19.5 Energy Stored in a Capacitor

Why Work is Required to Charge a Capacitor

When a capacitor is being charged, positive charges are transferred from the negative plate to the positive plate. Each successive charge element must be pushed against the repulsion of charges already accumulated on the plates. As more charge builds up, the potential difference $V$ between the plates increases, so progressively more work is needed to transfer each additional charge. This work is stored as electric potential energy in the capacitor.

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Derivation of Energy Stored

At any instant during charging, if the charge on the capacitor is $q$ and the potential difference is $v=q/C$, the small amount of work done to transfer an additional charge $dq$ is:

$dW=vdq=\frac{q}{C}dq$

Integrating from $q=0$ to $q=Q$ (the final charge):

$U={\int }_0^Q\frac{q}{C}dq=\frac{1}{C}\cdot \frac{Q^2}{2}=\frac{Q^2}{2C}$

Using $Q=CV$, this can be written in three equivalent forms:

$\boxed{U=\frac{1}{2}QV=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}}$

The factor of $\frac{1}{2}$ arises because the average potential difference during charging is $\frac{0+V}{2}=\frac{V}{2}$.

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Graphical Interpretation

If we plot charge $Q$ against potential difference $V$, we get a straight line through the origin (since $Q=CV$). The area under the $Q$–$V$ graph (a triangle) equals:

$\text{Area}=\frac{1}{2}\times Q\times V=U$

This confirms that the energy stored equals $\frac{1}{2}QV$.

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Energy Stored in the Electric Field

The energy stored in a capacitor resides in the electric field between the plates. For a parallel plate capacitor with plate area $A$ and separation $d$:

• Volume of field region: $\text{Vol}=A\cdot d$
• Electric field: $E=V/d$, so $V=Ed$
• Capacitance: $C={ϵ}_{0}A/d$

Substituting into $U=\frac{1}{2}C{V}^2$:

$U=\frac{1}{2}\cdot \frac{{ϵ}_{0}A}{d}\cdot (Ed{)}^2=\frac{1}{2}{ϵ}_0{E}^2\cdot (Ad)$

The energy density $u$ (energy per unit volume) is therefore:

$\boxed{u=\frac{U}{\text{Vol}}=\frac{1}{2}{ϵ}_0{E}^2}$

With a dielectric of relative permittivity ${ϵ}_r$:

$u=\frac{1}{2}{ϵ}_0{ϵ}_r{E}^2$

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Worked Example

Problem: A $10\mu \text{F}$ capacitor is charged to $100\text{V}$. Find the energy stored.

Solution: $U=\frac{1}{2}C{V}^2=\frac{1}{2}\times (10\times 10^{-6})\times (100{)}^2$ $U=\frac{1}{2}\times 10^{-5}\times 10^4=0.05\text{J}$

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Uses of Capacitors in Household Appliances

Capacitors are found in many everyday devices. Key examples include:

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Summary of Formulas