Question

7. If the distance between two point charges is increased by 3%, then
calculate the % decrease in the force between them.
Ans. 6% approximately.​

Answer 1

Explanation:

So let the magnitude of the two point charges be Q_1 and Q_2 and the distance between them be R.

Now, Case 1-

Initial Force=K*Q_1*Q_2*(R)^(-2) —— (1)

K is Culoumb’s Constant.

Now, Case 2-

Final Distance= R+3*R/100= 1.03*R

Final Force= K*Q_1*Q_2*(10^4)/10609*R^2 —— (2)

Now, equation 1 - equation 2

∆F= K*Q_1*Q_2*609/10609*R^2

So, the percentage of decreased Force is= 609*Q_1*Q_2*R^2*k/10609*R^2*Q_1*Q_2*K*100= 5.74% (approx)

So, the decrease in Force is approximately 5.74%( closed to 6%)

hope this answer helped you