Question

The four capacitors, each of 25 μ F are connected as shown
in fig. The dc voltmeter reads 200 V. The charge on each
plate of capacitor is
(a) ± 2´10⁻³C (b) ± 5´10⁻³C
(c) ± 2´10⁻²C (d) ± 5´10⁻²C

Answer 1

Answer:

From the figure two capacitors are connected in parallel and such two combinations are connected in series.

From the figure two capacitors are connected in parallel and such two combinations are connected in series.Now Voltmeter is connected parallel to one of the parallel combination of capacitor. Hence Potential difference across each capacitor in a parallel combination will be same

From the figure two capacitors are connected in parallel and such two combinations are connected in series.Now Voltmeter is connected parallel to one of the parallel combination of capacitor. Hence Potential difference across each capacitor in a parallel combination will be samei.e 200 V.

From the figure two capacitors are connected in parallel and such two combinations are connected in series.Now Voltmeter is connected parallel to one of the parallel combination of capacitor. Hence Potential difference across each capacitor in a parallel combination will be samei.e 200 V.Now C=Q/V

From the figure two capacitors are connected in parallel and such two combinations are connected in series.Now Voltmeter is connected parallel to one of the parallel combination of capacitor. Hence Potential difference across each capacitor in a parallel combination will be samei.e 200 V.Now C=Q/VTherefore Q=CV

From the figure two capacitors are connected in parallel and such two combinations are connected in series.Now Voltmeter is connected parallel to one of the parallel combination of capacitor. Hence Potential difference across each capacitor in a parallel combination will be samei.e 200 V.Now C=Q/VTherefore Q=CVQ=25×10-6 x 200 here Q=25×10-6 and V=200 V given

From the figure two capacitors are connected in parallel and such two combinations are connected in series.Now Voltmeter is connected parallel to one of the parallel combination of capacitor. Hence Potential difference across each capacitor in a parallel combination will be samei.e 200 V.Now C=Q/VTherefore Q=CVQ=25×10-6 x 200 here Q=25×10-6 and V=200 V givenQ=5 × 10-3 C.

From the figure two capacitors are connected in parallel and such two combinations are connected in series.Now Voltmeter is connected parallel to one of the parallel combination of capacitor. Hence Potential difference across each capacitor in a parallel combination will be samei.e 200 V.Now C=Q/VTherefore Q=CVQ=25×10-6 x 200 here Q=25×10-6 and V=200 V givenQ=5 × 10-3 C.Since Another combination is similar to voltmeter combination and all the capacitors equal. Charge across each capacitor will be same.

Explanation:

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