Courses for Kids
Free study material
Offline Centres
Store Icon


Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

(a) while the voltage supply remained connected.
(b) after the supply was disconnected.

Last updated date: 13th Jun 2024
Total views: 62.1k
Views today: 0.62k
62.1k+ views

Hint: Consider the effect of the mica sheet on the electric field between the plates and the capacitance of the capacitor.


Initial capacitance, $ C = 18\text{pF} $

Charge on each capacitor plate, $ q = 1.8 \times 10^{-9} \text{C} $

Dielectric constant of mica, $ k = 6 $

Area of each plate, $ A = 6 \times 10^{-3} \text{m}^2 $

Distance between plates, $ d = 3\text{mm} $

(a) With voltage supply still connected:

The capacitance with the mica sheet inserted is defined as:

$C' = k \times C$

Plugging in the provided values:

$C' = 6 \times 18\text{pF}$

This results in:

$C' = 108\text{pF}$

So, when the mica sheet is inserted while the voltage supply remains connected, the capacitance increases to 108pF. However, the potential difference (voltage) across the capacitor remains unchanged.

(b) After disconnecting the supply:

Again, the new capacitance is: $C' = 6 \times 18\text{pF} = 108\text{pF}$
But now, the charge remains constant at  $ q = 1.8 \times 10^{-9} \text{C} $. Using the formula: $q = C' \times V'$

Where V′ is the new potential difference. Rearranging for V′ we have:

$V' = \frac{q}{C'} = \frac{1.8 \times 10^{-9}}{108 \times 10^{-12}} = 16.6\, \text{V}$

Thus, after disconnecting the supply, while the capacitance is still 108pF, the potential difference across the capacitor drops to 16.6V.

Note: The insertion of the mica sheet has a significant impact on the capacitor's behavior. The mica sheet reduces the electric field, increases the capacitance, and changes the potential difference. These changes are important to consider when designing and using capacitors.

Students Also Read