# A parallel-plate

Given –

Plate area 20 cm2 = 0.002m2

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 s4 A2

From1),

Capacitance when distance d = 0.001m, C1 Substituting values, = 2)

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

2.0 mm = 0.002m, then capacitance C2 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 cm2 = 0.002m2

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

pulling apart.

Substituting values, from 7),     Rate this question :

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