Question

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
(A) The electric field in capacitor
(B) The charge on the capacitor
(C) The potential difference between the plates
(D) The stored energy in capacitor

Answer 1

Hint :Check each of the options using the concepts asked in the options. Use the formula for electric field due to a parallel plate capacitor, conservation of charge potential difference between the plates of a capacitor, stored energy in a capacitor.

Complete Step By Step Answer:
We have given here a parallel plate capacitor and asked if inserting a dielectric slab changes one of the quantities.
So, let’s first check the first option (A).
We know that the electric field due to a plate with charge density $ \sigma $ and total charge $ Q $ having distance between the plate $ d $ is given by , $ \dfrac{\sigma }{\varepsilon } $ , $ \varepsilon $ being the permittivity of the dielectric placed between.
So, here electric field is dependent on the dielectric constant $ \kappa = \dfrac{\varepsilon }{{{\varepsilon _0}}} $ where $ {\varepsilon _0} $ is the permittivity of the vacuum. So, since $ \sigma $ is constant , hence inserting the dielectric slab will change the electric field. Hence option (A) is not correct.
Check option (B): We know, conservation of charge states that total electric charge in an isolated system never changes. Here, the capacitor is kept isolated from the surroundings, hence charge cannot change by any means. So, the total charge of the system will be still the same as before inserting the dielectric slab.
Hence option ( B ) is correct.
Now let’s check option (C ):We know that the potential difference of a parallel plate charged capacitor is given by, $ V = \dfrac{{Qd}}{{\varepsilon A}} $ ,where $ A $ is the area of the each parallel plate.
 Since, total charge , the distance between the plates $ d $ and area $ A $ is constant then only changing the dielectric constant will change the potential difference.
Hence, inserting the dielectric slab will change the potential difference between the plates.
Hence option ( C ) is incorrect.
Now check option (D) : We know that the electrostatic energy stored in a parallel plate capacitor with charge $ Q $ is given by,
 $ E = \dfrac{1}{2}C{V^2} $ , where $ V $ is the potential difference between the plates and $ C $ is the capacitance.
Since, the capacitance is given by $ C = \dfrac{{\varepsilon A}}{d} $ ,changes with changing the dielectric constant. Also the potential difference between the plates changes, if we put the value of these two we get the energy term as,
 $ E = \dfrac{1}{2}\dfrac{{\varepsilon A}}{d}{\left( {\dfrac{{Qd}}{{\varepsilon A}}} \right)^2} $
 $ \Rightarrow E = \dfrac{1}{2}\dfrac{{{Q^2}d}}{{\varepsilon A}} $ , Which changes with the change in the dielectric constant. Hence option ( D) is also incorrect.
Therefore, only option ( B) is correct.

Note :
Remember conservation of energy states that the total energy of a system is constant. So, here the energy changes due to the placement of the slab and you can see placing the dielectric slab increases the energy of the system as $ {\varepsilon _0} > \varepsilon $ since, dielectric constant is always greater than 1.So, $ \dfrac{1}{{{\varepsilon _0}}} < \dfrac{1}{\varepsilon } $ $ \therefore E > {E_0} $ (as there was no material between the plates ) .Hence, The extra energy comes from the work done on the system due to placing the slab between the plates, is then converted into potential energy, which is stored in the system as electrostatic energy.