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Two identical condensers $M$ and $N$ are connected in series with a battery. The space between the plates of $M$ is completely filled with a dielectric medium of dielectric constant 8 and a copper plate of thickness $\frac{d}{2}$ is introduced between the plates of $N$. ( $d$ is the distance between the plates). Then potential differences across $M$ and $N$ are, respectively, in the ratio
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The correct answer is:
1 : 4
Let the capacitance of condensor, $M=C$
The new capacitance of condensor, $C_M=8 C$
The capacitance of condensor, $N=C$
The new capacitance of condensor, $C_N=\frac{C^{\prime}}{2}$
where $\quad C^{\prime}=\frac{A \varepsilon_0}{d / 4}=\frac{4 A \varepsilon_0}{d}=4 C$
$\therefore \quad C_N=\frac{C^{\prime}}{2}=\frac{4 C}{2}=2 C$
Since the two condensers are connected in series
$\begin{aligned}
\therefore \quad \quad V_M: V_N & =\frac{1}{C_M}: \frac{1}{C_N} \\
& =\frac{1}{8 C}: \frac{1}{2 C}=1: 4 \\
\Rightarrow \quad V_M: V_N & =1: 4
\end{aligned}$
The new capacitance of condensor, $C_M=8 C$
The capacitance of condensor, $N=C$
The new capacitance of condensor, $C_N=\frac{C^{\prime}}{2}$
where $\quad C^{\prime}=\frac{A \varepsilon_0}{d / 4}=\frac{4 A \varepsilon_0}{d}=4 C$
$\therefore \quad C_N=\frac{C^{\prime}}{2}=\frac{4 C}{2}=2 C$
Since the two condensers are connected in series
$\begin{aligned}
\therefore \quad \quad V_M: V_N & =\frac{1}{C_M}: \frac{1}{C_N} \\
& =\frac{1}{8 C}: \frac{1}{2 C}=1: 4 \\
\Rightarrow \quad V_M: V_N & =1: 4
\end{aligned}$
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