Question

A capacitor is discharged such that the charge remains on it is half of the initial value, then the ratio of the initial energy to the final energy of the capacitor is:
1. 4 : 1
2. 2 : 1
3. 1 : 4
4. 1 : 2

Answer 1

Correct Answer - Option 1 : 4 : 1

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
  • In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​​

• The energy stored in the capacitor is given as,

\(\Rightarrow U=\frac{1}{2}QV=\frac{1}{2}CV^2=\frac{Q^2}{2C}\)

Energy density:

• It is defined as the energy stored per unit volume of space between the plates.
• Energy density u between the plates is given as,

\(\Rightarrow u =\frac{1}{2}\epsilon_oE^2\)

Where C = capacitance of the capacitor, Q = charge on the plates, V = potential difference between the plates, and E = electric field intensity between the plates

CALCULATION:

Let initial charge on the capacitor is (Q_I) = Q, the initial energy stored in the capacitor is U_I and the final energy stored in the capacitor is UF.

So the final charge on the capacitor is, Q_F = \(\frac{Q}{2}\)

• We know that the energy stored in the capacitor is given as,

\(\Rightarrow U=\frac{Q^2}{2C}\)     -----(1)

By equation 1 the initial energy stored in the capacitor is given as,

\(\Rightarrow U_I=\frac{Q_I^2}{2C}\)

\(\Rightarrow U_I=\frac{Q^2}{2C}\)     -----(2)

By equation 1 the final energy stored in the capacitor is given as,

\(\Rightarrow U_F=\frac{Q_F^2}{2C}\)

\(\Rightarrow U_F=\left ( \frac{Q}{2} \right )^2\times\frac{1}{2C}\)

\(\Rightarrow U_F=\frac{1}{4}\times\frac{Q^2}{2C}\)     -----(3)

By equation 2 and equation 3,

\(\Rightarrow U_F=\frac{U_I}{4}\)

\(\Rightarrow \frac{U_I}{U_F}=\frac{4}{1}\)

• Hence, option 1 is correct.

Parallel plate capacitor:

• A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​
  • The electric field intensity at the outer region of the parallel plate capacitor is always zero whatever be the charge on the plate.
  • The electric field intensity in the inner region between the plates of a parallel plate capacitor remains the same at every point.
  • When the dielectric medium is filled in the space between the plates of the parallel plate capacitor, its capacitance increases.
• The electric field intensity in the inner region between the plates of a parallel plate capacitor is given as,

For Vacuum	For Dielectric Medium
\(E=\frac{σ}{\epsilon_o}=\frac{Q}{A\epsilon_o}\)	\(E'=\frac{σ}{\epsilon_oK}=\frac{Q}{A\epsilon_oK}\)

• The potential difference between the plates is given as,

For Vacuum	For Dielectric Medium
\(V=\frac{Qd}{A\epsilon_o}\)	\(V'=\frac{Qd}{A\epsilon_oK}\)

• The capacitance C of the parallel plate capacitor is given as,

For Vacuum	For Dielectric Medium
\(C=\frac{Q}{V}=\frac{A\epsilon_o}{d}\)	\(C'=\frac{Q'}{V'}=\frac{A\epsilon_oK}{d}\)

​Where A = area of the plates, d = distance between the plates, Q = charge on the plates, σ = surface charge density, E = electric field between the plates, and K = dielectric constant