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Magnetic Flux and Faraday's Law

A part of cir...

Question

A part of circuit is given here if $V_{A} - V_{B} = 4 V$ then find charge in 4mF capacitor

Zero

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$2 μ C$

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$8 μ C$

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$4 μ C$

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Solution

The correct option is A
Zero

$V_{A} - V_{B} - 6 + 2 = \frac{q}{C_{e q}} 4 - 6 = 2 \frac{q}{C_{e q}} q = 0$

Suggest Corrections

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4 μ C

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V_{a} - V_{b} = + 4 V

V_{a} - V_{b}

+ 4 V

V_{A} - V_{B} = 4 V

X

V_{B} - V_{A} = 4 V

X

V_{A} - V_{B}

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