Question

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge of Object 1 is one-third the original value AND the charge of object 2 is doubled AND the distance then the new electrostatic force will be _____ unit

Answer 1

Answer:

F'=(8/3)F

Explanation:

to find the change in the force you take into account that the electric force is given by:

$F=k\frac{q_1q_2}{(18.0u)^2}=k\frac{q_1q_2}{324.0}u^2$

However, if q1'=1/3*q, q2'=2*q2 and the distance is halved, that is 18/=9.o unit:

$F'=k\frac{q_1'q_2'}{(9u)^2}=k\frac{(1/3)q_1(2)q_2}{81.0u^2}=\frac{2}{3}k\frac{q_1q_2}{81.0u^2}$

if you multiply this result by 4 and divide by 4 you get:

$F'=\frac{8}{3}k\frac{q_1q_2}{324.0u^2}=\frac{8}{3}F$

hence, the new force is 8/3 of the previous force F.