Chapter 19: Q65P (page 801)

Work and energy with a capacitor: A capacitor with capacitance $C$ has an amount of charge q on one of its plates, in which case the potential difference across the plates is $ΔV = q / C$ (definition of capacitance). The work done to add a small amount of charge dq when charge the capacitor is $dq (ΔV) = dq (q / C)$ . Show by integration that the amount of work required to charge up the capacitor from no charge to final charge Q is $\frac{1}{2} (Q^{2} / C)$ . Since this is the amount of work required to charge the capacitor, it is also the amount of energy stored in the capacitor. Substituting $Q = CΔV$ , we can also express the energy as $\frac{1}{2} C {(ΔV)}^{2}$ .

Short Answer

Expert verified

The amount of work required to charge up capacitor is $\frac{1}{2} (\frac{Q^{2}}{C})$ and the amount of energy stored in the capacitor is $\frac{1}{2} C {(Δ V)}^{2}$ .

Step by step solution

Identification of given data

The capacitance of the capacitor is $C$

The potential difference across the plates is $Δ V = \frac{q}{C}$ .

The initial charge of the capacitor is $q_{i} = 0$ .

The final charge of the capacitor is $q_{f} = Q$ .

The charge on the capacitor is $Q = C Δ V$ .

Conceptual Explanation

The electric work is the effect.

Determination of amount of work required to charge the capacitor

The amount of work to add a small charge to capacitor is given as:

$\begin{array}{rcl} d W & = & d q (Δ V) \\ d W & = & d q (\frac{q}{C}) \\ W & = & \frac{1}{C} \int_{q_{i}}^{q_{f}} (q) d q \end{array}$

Substitute all the values in the above equation.

$\begin{array}{rcl} d W & = & \frac{1}{C} \int_{0}^{Q} (q) d q \\ W & = & \frac{1}{C} {(\frac{q^{2}}{2})}_{0}^{Q} \\ W & = & \frac{1}{C} (\frac{Q^{2}}{2} - \frac{{(0)}^{2}}{2}) \\ W & = & \frac{1}{2} (\frac{Q^{2}}{C}) \end{array}$

The amount of energy stored in the capacitor is given as:

$\begin{array}{rcl} U & = & W \\ W & = & \frac{Q^{2}}{2 C} \end{array}$

Substitute $Q = C Δ V$ in the above equation.

$\begin{array}{rcl} W & = & \frac{{(C Δ V)}^{2}}{2 C} \\ W & = & \frac{1}{2} C {(Δ V)}^{2} \end{array}$

Therefore, the amount of work required to charge up capacitor is $\frac{1}{2} (\frac{Q^{2}}{C})$ and the amount of energy stored in the capacitor is $\frac{1}{2} C {(Δ V)}^{2}$ .