First slide

Capacitance

Question

In the given circuit, a charge of +80 $μ$ C is given to the upper plate of the 4 $μ$ F capacitor. Then in the steady state, the charge on the upper plate of the 3 $μ$ F capacitor is :

Easy

A

+32 $μ$ C
B

+40 $μ$ C
C

+48 $μ$ C
D

+80 $μ$ C

Solution

$q_{3} = \frac{C_{3}}{C_{2} + C_{3}} \cdot Q = \frac{3}{3 + 2} \times 80 = \frac{3}{5} \times 80 = 48 μC$

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