Question

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

• A. $+32\mu C$
• B. $+40\mu C$
• C. $+48\mu C$
• D. $+80\mu C$

Answer 1

The correct option is C $+48\mu C$

The total charge on plate is $80\mu C$. If $q_B$ and $q_C$ charges on plates B and C then

$q_B+q_C=80....\left(1\right)$

As the capacitors B and C are in parallel so potential across both are equal .

i.e, $\frac{q_B}{C_B}=\frac{q_C}{C_C}$

or $\frac{q_B}{2}=\frac{q_C}{3}$

using (1), $\frac{80-q_C}{2}=\frac{q_C}{3}$

or $240-3q_C=2q_C\Rightarrow q_C=48\mu C$

Now for sign of charge: as lower plate of C is connected to ground so upper plate of C should be positive.

Thus, charge on upper plate of $3\mu F$ is $+48\mu C$