Question

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

(A). $\dfrac{5Q}{3}$
(B). $\dfrac{4Q}{3}$
(C). $\dfrac{3Q}{2}$
(D). None of these

Answer 1

Hint:The figure shows three plates connected together which can be broken into two capacitors. The plates of the capacitors are connected with each other; this means that they are connected in parallel. In parallel, the potential drop on the capacitors is the same. Using formula for capacitance we can find a relation between change and capacitance and from the distance between plates, we can determine the relation between both capacitors and then calculate charge.

Formulas used:
$C=\dfrac{Q}{V}$
$\dfrac{{{q}_{1}}}{{{C}_{1}}}=\dfrac{{{q}_{2}}}{{{C}_{2}}}$
$C=\dfrac{{{\varepsilon }_{0}}A}{d}$
$\dfrac{{{C}_{1}}}{{{C}_{2}}}=\dfrac{{{d}_{2}}}{{{d}_{1}}}$

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

The above plates can be divided into two capacitors as shown

As we can see, both plates of the two capacitors are connected together; this means that they are connected in parallel. The potential difference is the same in parallel combination. 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)
Here, ${{\varepsilon }_{0}}$ is the permittivity of free space
$A$ is the area of cross section between the plates
$d$ is the distance between the plates
${{\varepsilon }_{0}}$ and $A$ are constants, therefore, form eq (3),
$\dfrac{{{C}_{1}}}{{{C}_{2}}}=\dfrac{{{d}_{2}}}{{{d}_{1}}}$
Substituting values from the figure in 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,
$\begin{align}
  & \dfrac{{{q}_{1}}}{{{C}_{1}}}=\dfrac{{{q}_{2}}}{{{C}_{2}}} \\
 & \Rightarrow \dfrac{{{q}_{1}}}{2{{C}_{2}}}=\dfrac{{{q}_{2}}}{{{C}_{2}}} \\
 & \therefore {{q}_{1}}=2{{q}_{2}} \\
\end{align}$
We know that,
$\begin{align}
  & {{q}_{1}}+{{q}_{2}}=2Q \\
 & \Rightarrow 2{{q}_{2}}+{{q}_{2}}=2Q \\
 & \Rightarrow 3{{q}_{2}}=2Q \\
 & \therefore {{q}_{2}}=\dfrac{2Q}{3} \\
\end{align}$
${{q}_{2}}$has the value ${{q}_{2}}=\dfrac{2Q}{3}$, therefore, ${{q}_{1}}$ will have the value-
$\begin{align}
  & {{q}_{1}}=2{{q}_{2}} \\
 & \Rightarrow {{q}_{1}}=2\times \dfrac{2Q}{3} \\
 & \therefore {{q}_{1}}=\dfrac{4Q}{3} \\
\end{align}$
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 can key is closed is
$\begin{align}
  & -{{q}_{2}}-(-{{q}_{1}}) \\
 & \Rightarrow \dfrac{-4Q}{3}+\dfrac{2Q}{3}=-\dfrac{2Q}{3} \\
\end{align}$
Therefore, the charge that flows through the wire is $\dfrac{2Q}{3}$.

Hence, the correct option is (D).

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 capacitors is the same. The permittivity of a material is its ability to store electrical energy in an electric field.