Question

A parallel plate capacitor has each plate area A and separation d. Charge Q is given to positive plate of capacitor. The force on each plate of capacitor is​

Answer 1

Answer:

$\huge\mathbb\red{Q²/4c}$

Explanation:

Initial capacitance C

1

=

d

Aϵ

0

Final capacitance C

2

=

1.5d

Aϵ

0

=

3d

2Aϵ

0

Since the capacitor was isolated, the charge stored remains the same.

Thus work done=

2C

2

Q

2

−

2C

1

Q

2

=

4C

Q

2

Answer 2

Answer:

Explanation:

Force between two plates of the parallel plate capacitor is given by ,

        $F=q.E$

To get force between two parallel  plate capacitor we have first find electric field E and charge is q.

Electric field by any one plate is given by

      E = σ/∈₀(i)

   where,  σ is the surface charge density.

                ∈₀ is the vaccum permitivity

we know that ,

  surface charge density is given by -

       σ = q/A   (ii)

By substituting (i) and (ii) ,

E =  σ/∈₀

  = (q/A)/∈₀

 = q/A∈₀

And force between two plates of parallel capacitor is given by ,

  F = q/.E

     = q.(q/A∈₀)

    = q²/A∈₀

Therefore , answer is $\frac{Q^{2} }{EoA}$.

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