Question

The plates of a parallel plate capacitor are charged up to 100V. After the battery is removed, a 2mm thick plate is inserted between the plates then to maintain the same potential difference. The distance between the plates is then increased by 1.6mm. What is the dielectric constant of the plate?

Answer 1

Hint: When we introduced a thick plate between the plates of the capacitor, therefore, the electric field value was reduced, and the capacitance of the parallel plate capacitor increases by a factor $k$, i.e. (dielectric constant).

Complete step by step solution:
The expression for calculating the dielectric constant is:
\[ \Rightarrow {V_{air}} = {V_{medium}}\]
Here, \[{V_{air}}\] is the potential difference between the plates with air as a dielectric, \[{V_{medium}}\] is the potential difference between the plates.
We will now simplify the above equation further and it becomes,
\[ \Rightarrow \dfrac{\sigma }{{{\varepsilon _0}}}d = \dfrac{\sigma }{{{\varepsilon _0}}}\left( {d + d' - t + \dfrac{t}{k}} \right)\]
Here, \[\sigma \] is the charge density, d is the distance between plates, d| is the increased distance between the plates, t is the thickness of the plates, and \[{\varepsilon _0}\] is the permittivity of free space.
We can see that here, after simplifying the above equation that,
\[ \Rightarrow d + d' - t + \dfrac{t}{k} = d\]
\[ \Rightarrow t = d + \dfrac{t}{k}\] ……. (I)
We will now substitute t=2mm and d’=1.6 mm to find k.
\[ \Rightarrow 2 = 1.6 + \dfrac{2}{k}\]
\[\therefore k = 5\]

Therefore, the dielectric constant of the plate is 5.

Note: When a dielectric is put inside the plates then the value of the electric field increases and the value of voltage decreases and hence the capacitance increases. The plates store the equal charge one before and after introducing the dielectric but with a lower voltage.
We must know that polarization is very good for a dielectric apart from being insulators. It is very good if a dielectric gets polarized. If a certain voltage is applied to a dielectric, then the polarization takes place which really affects the electric field inside it.