Question

Force of attraction between two point charges $Q$ and $- Q$ separated by $d\,metre$ is ${F_e}$ . When these charges are placed on two identical spheres of radius $R = 0.3d$ whose centres are $d\,metre$ apart, the force of attraction between them is:
A. Greater than ${F_e}$
B. Equal to ${F_e}$
C. Less than ${F_e}$
D. None of these

Answer 1

To solve this question, first we will calculate the actual distance after placing both the charges on two identical spheres. And, as we know Force is inversely proportional to the distance. If the distance will decrease then the force of attraction will increase and vice versa.

Complete step by step answer:
In the given question, the force of attraction is increasing. As the given; the distance between both the point charges is $d\,metre$ .And, the force of attraction between them is ${F_e}$.Now, when the given charges are placed on two identical spheres of radius $R = 0.3d$ , the centre of radius of both the spheres is the same as the linear earlier distance.
Now, the distance between both the sphere $= d - (0.3 + 0.3)d = d - 0.6d = 0.4d$
As we can see that the distance between both the charges $Q$ and $- Q$ is decreasing.
And, we know that Force is inversely proportional to the square of distance:
${F_e}\alpha \dfrac{1}{{{d^2}}}$
So, as we calculated, the distance is decreasing, then the force is increasing.

Hence, the correct option is A.

Note: The relationship between distance and force is that, when using a simple machine, to move an object it requires less force to move an object a farther distance when using a simple machine than moving an object without a simple machine.