Question

What is the net charge stored by the positive plates of the capacitors in the circuit shown below ?

• A. $250\mu \text{C}$
• B. $200\mu \text{C}$
• C. $400\mu \text{C}$
• D. $100\mu \text{C}$

Answer 1

The correct option is B $200\mu \text{C}$

Let, $V_A=0\text{V};V_B=4\text{V}$

So $V_A=V_C=0\text{V}$

And, $V_B=V_D=V_E=4\text{V}$

So the potential difference across both capacitors is the same and that is $4\text{V}.$

It means both capacitors are in parallel combination & the circuit can be redrawn as shown below.


Now, equivalent capacitance of above combination is $C_{eq}$

$\therefore C_{eq}=C_1+C_2=10+40=50\mu \text{F}$

So, net charge on plate is given by

$Q=C_{eq}V=50\times {10}^{-6}\times 4=200\times {10}^{-6}\text{C}$

$\therefore Q=200\mu \text{C}$

Hence, option $\left(b\right)$ is correct.

Key concept- The voltage across capacitors in parallel combination are equal.