Question

2. A mobile -4C charge, initially at rest, is located 3.5m away from a fixed, stationary charge of value+250C. Determine the change in potential energy when the mobile charge is brought to a location 1.2maway from the fixed charge.

Answer 1

Given:

The charge is q1 = -4C

The charge is q2 = 250 C

The initial distance is r1 = 3.5 m

The final distance is r2 = 1.2 m

To find the change in potential energy.

Explanation:

The change in potential energy can be calculated by the formula

$\begin{gathered} \Delta U=\frac{kq1q2}{r2}-\frac{kq1q2}{r1} \\ =kq1q2(\frac{1}{r2}-\frac{1}{r1}) \end{gathered}$

Here, the constant is

$k=\text{ 9}\times10^9\text{ N m}^2\text{ /C}^2$

On substituting the values, the change in potential energy will be

$\begin{gathered} \Delta U=9\times10^9\times4\times250(\frac{1}{1.2}-\frac{1}{3.5}) \\ =4.92\times10^{12}\text{ J} \end{gathered}$