Question 668: Electrostats & Capacitance

Question

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dieletric constant K, which  can just fill the air gap of the capacitor, is now inserted in it . Which of the following is incorrect ?

Accepted Answer

q = CV  $$ \Rightarrow $$  V = q/C
Due to dielectric insertion, new capacitance C
2
= CK
Initial energy stored in capacitor, $${U_1} = {{{q^2}} \over {2C}}$$
Final energy stored in capacitor, $${U_2} = {{{q^2}} \over {2KC}}$$
Change in energy stored, $$\Delta $$U = U
2
Ã¢ÂÂ U
1
$$\Delta U = {{{q^2}} \over {2C}}\left( {{1 \over K} - 1} \right) = {1 \over 2}C{V^2}\left( {{1 \over K} - 1} \right)$$
New potential difference between plates
$$V' = {q \over {CK}} = {V \over K}$$

Answer

The change in energy stored is $${1 \over 2}C{V^2}\left( {{1 \over K} - 1} \right)$$

Answer

The charge on the capacitor is not conserved.

Answer

The potential difference between the plates decreases K times.

Answer

The energy stored in the capaciotor decreases K times.

Electrostats & Capacitance

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dieletric constant K, which can just fill the air gap of the capacitor, is now inserted in it . Which of the following is incorrect ?

Choose the correct answer: