Given:
The charge of X is
![q_x=\text{ 7.5}*10^(-9)\text{ C}](https://img.qammunity.org/2023/formulas/physics/college/4g09kvoujqhco3fnf114nwdgjlywwt83ip.png)
The charge of Y is
![q_y=-3.5*10^(-9)\text{ C}](https://img.qammunity.org/2023/formulas/physics/college/rim7x983k9ojlbkmry60b4ccvisvhdq3rv.png)
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
![q_y^(\prime)=-6.5\text{ }*10^(-9)\text{ C}](https://img.qammunity.org/2023/formulas/physics/college/72gvpd7mysi67mj90syk4u7pwlt50zpyf3.png)
and the distance between the charges is the same.
Step-by-step 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
![F=k(|q_x||q_y|)/(r^2)](https://img.qammunity.org/2023/formulas/physics/college/nts1sjmtqis1b0lliefpozl4gy7jm0ra6m.png)
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
![F\propto\text{ \mid q}_y|](https://img.qammunity.org/2023/formulas/physics/college/ga3fs0pumdo052kt7jb957hpc0508vwmyl.png)
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.