Two charges + q and -q are placed at a distance a in a uniform electric field. The dipole moment of the combination is 2qa cos θ i^ + sin θ j^, where, θ is the angle between the direction of the field and the line joining the two charges. Which of the following statement(s) is/are correct ?

  • The torque exerted by the field on the dipole vanishes

  • The net force on the dipole vanishes

  • The torque is independent of the choice of coordinates

  • The net force is independent of a


B.

The net force on the dipole vanishes

C.

The torque is independent of the choice of coordinates

D.

The net force is independent of a

The situation can be diagrammatically as,

             

The dipole moment,

    P = 2qa (cos θ i^ + sin θ j^)

The net force on the dipole is always zero while net torque on the dipole is not zero.


Advertisement

Two charges of magnitudes – 2Q and + Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?        


Using gauss law, we have




 

Electric flux is given by, 

4786 Views

Advertisement

A charge q is placed at one corner of a cube. The electric flux through any of the three faces adjacent to the charge is zero. The flux through any one of the other three faces is

  • q/3ε0

  • q/6ε0

  • q/12ε0

  • q/24ε0


D.

q/24ε0

Consider the diagram

           

In above the flux coverage of three face is shown in figure which is like a quadrant in any plane. So, flux will be

      ϕ = 14 × q6ε0 = q24ε0


What is the electric flux through a cube of side 1 cm which encloses an electric dipole? 

Given, a cube encloses an electric dipole.

So, the total charge enclosed by the cube is zero.

That is, Q= 0.

Now, using the formula for electric flux as per Gauss’s law of electrostatics, we have

2873 Views

For any charge configuration, equipotential surface through a point is normal to the electric field. Justify. 


Work done (W) in moving a test charge along an equipotential surface is zero.

Work done is given by, 


F is the electric force and s is the magnitude of displacement.

For non-zero displacement, this is possible only when cos   is equal to 0.
 
         = 90° 

Thus, the force acting on the point charge is perpendicular to the equipotential surface.

Electric field lines give us the direction of the electric force on a charge.

Thus, for any charge configuration, an equipotential surface through a point is normal to the electric field.

2675 Views

Advertisement