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Question:

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 μF and 9 μF capacitors), at a point distant 30 m from it, would equal.

480 N/C

240 N/C

360 N/C

420 N/C

Solution:

Equivalent capacitance in the above branch will be 4(9+3)/(4+9+3) μF = 3 μF. Total charge in above branch will be Q=CV=24 μC. This charge resides on the 4 μF capacitor and 12 μF (combination of 3 μF and 9 μF) capacitor. Now, voltage across 12 μF combination of capacitors is given by V=Q/C=24/12=2V. This is the same as the voltage across 9 μF capacitor. Hence, charge on 9 μF capacitor is Q=CV=9×2=18 μC. From above total charge on 4 μF and 9 μF capacitors is 24+18=42 μC. Now, by coulomb's law, E=kQr²; E=9×10⁹×42×10⁻⁶/30²=420 N/C