Answer:
a. [tex]1.019 \times 10^{-7} N[/tex]
b. Repulsive force
Explanation:
The computation is shown below:
a. The magnitude of the force is
Given that
Charge = 7.52nc
Location = 1.65m
Point charge = 4.10nC
Based on the above information,
As we know that
[tex]F= \frac{kqQ}{r^2} \\\\= \frac{9\times10^{9} \times 7.52 \times 10^{-9} \times 4.1 \times 10^{-9}}{1.652} \\\\= 1.019 \times 10^{-7} N[/tex]
b. Now as it can be seen that both contains the positive charges so this represents the repulsive force and the same is to be considered
