Question

. Two charges, Q1 and Q2 , are separated by a certain distance R. If the magnitudes of the charges are halved, and their separation is also halved, then what happens to the electrical force between these charges

Answer 1

Answer: Magnitude of electrical force stays the same.

Explanation:

Equation:

$F_{e} =k\frac{Q_{1}Qx_{2} }{r^{2} }$

Since the magnitude of each charge is halved.

and

the separation is halved.

$F_{e} =k\frac{(.5Q_{1}*.5Q_{2} }{(.5r)^{2} }$

$F_{e} =k\frac{.25*Q_{1}Qx_{2} }{.25*r^{2} }$

Cancel out .25 on the numerator and denominator. Leaving the original equation.