Question

Five identical capacitor plates, each of area $\text{A}$, are arranged such that adjacent plates are at $d$ distance apart, Plates are connected to a source of e.m.f $E$ as shown in the figure. Match the following


Column-I Column-II

$\text{(A)}\text{Charge on plate}1$	$\text{(P)}\frac{-{\epsilon }_{0}AE}{d}$
$\text{(B)}\text{Charge on plate}4$	$\text{(Q)}\frac{+{\epsilon }_{0}AE}{d}$
$\text{(C)}\text{Potential difference between plates 1 and 2}$	$\text{(R)}\text{Zero}$
$\text{(D)}\text{Potential difference between plates 1 and 5}$	$\text{(S)}E$

• A. $A\to Q;B\to P;C\to S;D\to R$
• B. $A\to P;B\to Q;C\to S;D\to R$
• C. $A\to Q;B\to P;C\to R;D\to S$
• D. $A\to P;B\to P;C\to R;D\to S$

Answer 1

The correct option is A $A\to Q;B\to P;C\to S;D\to R$


From the figure, cells $2,3$ and $4$ are grounded and cells $1$ and $5$ are kept at a potential of $E$.

So, the electric field between the cells $2-3$ and $3-4$ will be zero as there is no potential difference.

Similarly, between cells $1-2$ the potential difference is $E$.

Thus, the charge on the plate$-1$ is $+CE$ and charge on plate$-4$ is $-CE$

Hence, option (a) is the correct alternative.