If the distance between the plates of a capac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.If the distance $d$ is doubled $(d' = 2d)$,the new capacita

Vedclass · Accepted Answer

$\frac{Q^2}{2C}$

Vedclass · Answer

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.If the distance $d$ is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2}$.The charge $Q$ remains constant as the capacitor is isolated.The initial energy stored is $U_i = \frac{Q^2}{2C}$.The final energy stored is $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/2)} = \frac{Q^2}{C}$.The work done by the external agent is equal to the change in potential energy: $W = U_f - U_i$.$W = \frac{Q^2}{C} - \frac{Q^2}{2C} = \frac{Q^2}{2C}$.

If the distance between the plates of a capacitor having capacity $C$ and charge $Q$ is doubled,then the work done will be:

Similar Questions

$A$ parallel plate capacitor has a uniform electric field '$E$' in the space between the plates. If the distance between the plates is '$d$' and the area of each plate is '$A$', what is the energy stored in the capacitor?

How much work is required to increase the voltage of a $6\,\mu F$ capacitor from $10\ V$ to $20\ V$?

$A$ piece of cloud having an area of $25 \times 10^6 \, m^2$ and an electric potential of $10^5 \, V$. If the height of the cloud is $0.75 \, km$,then the energy of the electric field between the earth and the cloud will be.....$J$

$A$ parallel plate capacitor having a plate separation of $2\, mm$ is charged by connecting it to a $300\, V$ supply. The energy density is.....$J/m^3$

Vedclass Test Series

Exam Paper Generator

Online Exam Module