Chapter 22, Problem 101A

To determine

To plot: a graph of $\Delta V$ as the capacitor is charged by adding 5 C to it.

To find: the voltage across the capacitor.

To find: the energy in joule.

To explain: whether the energy equal to the total charge times the final potential difference or not.

Answer to Problem 101A

The required graph is shown. The voltage across the capacitor is 5.0 V. energy stored in a capacitor is 12.5 J and it is half of the product of charge and potential value.

Explanation of Solution

Given:

The value of the capacitor is $C=1\text{F}$ .

Charge to be added is $q=5\text{}\text{C}$ .

Formula used:

The energy needed to increase the potential difference of a charge $\left(q\right)$ is represented by $E=q\Delta V$  ......(1)

The expression for potential difference can be expressed as

  $\Delta V=\frac{q}{C}$  ......(2)

Calculation:

The potential difference in a capacitor is equal to the ratio of charge and the capacitance of the capacitor that is,

  $V=\frac{q}{C}$  ......(3)

Here, $q$ is the charge and $C$ is the capacitance of the capacitor.

Substitute the values of $C$ and $q$ in equation (3)

  $\begin{array}{l}V=\frac{5.0\text{C}}{1.0\text{F}}\\ \boxed{V=5.0\text{V}}\end{array}$

Therefore, the potential difference in a supercap is 5.0 V.

Below table shows the data of the charge and the voltage:

Charge (C)	Potential difference (V)
1.0	1.0
2.0	2.0
3.0	3.0
4.0	4.0
5.0	5.0

The graph between charge and the potential difference is shown in the following figure. Here, the charge is taken along the $x$ axis and the potential difference is taken along the $y$ axis.

  

The area under the curve is equal to the energy stored in the capacitor. Hence, the area under the curve is equal to the area of the triangle.

From the above graph, the base of the triangle is 5.0 C and the height of the triangle is 5.0 V.

Substitute the values of base and height in the equation in the equation of the area of the triangle formula,

  $\begin{array}{c}\text{Area}=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)\\ =\frac{1}{2}\left(5.0\text{C}\right)\left(5.0\text{V}\right)\\ =12.5\text{J}\end{array}$

Rounding off to two significant figures, the energy stored in a capacitor is 13 J.

The total energy stored in a capacitor is not equal to the charge times the final potential difference.

The product of the charge and final potential difference is,

  $\left(5.0\text{C}\right)\left(5.0\text{V}\right)=25\text{J}$

This value is approximately half of the energy stored in a capacitor. So, each charge could require the same amount of maximum charge to place it on the capacitor.

Conclusion:

Hence, the required graph is shown. The voltage across the capacitor is 5.0 V. Energy stored in a capacitor is 12.5 J and it is half of the product of charge and potential value.