Question

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

Short answer

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

Step-by-step solution

Identification of given data

The capacitance of the capacitor is$C$

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

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$Q=C\Delta V$.

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:

$\begin{array}{rcl}dW& =& dq\left(\Delta V\right)\\ dW& =& dq\left(\frac{q}{C}\right)\\ W& =& \frac{1}{C}\underset{q_i}{\overset{q_f}{\int }}\left(q\right)dq\\ & & \end{array}$

Substitute all the values in the above equation.

$\begin{array}{rcl}dW& =& \frac{1}{C}\underset{0}{\overset{Q}{\int }}\left(q\right)dq\\ W& =& \frac{1}{C}{\left(\frac{q^2}{2}\right)}_0^Q\\ W& =& \frac{1}{C}\left(\frac{Q^2}{2}-\frac{{\left(0\right)}^2}{2}\right)\\ W& =& \frac{1}{2}\left(\frac{Q^2}{C}\right)\\ & & \end{array}$

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

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

Substitute $Q=C\Delta V$ in the above equation.

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

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