Question

An attracting force of 5N is acting between two charges of $ +2\mu $ C and $ -2\mu $ C placed at some distance apart. If the charges are naturally touched and placed again at the same distance. What will be the force between them?

Answer 1

Hint: This question can be solved by using coulomb’s Law. According to this, the force of interaction between any two points charges is directly proportional to the product of charge and the inversely proportional to square of distance between them.
i.e.
 $ F=K\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} $
Here, K is electrostatic constant, r is the distance between the point charges

Complete Step By Step Solution
When two charges q1= $ 2\mu $ C and q2 $ =-2\mu $ C are at the distance r, then the force of attraction is 5N.

When both the charges are naturally touched, then charge will flow from Negative charged sphere to positively charged sphere charge will distribute equally.
 After touching, charge on each sphere
 $ =\dfrac{-2\mu C+2\mu C}{2} $
 $ =\dfrac{0}{2}=0 $
Then $ q_1 $ = $ q_2 $ = 0
Hence, force of attraction is given by
 $ F=K\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} $
= K(0)
F = 0
So, Net force is zero.

Note
1) Coulomb’s Law has been verified over distance ranging from atomic dimension to macroscopic distance
2) The charging by conduction process involves touching of a charged sphere to a conductive material. This way, the charges are transferred from charged material to the conductor. This method is useful for charging conductors
3) The induction charging is a charging method that charges an object without actually touching the object to any other object.