Question

The following information from Problem 2 is provided to help with answering Problem 3

The plates of a parallel plate capacitor each have an area of 0.40 m² and are separated by a distance of 0.02m. They are charged until potential difference between the plates is 3000 V. The charged capacitor is then disconnected from the battery. Suppose that a dielectric sheet is inserted to completely fill the space between the plates and the potential difference between the plates drops to 1000 V.

Problem 3

The capacitor in Problem 2 remains connected to a 3000 V battery while the same dielectric sheet is inserted to fill the gap of the capacitor.

a. What is the capacitance of the system after the dielectric is inserted?

b. Determine the amount of the charge on either capacitor plate in the new situation.

c. What is the magnitude of the electric field in the new situation?

d. How much energy is stored in the capacitor in the new situation?

Answer 1

Information from problem 2 :

The capacitance (C) of a parallel plate capacitor having plate area ' A' and separation between the plates ' d ' is given by

  $\dpi{100} C = \frac{A \epsilon}{d} = \frac{A K \epsilon _{0}}{d}$

$\dpi{100} \epsilon$ is permittivity of the medium

$\dpi{100} \epsilon_{0} = 8.854 \times 10^{-12} \, \, F/m$ is permittivity of free space

K is dielectric constant

When there were no any dielectric medium between the plates

A = 0.40 m²

d = 0.02 m

K = 1 (As there was no any dielectric medium between the plates)

So,

$\dpi{100} C_{0} = \frac{0.40\times 8.85 \times 10^{-12}}{0.02} = 177 \times 10^{-12} \, \, F = 0.177 \, \, nF$

The charge on any plate of capacitor is given by

  $\dpi{100} Q_{0} = C_{0}V_{0}$

Given, $\dpi{100} V_{0} = 3000 \, \, V$ (dielectric absent )

So. charge on the capacitor was

  $\dpi{100} Q_{0} = 177 \times 10^{-12} \times 3000 = 531 \times 10^{-9} \, \, C = 531 \, \, nC$

After inserting dielectric sheet between the plates :

NOTE : A capacitor with a dielectric stores the same charge as one without a dielectric, but at a lower voltage.

  $\dpi{100} C = \frac{A K \epsilon _{0}}{d} = K C_{0}$

And $\dpi{100} Q_{0} = CV = KC_{0}V$

Given, the new potential difference across the plates of capacitor is 1000 V .

So,   $\dpi{100} Q_{0} = KC_{0} \times 1000$

or, $\dpi{100} 531\times 10^{-9} = K \times 0.177 \times 10^{-9} \times 1000$

or,   $\dpi{100} K = 3$

So, the value of dielectric constant of the material is 3 .

Problem 3 :

If the capacitor with the dielectric remains connected to a 3000 V battery.

(a) The new capacitance of the capacitor will be

  $\dpi{100} C = KC_{0} = 3 \times 0.177 \times 10^{-9}= 0.531 \times 10 ^{-9} \, \, F = 0.531 \,\,nF$

(b) We have

$\dpi{100} Q = CV$

Given V = 3000 V

So, charge on the plates of capacitor will be

$\dpi{100} Q = 0.531 \times 10^{-9} \times 3000 = 1593 \times 10^{-9} \, \, C = 1593 \,\,nC$

(c) Electric field in the new situation will be

$\dpi{100} E = \frac{Q}{AK \epsilon _{0}} = \frac{1593 \times 10^{-9}}{0.40 \times 3 \times 8.85 \times 10^{-12}} = 1.5 \times 10^{5} \, \, V/m$

(d) The energy stored in the capacitor is given by

  $\dpi{100} U = \frac{1}{2} \, CV^{2} = 0.5 \times 0.531 \times 10^{-9} \times (3000)^{2} = 2.4 \times 10^{-3} \, \, Joule$

For any doubt please comment and please give an up vote . Thank you.