Given:
The charge of X is
[tex]q_x=\text{ 7.5}\times10^{-9}\text{ C}[/tex]
The charge of Y is
[tex]q_y=-3.5\times10^{-9}\text{ C}[/tex]
The distance between the charges is r = 80 cm.
To find whether there is an increase or decrease in the force of attraction/repulsion if the new charge of y is
[tex]q_y^{\prime}=-6.5\text{ }\times10^{-9}\text{ C}[/tex]
and the distance between the charges is the same.
Explanation:
The charges q_x and q_y will have the force of attraction as they have opposite charges.
The magnitude of the force is given by the formula
[tex]F=k\frac{|q_x||q_y|}{r^2}[/tex]
Here, the k is the electrostatic constant.
The distance between the charges is the same and the charge on X is also the same.
Thus, the force
[tex]F\propto\text{ \mid q}_y|[/tex]
The new charge has a magnitude more than the old charge on Y.
Thus, the force of attraction will increase due to the increase in the magnitude of the charge.
Final Answer: The electrical force of attraction will increase.
