Question

The capacity of a parallel plate condenser is\[5\mu F\]. When a glass plate is placed between the plates of the condenser, its potential difference reduces to \[\dfrac{1}{8}\]of the original value. The magnitude of the relative dielectric constant of glass is
A. 2
B. 6
C. 7
D. 8

Answer 1

Hint: We will consider the equations of the potential difference and the capacitance before and after inserting the glass plate. Then, comparing these equations, we will find the equation that relates to the original potential difference, the dielectric constant and the final potential difference.
Formula used:
\[V=\dfrac{Q}{C}\]

Complete answer:
From the given information, we have the data as follows.
The capacity of a parallel plate condenser is\[5\mu F\]. When a glass plate is placed between the plates of the condenser, its potential difference reduces to \[\dfrac{1}{8}\]of the original value.
Let the potential difference across the condenser before inserting the glass plate be V. Then, the potential difference is given as follows.
\[V=\dfrac{Q}{C}\]……. (1)
Where Q is the charge enclosed by the condenser and C is the capacitance of the condenser.
Here, the capacitance of the condenser is given as follows.
\[C=\dfrac{{{\varepsilon }_{0}}A}{d}\]…… (2)
Where A is the area of the parallel plates and d is the difference between the parallel plates.
Let the potential difference across the condenser after inserting the glass plate be V’. Then, the potential difference is given as follows.
\[V'=\dfrac{Q}{C'}\]…… (3)
Where Q is the charge enclosed by the condenser and C is the capacitance of the condenser.
Here, the capacitance of the condenser is given as follows.
\[C'=k\dfrac{{{\varepsilon }_{0}}A}{d}\]…… (4)
Where k is the dielectric constant of the medium inserted between the plates, A is the area of the parallel plates and d is the difference between the parallel plates.
Using equation (2), this capacitance can be expressed as follows.
\[C'=kC\]…… (5)
Now substitute the equation (5) in equation (3).
\[V'=\dfrac{Q}{kC}\]
Substitute the equation (1) in the above equation.
\[V'=\dfrac{1}{k}V\]….. (a)
The potential difference reduces to \[\dfrac{1}{8}\]of the original value. So,
\[V'=\dfrac{1}{8}V\]……. (b)
Upon comparing the equations (a) and (b), we get the value of the magnitude of the relative dielectric constant of glass as 8.
\[\therefore k=8\]
\[\therefore \] The magnitude of the relative dielectric constant of glass as 8.

Thus, option (D) is correct.

Note:
The condenser is the capacitor. This is a direct question. If they ask for the capacitance value, then, the final capacitance values will be dielectric constant times the original capacitance value.