A parallel plate capacitor having capacity C_ | vedclass

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

Question

(D) Case $1$: Battery is disconnected. The charge $Q$ remains constant.$Q = Q_0 = C_0 V_0$New separation $d' = 2d$. The new capacitance is $C' = \frac

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(D) Case $1$: Battery is disconnected. The charge $Q$ remains constant.$Q = Q_0 = C_0 V_0$New separation $d' = 2d$. The new capacitance is $C' = \frac{\epsilon_0 A}{d'} = \frac{\epsilon_0 A}{2d} = \frac{C_0}{2}$.The energy stored is $E_1 = \frac{Q^2}{2C'} = \frac{(C_0 V_0)^2}{2(C_0/2)} = C_0 V_0^2$.Case $2$: Battery remains connected. The potential $V$ remains constant.$V = V_0$New capacitance $C' = \frac{C_0}{2}$.The energy stored is $E_2 = \frac{1}{2} C' V^2 = \frac{1}{2} 	imes (\frac{C_0}{2}) 	imes V_0^2 = \frac{1}{4} C_0 V_0^2$.Ratio: $\frac{E_2}{E_1} = \frac{\frac{1}{4} C_0 V_0^2}{C_0 V_0^2} = 0.25$.

Similar Questions

When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

If two charges $q_1$ and $q_2$ are separated by a distance $d$ and placed in a medium of dielectric constant $K$, what will be the equivalent distance between the charges in air for the same electrostatic force?

When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery,then the charge on the plates relative to the earlier charge:

Two parallel plates of area $A$ are separated by two different dielectrics as shown in the figure. Find the resultant capacitance.

An air capacitor is connected to a battery. The effect of filling the space between the plates with a dielectric is to increase:

Vedclass Test Series

Exam Paper Generator

Online Exam Module