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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

A
240 N/C
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B
360 N/C
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C
420 N/C
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D
480 N/C
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Solution

The correct option is C 420 N/CEquivalent 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. ∵4μF and 12μF are in series , charge on them is same. Now, voltage across the 12 μF combination of capacitors is given by V=Q/C=24/12=2 V. 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 the above, total charge on 4 μF and 9 μF capacitors is 24+18=42 μC Now, by Coulomb's law, E=kQr2 E=9×109×42×10−6302=420 N/C

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