91影视

• Value of point charges is 5.00 N

Coulomb stated that when two-point charges are separated by some distance, they experience some force of attraction or repulsion. This force of attraction or repulsion is known as electrostatic force.

Initially, when the charges q and Q are separated by some distance r the electrostatic force between them is,

$F=\frac{KqQ}{r^2}..........(1.1)$

Here, F is the electrostatic force (F = 5.00 N) , and K is the electrostatic force constant.

When the separation between the charges increased by a factor of three $\left(r^{\mathbf{\text{'}}}=3r\right)$. The new electrostatic force (F^') between the point charges is,

$$\begin{gathered} F^{\text{'}}=\frac{KqQ}{r^{\text{'}2}} \\ =\frac{KqQ}{{\left(3r\right)}^2} \\ =\frac{KqQ}{9r^2} \end{gathered}$$

Using equation (1.1),

$F^{\text{'}}=\frac{F}{9}$

Substituting 5.00 N for F,

$$\begin{gathered} F^{\text{'}}=\frac{5.00N}{9} \\ =0.556N \end{gathered}$$

Hence, when the separation becomes three times of the original, the new force will be 0.556 N.