Question

Five similar condenser plates, each of area $A$ are placed at equal distance $d$ apart and are connected to source voltage $V$ as shown in the following diagram. The charge on plates $1$ and $4$ will be:

• A. $\frac{{ϵ}_{o}A}{d},\frac{-2{ϵ}_{o}A}{d}$
• B. $\frac{{ϵ}_{o}AV}{d},\frac{-2{ϵ}_{o}AV}{d}$
• C. $\frac{-{ϵ}_{o}AV}{d},\frac{-3{ϵ}_{o}AV}{d}$
• D. $\frac{{ϵ}_{o}AV}{d},\frac{-4{ϵ}_{o}AV}{d}$

Answer 1

The correct option is B $\frac{{ϵ}_{o}AV}{d},\frac{-2{ϵ}_{o}AV}{d}$

As we know that the plates connected together by the same conducting wire will be at the same potential. Also, the pair of plates facing each other will form a parallel plate capacitor.

So, redrawing the equivalent circuit after voltage distribution




All above four capacitors are in parallel combination and potential difference across them is $V$.

Since, plate no. $1$ is involved in one capacitor, therefore charge on plate $1$ is;

$Q=CV$

$\Rightarrow Q=\frac{A{ϵ}_{o}V}{d}=\frac{A{ϵ}_{o}V}{d}$

Since plate no. $4$ is serving as negative plate for two capacitors (i.e., repeated twice). Hence, the charge on plate $4$ will be;

$Q^{\prime \prime }=-2Q$

$\Rightarrow Q^{\prime \prime }=\frac{-2A{ϵ}_{o}V}{d}$

Hence, option $\left(b\right)$ is correct.

Why this question? Tip: In such problems always simplify the given circuit using ''same potential technique'' and then observe the share of plates in the capacitor circuit.