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…

Question

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

Aicharya · Accepted Answer

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 $\quad C^{\prime}=\frac{A \varepsilon_0}{d / 4}=\frac{4 A \varepsilon_0}{d}=4 C$ $	herefore \quad C_N=\frac{C^{\prime}}{2}=\frac{4 C}{2}=2 C$ Since the two condensers are connected in series $\begin{aligned} 	herefore \quad \quad V_M: V_N & =\frac{1}{C_M}: \frac{1}{C_N} \ & =\frac{1}{8 C}: \frac{1}{2 C}=1: 4 \ \Rightarrow \quad V_M: V_N & =1: 4 \end{aligned}$

Two identical condensers $M$ and $N$ are connected in series with a battery. The space between the plates of…

Solution