Q. 445.0( 2 Votes )

# A parallel-plate capacitor having plate area 20 cm^{2} and separation between the plates 1.00 mm is connected to a battery of 12.0 V. The plates are pulled apart to increase the separation to 2.0 mm.

a) Calculate the charge flown through the circuit during the process.

b) How much energy is absorbed by the battery during the process?

c) Calculate the stored energy in the electric field before and after the process.

d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart.

e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.

Answer :

Given –

Plate area 20 cm^{2} = 0.002m^{2}

Separation between the plates 1.00 mm = 0.001m

Battery Voltage = 12.0 V

We know capacitance, C

1)

Where,

A= Plate Area

d= separation between the plates,

∈_{0} = Permittivity of free space

= 8.854 × 10^{-12} m^{-3} kg^{-1} s^{4} A^{2}

From1),

Capacitance when distance d = 0.001m, C_{1}

Substituting values,

= 2)

When The plates are pulled apart to increase the separation to –

2.0 mm = 0.002m, then capacitance C_{2} becomes,

Substituting values

= 3)

From 1) and 2)

4)

We know capacitance in terms of voltage is given by –

Q = CV 5)

Where

Q= charge stored on the capacitor

C= capacitance of the capacitor

V = applied voltage

Given applied v = 12V

From 2) and 3) and 5)

=

= 6)

And

7)

a) The charge flown through the circuit during the process –

From 6) and 7)

=

=

= 8)

b)Energy absorbed by the battery during the process-

We know Energy E is given by -

E=QV 9)

Where

Q = charged present on the surface

V= Applied voltage

From 8),

Applied voltage V = 12V

From 9),

Energy absorbed,

=

c)Stored energy in the electric field before and after the process

We know that stored energy in the electric field,

Before process, the energy stored -

From 2),

10)

From 3), After process, the energy stored will become

11)

d) The work done by the person pulling the plates apart.

We know, work done, W

12)

Where, f = force

d = displacement

We, know in parallel plate capacitor, the force between the plates is given by.

13)

Where, Q = charge enclosed,

σ = surface charge density,

σ, surface charge density is given by ,

From 12) and 13)

Work done,

Given, Plate area 20 cm^{2} = 0.002m^{2}

Separation between the plates changed to, 2.00 mm = 0.002 m by

pulling apart.

Substituting values, from 7),

Rate this question :

A parallel-plate capacitor has plate area 20 cm^{2}, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy for the capacitor 8.9 μs after the connections are made.

How many time constants will elapse before the energy stored in the capacitor reaches half of its equilibrium value in a charging RC circuit?

HC Verma - Concepts of Physics Part 2