## Section37.3Energy in Capacitors

A capacitor is an energy-storing device. By storing charges separated by a distance, the capacitor essentially stores energy in the potential energy of the charges, or equivalently in the electric field of the space between plates.

One way to easily figure out the energy stored in a capacitor is to use energy conservation in the discharging circuit. Connect a charged capacitor to a resistor $R$ and let current flow in the simple RC-circuit and determine the net energy dissipated in the resistor. When current $I(t)$ passes through a resistor, the instantaneous power-dissipation rate, $P(t) = I(t)^2 R\text{,}$ as we have seen before.

We have already found the following expression for current at instant $t$ in discharging circuit in which the initial charg eon the capacitor is $Q_0\text{.}$

\begin{equation*} I(t) = \dfrac{Q_0}{RC}\, \exp{\left( -t/RC\right ) }. \end{equation*}

Therefore, instantaneous power dissipated in the resistor will be

$$P(t) = \dfrac{Q_0^2}{RC^2}\, \exp{\left( -2t/RC\right ) }.\label{eq-instantaneous-power-in-discharging-capacitor}\tag{37.3.1}$$

This equation tells us the rate at which energy is being discharged by the capacitor. Therefore, energy discharged between $t$ and $t+\Delta t$ will be

\begin{equation*} \Delta U = P(t)\Delta t. \end{equation*}

Therefore, integrating Eq. (37.3.1) from $t=0$ to $t=T$ will give us the amount of energy discharged during this time.

$$U = \int_0^T P(t)\; dt.\tag{37.3.2}$$

You can show that this leads to

$$U = U_0\left( 1 - e^{-2T/RC}\right),\tag{37.3.3}$$

where $U_0$ is the energy in the capacitor at time $t=0\text{,}$ as we will interpret below.

\begin{equation*} U_0 = \frac{1}{2}Q_0 V_0. \end{equation*}

By taking $T=\infty$ we get all energy that was dissipated in the resistor.

\begin{equation*} E_\text{dissipated} = \dfrac{1}{2}\dfrac{Q_0^2}{C}. \end{equation*}

But, by conservation of energy in the circuit, this energy must have been originally stored in the capacitor. Therefore, a capacitor of capacitance $C$ charged to $Q_0$ stores the following energy. Since this energy is potential energy, we use symbol $U$ for it.

\begin{equation*} U_\text{in capacitor} = \dfrac{1}{2}\dfrac{Q_0^2}{C}. \end{equation*}

By using the capacitor formula, $Q=CV\text{,}$ we can write this in other forms.

$$U_\text{in capacitor} = \dfrac{1}{2}\dfrac{Q_0^2}{C} = \dfrac{1}{2}Q_0V_0 = \dfrac{1}{2}CV_0^2.\tag{37.3.4}$$

(a) A capacitor of capacitance $2.0\text{ mF}$ is charged by a $1.5\text{-V}$ battery. How much total energy will be released when this capacitor is discharged?

(b) If discharging circuit has resistance $5.0\;\Omega\text{,}$ how much time will it take to get $70\%$ of the energy?

Hint

(a) Use total energy formula. (b) Use finite time formula.

(a) $2.25\text{ mJ}\text{,}$ (b) $1.78\text{ ms}\text{.}$

Solution 1 (a)

(a) The energy released will be equal to the energy stored in the capacitor. Since $C$ and $V_0$ are given, we use the formula using these.

\begin{equation*} U_0 = \frac{1}{2}CV_0^2 = \frac{1}{2}\times 2.0\text{ mF}\times (1.5\text{ V})^2 = 2.25\text{ mJ}. \end{equation*}
Solution 2 (b)

(b) Energy released in time $t=0$ to $t=T$ is

\begin{equation*} U = U_0 \left( 1 - e^{-2t/RC}\right). \end{equation*}

We are given the following.

\begin{equation*} \frac{U}{U_0} = 0.3,\ \ RC = 5\times 2\times 10^{-3} = 0.01\text{ s}. \end{equation*}

Therefore, we have the following equation to solve.

\begin{equation*} 0.3 = 1 - e^{-200\,t}. \end{equation*}

We first isolate the exponential on one side and take natural log to solve this.

\begin{equation*} t = -\frac{ \ln\;0.7 }{200} = 1.78\text{ ms}. \end{equation*}