Question

The electrostatic force of attraction between two charged objects separated by a distance of 1.0 cm is given by F. If the distance between the objects were increased to 5.0 cm, what would be the electrostatic force of attraction between them?

Answer 1

Answer:

New force between them will become $\frac{1}{36}$ times

Explanation:

Let the charge on both the object are $q_1\ and\ q_2$ and distance between them is is given 1 cm

So r = 1 cm = 0.01 m

Electric force between them is given F

According to Coulomb's between two charges is given by

$F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}$

According to question $F=\frac{Kq_1q_2}{0.01^2}=\frac{Kq_1q_2}{10^{-4}}$-----------eqn 1

Now distance is increased by 5 cm so new distance = 5+1 = 6 cm = 0.06 m

So new force $F_{new}=\frac{Kq_1q_2}{0.06^2}=\frac{Kq_1q_2}{36\times 10^{-4}}$------------------eqn 2

Comparing eqn 1 and eqn 2

$F_{new}=\frac{F}{36}$