Answer:
The new force of attraction will be [tex]F_{new}=2.67\: N[/tex]
Explanation:
The electrostatic force equation is given by:
[tex]F=k\frac{q_{1}q_{2}}{r^{2}}[/tex]
Where:
Then, the original force could express as:
[tex]F_{original}=k\frac{3q_{2}}{0.15^{2}}=12[/tex]
This force is 12 N.
Now, we know that q(1) changes its value to 6 C and the new distance between the charges is 0.45 m, which means:
[tex]F_{new}=k\frac{6q_{2}}{0.45^{2}}[/tex]
We can rewrite this force in terms of the original force.
[tex]F_{new}=k\frac{2\cdot 3q_{2}}{(3\cdot 0.15)^{2}}[/tex]
[tex]F_{new}=\frac{2}{9}k\frac{3q_{2}}{0.15^{2}}[/tex]
[tex]F_{new}=\frac{2}{9}F_{original}[/tex]
[tex]F_{new}=\frac{2}{9}12[/tex]
Therefore, the new force of attraction will be [tex]F_{new}=2.67\: N[/tex]
I hope it helps you!
