Question

You'll learn in Chapter 25 that the potential energy of two electric charges is inversely proportional to the distance between them. Two charges $30\text{nm}$apart have $1.0\text{J}$of potential energy. What is their potential energy if they are $20\text{nm}$apart?

Short answer

The energy when the charges are $20\text{nm}$apart is $1.5\text{J}$.

Step-by-step solution

Step.1.

The potential energy of two electric charges is inversely proportional to distance,

Mathematically,

$E\propto \frac{1}{r}$

That is for different position the ratio of potential energy can be expressed as,

$\frac{E_1}{E_2}=\frac{r_2}{r_1}$

Here $E_1$is the potential energy when the charge is distance $r_1$apart, and $E_2$is the potential energy when the charge is distance$r_2$ apart.

Step.2.

The potential energy $\left(E_2\right)$when the charge is at $r_2=20\text{nm}$apart is calculated as follows. Rearrange the expression of potential energy to find the value of potential

$E_2=E_1\left(\frac{r_1}{r_2}\right)$

Substitute $1.0\text{J}$for $E_1,20.0\text{nm}$for $r_2$and $30.0\text{nm}$for $r_1$in the above expression,

$\begin{array}{l}{E}_2=(1.0\text{J})\left(\frac{30.0\text{nm}}{20.0\text{nm}}\right)\\ =1.5\text{J}\end{array}$

Hence the energy when the charges are $20\text{nm}$apart is $1.5\text{J}$.