Question

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

Answer 1

Answer:

(c)

Explanation:

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

i.e 200 V.

Now C=Q/V

Therefore Q=CV

Q=25×10-6 x 200 here Q=25×10-6 and V=200 V given

Q=5 × 10-3 C.

Since Another combination is similar to voltmeter combination and all the capacitors equal. Charge across each capacitor will be same.

Answer 2

Answer:

The charge on each plate of the capacitor is ±5×10⁻³C i.e.option(b).

Explanation:

• The two capacitors on the left side are wired in parallel, with a potential difference of V between them.
• Similarly, the capacitors on the right side are in parallel, with a potential of V across each.

And for finding the charge on the capacitor is calculated as,

$Q=\pm CV$       (1)

Where,

Q=charge on the capacitor

C=capacitance of the capacitor

V=potential difference across the plates of the capacitor

From the question we have,

C=25μF=25×10⁻⁶F

The potential difference across the plates of the capacitor=200volt

By placing the required values in equation (1) we get;

$Q=\pm 25\times 10^-^6\times 200$

$Q=\pm 5\times 10^-^3\ Coulomb$

Hence, The charge on each plate of the capacitor is ±5×10⁻³C i.e.option(b).

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