Question

A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E, and W denote respectively, the magnitude of charge on each plate, electric field between the plates (after the slab is inserted), and work done on the system in question, in the process of inserting the slab, then which is wrong?
A. \[Q = \dfrac{{{\varepsilon _0}AV}}{d}\]
B. \[Q = \dfrac{{{\varepsilon _0}KAV}}{d}\]
C. \[E = \dfrac{V}{{Kd}}\]
D. \[W = \dfrac{{{\varepsilon _0}A{V^2}}}{{2d}}\left( {1 - \dfrac{1}{K}} \right)\]

Answer 1

Hint:The charge remains the same in the series combination. When we insert the dielectric material of dielectric constant K, the capacitance of the capacitor becomes K times its initial capacitance. The potential across the capacitor is the ratio of the charge to the capacitance.

Formula used:
Capacitance, \[C = \dfrac{{{\varepsilon _0}A}}{d}\]
Here, \[{\varepsilon _0}\] is the permittivity of the medium, A is the cross-sectional area of the plates and d is the distance of separation of the plates.
Electric field, \[E = \dfrac{V}{d}\]
Here, V is the potential difference across the plates of the capacitor.
Potential energy stored in the capacitor, \[U = \dfrac{1}{2}C{V^2}\]

Complete step by step answer:
We express the initial capacitance of the parallel plate capacitor as,
\[{C_0} = \dfrac{{{\varepsilon _0}A}}{d}\] …… (1)
 Here, \[{\varepsilon _0}\] is the permittivity of the medium, A is the cross-sectional area of the plates and d is the distance of separation of the plates.

We have the initial charge stored on the plates of capacitor is given as,
\[{Q_0} = {C_0}V\]
 Using equation (1) in the above equation, we get,
\[{Q_0} = \dfrac{{{\varepsilon _0}AV}}{d}\] ……. (2)

This is the initial charge stored on the plates of the capacitor. When the battery is disconnected, the charge has to go nowhere. Therefore, the charge on the capacitor will remain the same when the dielectric slab is inserted between the plates. Therefore, we can express the final charge stored on the plates of capacitor as,
\[Q = {Q_0} = \dfrac{{{\varepsilon _0}AV}}{d}\] ……. (3)

We know that when we insert a dielectric slab between the plates of the capacitor, the capacitance changes according to the expression,
\[C = K{C_0} = \dfrac{{{\varepsilon _0}KA}}{d}\] ……. (4)

Let’s express the potential across the plates of capacitor after the insertion of slab as follows,
\[V' = \dfrac{Q}{C}\]
Using equation (3) and (4) in the above equation, we get,
\[V' = \dfrac{{{Q_0}}}{{K{C_0}}}\]

But we have, \[V = \dfrac{{{Q_0}}}{{{C_0}}}\]. Therefore, the above equation becomes,
\[V' = \dfrac{V}{K}\] ……. (5)
We have the expression for electric field produced across the plates of capacitor,
\[E = \dfrac{{V'}}{d}\]
Using equation (5) in the above equation, we get,
\[E = \dfrac{V}{{Kd}}\] …… (6)

Let’s express the initial potential energy stored in the capacitor as,
\[{U_0} = \dfrac{1}{2}{C_0}{V^2}\]
Using equation (1) in the above equation, we get,
\[{U_0} = \dfrac{1}{2}\left( {\dfrac{{{\varepsilon _0}A}}{d}} \right){V^2}\]
\[ \Rightarrow {U_0} = \dfrac{{{\varepsilon _0}A{V^2}}}{{2d}}\] ……. (7)

Let’s express the final potential energy stored in the capacitor as,
\[{U_f} = \dfrac{1}{2}C{V'^2}\]
Using equation (4) and (5) in the above equation, we get,
\[{U_f} = \dfrac{1}{2}\left( {\dfrac{{{\varepsilon _0}KA}}{d}} \right){\left( {\dfrac{V}{K}} \right)^2}\]
\[ \Rightarrow {U_f} = \dfrac{{{\varepsilon _0}A{V^2}}}{{2Kd}}\] …… (8)

We know that the work done is the decrease in the potential energy. Therefore,
\[W = - \Delta U\]
\[ \Rightarrow W = {U_0} - {U_f}\]
Using equation (7) and (8) in the above equation, we get,
\[W = \dfrac{{{\varepsilon _0}A{V^2}}}{{2d}} - \dfrac{{{\varepsilon _0}A{V^2}}}{{2Kd}}\]
\[ \therefore W = \dfrac{{{\varepsilon _0}A{V^2}}}{{2d}}\left( {1 - \dfrac{1}{K}} \right)\] …… (9)
Therefore, from equation (3), (6) and (9), the options A, C and D are correct.

So, the correct answer is option B.

Note:On inserting the dielectric material between the plates of the capacitor, the capacitor behaves as two capacitors are connected in series. We know that the voltage drops in the series combination. Therefore, the potential difference across the capacitor changes on inserting the dielectric material. According to conservation of charge, the charge in the series combination remains the same. Thus, you can think as two capacitors are connected in the series and that’s why the charge is remaining the same.