Question

Four metallic plates are placed as shown in the figure. Plate 2 is given a charge Q whereas all other plates are uncharged. Plates 1 and 4 are earthed. The area of each plate is the same.

1. The potential difference between plates 1 and 2 is?

A) $\dfrac{3Qd}{2{{\varepsilon }_{0}}A}$

B) $\dfrac{Qd}{{{\varepsilon }_{0}}A}$

C) $\dfrac{3Qd}{4{{\varepsilon }_{0}}A}$

D) $\dfrac{3Qd}{{{\varepsilon }_{0}}A}$

2. The charge appearing on the right side of plate 4 is?

A) Zero

B) $\dfrac{Q}{4}$

C)$\dfrac{-3Q}{4}$

D)$\dfrac{-Q}{2}$

Answer 1

In the given diagram, we can see four metallic plates with some gap between them. When two metal plates with a gap act as a parallel plate capacitor. And if we have $+Q$ charge on one plate of the capacitor, then $-Q$ charge will be induced on the other plate. We have to remember charge is not stored in capacitors instead charge imbalances are stored in capacitors. Using this concept we can solve the problem.

**
**

Formula used:

1. $Q=CV$

This means the charge $(Q)$ on the plates is directly proportional to the voltage difference ($V$) between the plates. And $C$ is the capacitance which will remain as a constant.

2. $C=\dfrac{{{\varepsilon }_{0}}A}{d}$

Here, $C$ is the capacitance of the plates, ${\varepsilon }_{0}$ is the permittivity of the free space, $A$ is the area of the plates and $d$ is the distance between the plates.

Complete step by step answer:

1. We need to find the potential difference between the plates 1 and 2. It is given that plate 2 is given a total charge $Q$. We can say that the total charge on a plate is the net amount of the charge present on both sides of the plate. So, let $x$ be the charge present on the right side of the plate 2 then $(Q-x)$ charge will be present on the left side of the plate.

As we know, two metallic plates with a gap between them act as parallel plate capacitors.

$\Rightarrow C=\dfrac{{{\varepsilon }_{0}}A}{d}$ (Here, we are assuming the gap is only filled with air).

And here we have four metallic plates which act as three parallel plate capacitors.

Let the capacitance between plate 1 and 2 be $C_1$ and Voltage be $V_1$.

Let the capacitance between plate 2 and 3 be $C_2$ and Voltage be $V_2$.

Let the capacitance between plate 3 and 4 be $C_3$ and Voltage be $V_3$.

In a capacitor, if the charge on one plate is given, then the other plate will be having the same charge with a negative symbol.

Representing all these in the given circuit diagram,

Now we need to find the potential difference between plate 1 and 2 which are separated by $d$ distance.

Applying the KVL to the above circuit, we get

$V_1+V_2 + V_3 = 0$

$\Rightarrow \dfrac{Q_1}{C_1} + \dfrac{Q_2}{C_2} + \dfrac{Q_3}{C_3}=0$

Substituting the formula of $C$

$\dfrac{Q_1 \times d}{{\varepsilon }_{0} \times A} + \dfrac{Q_2 \times 2d}{{\varepsilon }_{0} \times A} + \dfrac{Q_3 \times d}{{\varepsilon }_{0} \times A}=0$

Substituting the corresponding charges,

$\dfrac{-(Q+x) \times d}{{\varepsilon }_{0} \times A} + \dfrac{x \times 2d}{{\varepsilon }_{0} \times A} + \dfrac{x \times d}{{\varepsilon }_{0} \times A}=0$

On simplification,

$(x-Q)d + x (2d) + x d = 0$

$\Rightarrow xd - Qd + 2xd + xd =0$

This implies that

$4xd-Qd=0$

$\Rightarrow 4x-Q=0$

On further simplification,

$\Rightarrow x=\dfrac{Q}{4}$ - (1)

From this value of $x$ we can find the value of all the other charges.

To find the voltage difference between plate 1 and 2,

$V_1= \dfrac{ (x-Q) \times d }{{\varepsilon }_{0} \times A}$

Substituting the value of $x$,

$V_1= \dfrac{(\dfrac{Q}{4}-Q) \times d}{{\varepsilon }_{0} \times A}$

$\Rightarrow V_1 =$ $\dfrac{3Qd}{4 {\varepsilon}_{0} A}$

Therefore, the potential difference between plate 1 and 2 is $\dfrac{3Qd}{4{\varepsilon}_{0} A}$. So, here option (C) is correct.

2. To find the charge on the right side of plate 4, we have to refer to the above diagram.

We can see that, on the right side of plate 4, charge is $x$

From equation (1) we got $x$ value.

We got $x= \dfrac{Q}{4}$.

So, here option (B) is correct.

Note: In this problem the space between the two plates of the capacitor was empty and we have assumed vacuum is between the two plates. But if space was filled with some dielectric then we will have to take into account the role of dielectric too. When the capacitor is connected to a battery then the charging process gets started. The charge moves from one plate of the capacitor to the other and so an electric field is produced in the space between the two plates.