Question

Three large plates are arranged as shown. How much charge will flow through the key $K$ if it is closed ?

Answer 1

The figure shows that these three plates of the capacitors are connected with each other, this means that these three plates are connected in parallel.As we know that in parallel combination, the potential difference on the capacitors is the same.

Formula used:
$C = \dfrac{Q}{V}$
[$C$=capacitance in Farads, $Q$= Charge in Coulombs, $V$=Potential difference in Volts]
$\dfrac{{{q_1}}}{{{C_1}}} = \dfrac{{{q_2}}}{{{C_2}}}$
$C = \dfrac{{{\varepsilon _0}A}}{d}$
[ $C$=capacitance in Farads,${\varepsilon _0}$ = permittivity of free space, $A$= area of one plate in square meters, $d$=distance between the two plates in meters]
$\dfrac{{{C_1}}}{{{C_2}}} = \dfrac{{{d_1}}}{{{d_2}}}$

Complete step by step answer:
Capacitors are the devices that store the charge on them. Their ability to store the charge on them is represented by capacitance. It is calculated as
$C = \dfrac{Q}{V}$ $- (1)$

The above plates can be divided into two capacitors, we can show them as

As we can see, both plates of the capacitors are connected together; this means that they are connected in parallel. In parallel combination, the potential difference is the same. Therefore,
${V_1} = {V_2}$
From eq $(1)$
$\dfrac{{{q_1}}}{{{C_1}}} = \dfrac{{{q_2}}}{{{C_2}}}$ -($2$)
The capacitance of a parallel plate capacitor depends on the dimensions of the capacitor as-
$C = \dfrac{{{\varepsilon _0}A}}{d}$ $- (3)$
${\varepsilon _0}$and A are constants, therefore, from eq $(3)$,
$\dfrac{{{C_1}}}{{{C_2}}} = \dfrac{{{d_1}}}{{{d_2}}}$
Substituting values from the above equation, we get,
$\dfrac{{{C_1}}}{{{C_2}}} = \dfrac{{2d}}{d}$
$\Rightarrow {C_1} = 2{C_2}$ -(4)
Substituting eq (4) in eq (2),we get
$\dfrac{{{q_1}}}{{{C_1}}} = \dfrac{{{q_2}}}{{{C_2}}}$
$\Rightarrow \dfrac{{{q_1}}}{{2{C_2}}} = \dfrac{{{q_2}}}{{{C_2}}}$
$\therefore {q_1} = 2{q_2}$

We know that,
${q_1} + {q_2} = 2Q$
$\Rightarrow 2{q_2} + {q_2} = 2Q$
$\Rightarrow 3{q_2} = 2Q$
$\therefore {q_2} = \dfrac{{2Q}}{3}$
${q_2}$has the value ${q_2} = \dfrac{{2Q}}{3}$, therefore, ${q_1}$will have the value-
${q_1} = 2{q_2}$
$\Rightarrow {q_1} = 2 \times \dfrac{{2Q}}{3}$
$\therefore {q_1} = \dfrac{{4Q}}{3}$
Therefore, the charge on the first capacitor is$\dfrac{{2Q}}{3}$ and the charge on the second capacitor is $\dfrac{{4Q}}{3}$. The charge that flows through the wire when the key is closed is
$- {q_1} - ({q_1}) = - {q_2} + {q_1}$
$\therefore \dfrac{{ - 4Q}}{3} + \dfrac{{2Q}}{3} = - \dfrac{{2Q}}{3}$

Therefore, the charge that flows through the wire when the key is closed is $\dfrac{{2Q}}{3}$.

Note: The negative charge on the charge that flows through the circuit indicates that this charge flows opposite to the flow of current. In series, the charge is the same on the capacitors, while in parallel, the potential drop on the capacitors is the same. The permittivity of a material is its ability to store electrical energy in an electric field.