Question

Grouping of Capacitors 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 4uF and 9uF capacitors), at a point distance 30 m from it, would equal [JEE (Main) 2016) ЗuF 4uF 9uF 11 HH TH 2uF + 8 V

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Answer 1

Answer:

 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: 240N/C.

hope that's enough hehe

Answer 2

Equivalent capacitance in the above branch will be

4+9+3

4(9+3)

μF=3μF.

Total charge in above branch will be Q=CV=24μC. This charge resides on the 4μF capacitor and 12μF (combination of 3 μF and 9μF) capacitor.

Now, voltage across 12μF combination of capacitors is given by V=Q/C=24/12=2V. This is the same as the voltage across 9μF capacitor.

Hence, charge on 9μF capacitor is Q=CV=9×2=18μC

From above total charge on 4μF and 9μF capacitors is 24+18=42μC