Question

Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between them is doubled, what does the force become?

Answer 1

Given:

The magnitude of each charge is q1 = q2 = 1 C

The distance between them is r = 1 m

To find the force when distance is doubled.

Explanation:

The new distance is

$\begin{gathered} r^{\prime}=\text{ 2r} \\ =2\times1 \\ =2\text{ }m \end{gathered}$

The force can be calculated by the formula

$F=k\frac{q1q2}{(r^{\prime})^2}$

Here, k is the constant whose value is

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

On substituting the values, the force will be

$\begin{gathered} F=9\times10^9\times\frac{1\times1}{(2)^2} \\ =2.25\times10^9\text{ N} \end{gathered}$