Electrostatics 5 Question 12
13. In the figure shown below, the charge on the left plate of the $10 \mu F$ capacitor is $-30 \mu C$. The charge on the right plate of the $6 \mu F$ capacitor is
(Main 2019, 11 Jan I)
(a) $+12 \mu C$
(b) $+18 \mu C$
(c) $-12 \mu C$
(d) $-18 \mu C$
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Solution:
- Applying the concept of charge conservation on isolated plates of $10 \mu F, 6 \mu F$ and $4 \mu F$. Since, $6 \mu F$ and $4 \mu F$ are in parallel, so total charge on this combination will be $30 \mu C$.
$\therefore$ Charge on $6 \mu F$, capacitor
$$ =\frac{C _1}{C _1+C _2} \quad q=\frac{6}{6+4} \times 30=18 \mu C $$
Since, the charge has been asked on the right plate of the capacitor. Thus, it would be $+18 \mu C$.
Alternative method
Let charge on $6 \mu F$ capacitor is $q \mu C$.
Now, $V$ at $6 \mu F=V$ at $4 \mu F$
$$ \begin{aligned} & \therefore \quad \frac{q}{6 \mu F}=\frac{30-q}{4 \mu F} \\ & (\because V=q / C) \\ & \Rightarrow \quad 4 q=-6 q+180 \\ & \Rightarrow \quad q=18 \mu C \text {. } \end{aligned} $$
