Question

The capacitance of a parallel plate capacitor is 10 uF. When a dielectric plate is introduced in between the plates, its potential becomes 1/4th of its original value. What is the value of the dielectric constant of the plate introduced? (A) 4 (B) 40 (C) 2.5 (D) none of the above​

Answer 1

Given: The capacitance of a parallel plate capacitor is 10 uF. When a dielectric plate is introduced in between the plates, its potential becomes 1/4th of its original value.

To find: The value of the dielectric constant of the plate introduced.

Solution:

• Capacitance of a parallel plate is its ability to store charge with it.
• According to the formula of capacitance,

        $C= \frac{K \epsilon_{o} A }{d}$

• Here, C is the capacitance, K is the dielectric constant, epsilon is the permittivity of free space, A is the srea of the plate and d is the distance between the two plates.

• This capacitance is equal to 10 μF, that is, 10 * 10⁻⁶  F.
• According to the question, potential after the introduction of dielectric plate is,

       $C=\frac{1}{4} * \frac{K \epsilon_{o} A }{d}$

• So, $C'=\frac{1}{4} * C$

• But here, C is 10 * 10⁻⁶  F.

• So, $C'=\frac{1}{4} * 10 *10^{-6}$

                   $= 2.5 \mu$

Therefore, the value of the dielectric constant of the plate introduced is 2.5.