Question

Derive an expression for the energy stored in a parallel plate capacitor. On charging a parallel plate capacitor to a potential v, the spacing between the plates is halved, and dielectric medium of = 10 is introduced between the plates, without disconnecting the d.C source. Explain, using suitable expressions, how the (i) capacitance, (ii) electric field and (iii) energy density of the capacitor change.

Answer 1

Energy stored in a parallel plate capacitor :

Explanation:

• Energy stored in a capacitor is the electrical potential energy, and it is related to the charge Q and voltage V on the capacitor.
• When applying the equation for electrical potential energy ΔPE = qΔV to a capacitor, ΔPE is the potential energy of a charge q going through a voltage ΔV.
• But the capacitor starts with zero voltage and gradually increases to its full voltage as it is charged.
• The first charged placed on a capacitor experiences a change in voltage ΔV = 0.
• The capacitor has zero voltage when uncharged.
• The final charge on a capacitor experiences ΔV = V, Now the capacitor now has its full voltage V on it.
• The average voltage on the capacitor during in the charging process is $\displaystyle \frac{V}{2}$ .

• So the total average voltage experienced by the full charge q is $\displaystyle \frac{V}{2}$ .

• Thus the energy stored in a parallel plate capacitor Ecap is

        $\displaystyle E_{\text{cap}}=\frac{QV}{2}$

where

Q is the charge on a parallel plate capacitor with a voltage V applied. (Note that the energy is not QV, but $\displaystyle \frac{QV}{2}$ .

• The charge and the voltage are related to the capacitance C of a capacitor by Q = CV, and so the expression for Ecap can be algebraically manipulated into three equivalent expressions:

        $\displaystyle{E}_{\text{cap}}=\frac{QV}{2}=\frac{CV^2}{2}$

        $\displaystyle \mathbf{E_{cap} =\frac{Q^2}{2C}}$

where

Q is the charge and V the voltage on a parallel plate capacitor C.

• The energy is measured in joules for a charge in coulombs, a voltage in volts, and capacitance in farads.