A capacitor is charged by a battery. The batt | vedclass

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

Question

(B) Let the initial capacitance be $C$ and the potential difference provided by the battery be $V$. The initial charge on the capacitor is $q = CV$.Th

Vedclass · Accepted Answer

decreases by a factor of $2$

Vedclass · Answer

(B) Let the initial capacitance be $C$ and the potential difference provided by the battery be $V$. The initial charge on the capacitor is $q = CV$.The initial electrostatic energy stored in the capacitor is $U_i = \frac{1}{2} CV^2$.When the battery is removed and an identical uncharged capacitor of capacitance $C$ is connected in parallel,the total charge $q$ is shared between the two capacitors. Since the capacitors are identical,the charge on each capacitor becomes $q' = q/2 = CV/2$.The common potential difference across the parallel combination is $V_c = \frac{q}{C_{eq}} = \frac{CV}{C + C} = \frac{V}{2}$.The final electrostatic energy of the system is the sum of the energy stored in both capacitors:$U_f = \frac{1}{2} C V_c^2 + \frac{1}{2} C V_c^2 = C V_c^2$.Substituting $V_c = V/2$:$U_f = C \left(\frac{V}{2}\right)^2 = C \left(\frac{V^2}{4}\right) = \frac{1}{4} CV^2$.Comparing the final energy with the initial energy:$U_f = \frac{1}{2} \left(\frac{1}{2} CV^2\right) = \frac{1}{2} U_i$.Thus,the total electrostatic energy of the resulting system decreases by a factor of $2$.

$A$ capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:

Similar Questions

$A$ capacitor of capacity $4 \, \mu F$ is charged to $80 \, V$ and another capacitor of capacity $6 \, \mu F$ is charged to $30 \, V$. When they are connected, the energy lost by the $4 \, \mu F$ capacitor is: (in $ \, mJ$)

$A$ capacitor of capacity $C_1$ is charged to potential $V_1$ and then disconnected. An uncharged capacitor of capacity $C_2$ is connected in parallel with $C_1$. The resultant potential $V_2$ is

$A$ capacitor of capacitance $50 \; pF$ is charged by a $100 \; V$ source. It is then connected to another uncharged identical capacitor. The electrostatic energy loss in the process is $\dots \; nJ$.

$A$ capacitor of capacitance $C_1 = 1 \ \mu F$ is charged using a $9 \ V$ battery. $C_1$ is then removed from the battery and connected to capacitors $C_2$ and $C_3$ of $2 \ \mu F$ and $3 \ \mu F$ respectively,as shown in the figure. Find the charge on $C_3$ after equilibrium is reached.

Two metal spheres of capacitance $C_1$ and $C_2$ carry some charges. They are put in contact and then separated. The final charges $Q_1$ and $Q_2$ on them will satisfy

Vedclass Test Series

Exam Paper Generator

Online Exam Module