Question

A capacitor is given a charge q. The distance between the plates of the capacitor is d. One of the plates is fixed and the other plate is moved away from the other till the distance between them becomes 2d. Find the work done by the external force.

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

A capacitor is given a charge q. The distance between the plates of the capacitor is d. One of the plates is fixed and the other plate is moved away from the other till the distance between them becomes 2d. Find the work done by the external force.

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Charge = q
• Distance between plates = d
• Next distance = 2d

$\large\underline{\underline{\sf To\:Find:}}$

• Work done by external force (W)

❏$\large{\boxed{\bf \blue{F=\dfrac{q^2}{2A\epsilon}}}}$

❏$\large{\boxed{\bf \blue{Work\:Done= F. dx} }}$

$\implies{\sf \int W = \int^{2d}_d F.dx }$

$\implies{\sf W = \dfrac{q^2}{2A\epsilon}×\int^{2d}_{d} dx }$

$\implies{\sf W = \dfrac{q^2}{2A\epsilon}×(2d-d) }$

$\implies{\bf \red{W=\dfrac{q^2d}{2A\epsilon}} }$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Work done by the external force is ${\bf \red{\dfrac{q^2d}{2A\epsilon}}}$.

Answer 2

GIVEN :

A capacitor is given a charge q.

The distance between the plates ,d. One of the plates is fixed, other is moved away, till the distance becomes 2d.

TO FIND :

Work done by the external force.in moving from d to 2d.

SOLUTION:

◆We know ,On a capacitor,

Force, F = q^2/(2AE° )

◆Where, q - charge,

A - Area of Capacitor,

E° - permittivity of medium

◆Work done, dW = F.dx

◆Work done in moving a parallel plate from d to 2d,

W = ∫ dW = ∫ F.dx

=∫ q^2/(2AE°).dx

= q^2/(2AE°) ∫ dx ( from d to 2d)

= q^2/(2AE°) .(2d-d)

= q^2d/(2AE°).

ANSWER:

WORK DONE ,W = q^2d/(2AE°).