The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become:

1.	7.20 N	2.	11.25 N
3.	22.50 N	4.	45.00 N

Two equally charged, identical metal spheres A and B repel each other with a force 'F'. The spheres are kept fixed with a distance 'r' between them. A third identical, but uncharged sphere C is brought in contact with A and then placed at the mid-point of the line joining A and B. The magnitude of the net electric force on C is 

(1) F

(2) 3F/4

(3) F/2

(4) F/4

The magnitude of electric field intensity E is such that, an electron placed in it would experience an electrical force equal to its weight is given by 

(1) mge

(2) $\frac{mg}{e}$

(3) $\frac{e}{mg}$

(4) $\frac{e^2}{m^2}g$

An uncharged sphere of metal is placed in between two charged plates as shown. The lines of force look like 

(1) A

(2) B

(3) C

(4) D

The figures below show regular hexagons, with charges at the vertices. In which of the following cases the electric field at the centre is not zero?

(1) 1

(2) 2

(3) 3

(4) 4

An electric charge q is placed at the centre of a cube of side a. The electric flux on one of its faces will be:

(1) $\frac{q}{6{\epsilon }_0}$

(2) $\frac{q}{{\epsilon }_0{a}^2}$

(3) $\frac{q}{4\pi {\epsilon }_0{a}^2}$

(4) $\frac{q}{{\epsilon }_0}$ 

Total electric flux coming out of a unit positive charge put in air is 

(1) ${\epsilon }_0$

(2) ${\epsilon }_0^{-1}$

(3) ${\left(4p{\epsilon }_0\right)}^{-1}$

(4) $4\pi {\epsilon }_0$ 

A cube of side l is placed in a uniform field E, where $E=E\hat{i}$. The net electric flux through the cube is

(1) Zero

(2) l²E

(3) 4l²E

(4) 6l²E

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is 

(1) $3q/{\epsilon }_0$

(2) $2q/{\epsilon }_0$

(3) $q/{\epsilon }_0$ 

(4) Zero

Consider the charge configuration and spherical Gaussian surface as shown in the figure. While calculating the flux of the electric field over the spherical surface, the electric field will be due to: 

(1) q₂ only

(2) Only the positive charges

(3) All the charges

(4) +q₁ and – q₁ only