Assertion: Charges are given to plates of two | vedclass

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Question

(C) For capacitor $C_1$,the charges on the plates are $q_1 = 2 \mu C$ and $q_2 = 4 \mu C$. The potential difference across $C_1$ is $V_1 = (q_2 - q_1)

Vedclass · Accepted Answer

If Assertion is correct but Reason is incorrect.

Vedclass · Answer

(C) For capacitor $C_1$,the charges on the plates are $q_1 = 2 \mu C$ and $q_2 = 4 \mu C$. The potential difference across $C_1$ is $V_1 = (q_2 - q_1) / (2C_1) = (4 - 2) / (2C_1) = 1 / C_1$.For capacitor $C_2$,the charges on the plates are $q_3 = 4 \mu C$ and $q_4 = 8 \mu C$. The potential difference across $C_2$ is $V_2 = (q_4 - q_3) / (2C_2) = (8 - 4) / (2 	imes 2C_1) = 4 / (4C_1) = 1 / C_1$.Since $V_1 = V_2$,there is no potential difference between the plates when the key $K$ is closed,so no charge flows.The net charge on the upper plate of $C_1$ is $2 \mu C$ (positive) and on the lower plate is $4 \mu C$ (positive). Thus,the Assertion is correct.The Reason is incorrect because,in an isolated capacitor,the plates carry equal and opposite charges,but when connected to external circuits or given arbitrary charges,the plates do not necessarily carry equal and opposite charges.

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