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If the charge on the capacitor is increased by 2 coulombs, the energy stored in it increase by $21 \%$. the original charge on the capacitor is
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Verified Answer
The correct answer is:
20 C
Concept: energy stored in capacitor is given by
$U=\frac{q^2}{2 c}$
Where, $q$ is charge and $c$ is capacitance.
Now, if change is increased by $\delta$, the energy stored is
$\Rightarrow U_n=\frac{(q+\delta)^2}{2 c}$
Given, $\delta=2 \mathrm{C}, U_n=1.21 \mathrm{U}$
$\begin{aligned}
& \therefore 1.21\left(\frac{q^2}{2 c}\right)=\frac{(q+\delta)^2}{2 c} \\
& \Rightarrow\left(\frac{q+\delta}{q^2}\right)^2=1.21 \\
& \Rightarrow\left(\frac{q+\delta}{q}\right)=1.1 \\
& \Rightarrow \frac{\delta}{q}=0.1
\end{aligned}$
$\Rightarrow q=\frac{\delta}{0.1}=20 \mathrm{C}$
$U=\frac{q^2}{2 c}$
Where, $q$ is charge and $c$ is capacitance.
Now, if change is increased by $\delta$, the energy stored is
$\Rightarrow U_n=\frac{(q+\delta)^2}{2 c}$
Given, $\delta=2 \mathrm{C}, U_n=1.21 \mathrm{U}$
$\begin{aligned}
& \therefore 1.21\left(\frac{q^2}{2 c}\right)=\frac{(q+\delta)^2}{2 c} \\
& \Rightarrow\left(\frac{q+\delta}{q^2}\right)^2=1.21 \\
& \Rightarrow\left(\frac{q+\delta}{q}\right)=1.1 \\
& \Rightarrow \frac{\delta}{q}=0.1
\end{aligned}$
$\Rightarrow q=\frac{\delta}{0.1}=20 \mathrm{C}$
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