A parallel plate capacitor having plate area | vedclass

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

Question

(D) Initial capacitance $C = \frac{\varepsilon_0 A}{d}$. Initial charge $q = CV = \frac{\varepsilon_0 A V}{d}$.Initial energy $U_i = \frac{q^2}{2C} =

Vedclass · Accepted Answer

$\frac{3 \varepsilon_0 A V^2}{2 d}$

Vedclass · Answer

(D) Initial capacitance $C = \frac{\varepsilon_0 A}{d}$. Initial charge $q = CV = \frac{\varepsilon_0 A V}{d}$.Initial energy $U_i = \frac{q^2}{2C} = \frac{1}{2} CV^2 = \frac{\varepsilon_0 A V^2}{2d}$.When the battery is disconnected,the charge $q$ remains constant.The new separation is $d' = 4d$,so the new capacitance is $C' = \frac{\varepsilon_0 A}{4d} = \frac{C}{4}$.Final energy $U_f = \frac{q^2}{2C'} = \frac{q^2}{2(C/4)} = \frac{4q^2}{2C} = 4U_i$.Work done $W = U_f - U_i = 4U_i - U_i = 3U_i$.Substituting $U_i = \frac{\varepsilon_0 A V^2}{2d}$,we get $W = 3 	imes \frac{\varepsilon_0 A V^2}{2d} = \frac{3 \varepsilon_0 A V^2}{2d}$.

$A$ parallel plate capacitor having plate area $A$ and separation $d$ is charged to a potential difference $V$. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between plates is:

Similar Questions

$A$ $6\,\mu F$ capacitor is charged from $10\,V$ to $20\,V$. The increase in energy will be:

If the charge on the capacitor is increased by $2 \text{ C}$, the energy stored in it increases by $21 \%$. The original charge on the capacitor is (in $\text{ C}$)

The energy of a charged capacitor resides in

$A$ 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 $A$,the energy stored in the capacitor is (where $\epsilon_{0}$ is the permittivity of free space).

Vedclass Test Series

Exam Paper Generator

Online Exam Module