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),

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