A particle having charge ${\mathrm{q}}_1$ exerts F electrostatic force on charge ${\mathrm{q}}_2$ at rest. If a particle having charge $\frac{q_1}{4}$ is placed midway between the line joining the two charges ${\mathrm{q}}_1<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{q}}_2$ then electrostatic force on ${\mathrm{q}}_2$ due to ${\mathrm{q}}_1$ will become/remain

1. 2F

2. $\frac{F}{2}$

3. F

4. zero

A charge q is to be divided on two small conducting spheres. What should be the value of charges on the spheres so that when placed at a certain distance apart, the repulsive force between them is maximum?

1. $\frac{q}{4}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>\frac{3q}{4}$

2. $\frac{q}{2}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>\frac{q}{2}$

3. $\frac{q}{3}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>\frac{q}{3}$

4. $\frac{q}{4}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>\frac{q}{4}$

The law, governing the force between electric charges is known as 

(1) Ampere's law

(2) Ohm's law

(3) Faraday's law

(4) Coulomb's law

F_g and F_e represents gravitational and electrostatic force respectively between electrons situated at a distance 10 cm. The ratio of F_g/ F_e is of the order of 

(1) 10⁴²

(2) 10

(3) 1

(4) 10^–43

Four charges are arranged at the corners of a square \(ABCD,\) as shown in the adjoining figure. The force on the positive charge \(Q\) kept at the centre \(O\) is: