 30}} = 420N{c^{ - 1}}$$
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