Question

Find the potential difference between points $A$ and $B$ of the system shown in Fig. if the emf is equal to $\epsilon =110V$ and the capacitance ratio $C_2/C_1=\eta =2.0$.

Answer 1

Let us distribute the charges, as shown in the figure.

Now, we know that in a closed circuit, $-△\phi =0$

So, in the loop, $DCFED$,

$\frac{q_1}{C_1}-\frac{q_2}{C_1}-\frac{q_2}{C_2}=\xi$ or, $q_1=C_1[\xi +q_2(\frac{1}{C_1}+\frac{1}{C_2})]....\left(1\right)$

Again in the loop $DGHED$,

$\frac{q_1}{C_1}+\frac{q_1+q_2}{C_2}=\xi ....\left(2\right)$

Using Eqs. (1) and (2), we get

$q_2[\frac{1}{C_1}+\frac{3}{C_2}+\frac{C_1}{C_2}]=-\frac{\xi C_1}{C_2}$

Now, ${\phi }_A-{\phi }_B=\frac{-q_2}{C_2}=\frac{\xi }{C_2^2/C_1}\frac{1}{[\frac{1}{C_1}+\frac{3}{C_2}+\frac{C_1}{C_2^2}]}$

or, ${\phi }_A-{\phi }_B=\frac{\xi }{[\frac{C_2^2}{C_1^2}+\frac{3C_2}{C_1}+1]}=\frac{\xi }{{\eta }^2+3\eta +1}=10V$.