Answer:
The magnitude of electric force on object B due to object A when they are 2r distance apart is F/4.
Explanation:
Given :
Electric charge on object A = +Q
Electric charge on object B = +4Q
When the objects A and B are r distance apart, the magnitude of electrostatic force exerted on B due to A is given by the relation :
[tex]F=\frac{1}{4\pi\epsilon_{0} } \frac{Q\times4Q}{r^{2} }[/tex] .....(1)
Now, the two objects are apart by distance of 2r, hence the force acting on the object B due to A is :
[tex]F_{1} =\frac{1}{4\pi\epsilon_{0} } \frac{Q\times4Q}{(2r)^{2} }[/tex]
[tex]F_{1} =\frac{1}{4\pi\epsilon_{0} } \frac{Q\times4Q}{4r^{2} }[/tex]
Substitute equation (1) in the above equation.
[tex]F_{1} =\frac{F}{4}[/tex]