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.

Step-by-step explanation:

The new distance is

`\begin{gathered} r^(\prime)=\text{ 2r} \\ =2*1 \\ =2\text{ }m \end{gathered}`

The force can be calculated by the formula

`F=k(q1q2)/((r^(\prime))^2)`

Here, k is the constant whose value is

`k=9*10^9\text{ N m}^2\text{ /C}^2`

On substituting the values, the force will be

`\begin{gathered} F=9*10^9*(1*1)/((2)^2) \\ =2.25*10^9\text{ N} \end{gathered}`