An electric dipole is placed in a uniform electric field. What will be the net electric force on the dipole?

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Ques: An electric dipole is placed in a uniform electric field. What will be the net electric force on the dipole?

Ans: The net force on electric dipole when it is placed in a uniform electric field is “zero”. 

Explanation: When an electric dipole is placed in a uniform electric field, the electric field exerts a force on each charge of the dipole. Since the charges in a dipole are of equal magnitude but opposite in sign, the force acting on them also becomes equal in magnitude but opposite in direction.

electric dipole placed in a uniform electric field

Electric dipole placed in a uniform electric field

The force on the positive charge of the dipole is in the direction of the electric field, and the force on the negative charge is in the opposite direction.

  • As a result, the two forces create a torque on the dipole, which tends to rotate it.
  • The force on +q is directed to the right and on −q is to the left.
  • The magnitude of forces are equal as the electric field is uniform. Hence, the net force is always zero.

However, the net electric force on the dipole is always zero because the electric field is uniform and the two forces on the charges are equal and opposite in direction. This means that the electric force on the dipole's center of mass is zero, and the dipole will not experience any translational acceleration in the electric field.

Therefore, the electric dipole experiences only a torque in a uniform electric field, but the net force is always zero.

CBSE CLASS XII Related Questions

  • 1.
    A point source is placed at the bottom of a tank containing a transparent liquid (refractive index \( n \)) to a depth H. The area of the surface of the liquid through which light from the source can emerge out is:

      • \( \frac{\pi H^2}{(n-1)} \)
      • \( \frac{\pi H^2}{(n^2-1)} \)
      • \( \frac{\pi H^2}{\sqrt{n^2-1}} \)
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    • 2.
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        • 0.31 V
        • 0.20 V
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      • 3.
        Two point charges of \( -5\,\mu C \) and \( 2\,\mu C \) are located in free space at \( (-4\,\text{cm}, 0) \) and \( (6\,\text{cm}, 0) \) respectively.
        (a) Calculate the amount of work done to separate the two charges at infinite distance.
        (b) If this system of charges was initially kept in an electric field \[ \vec{E} = \frac{A}{r^2}, \text{ where } A = 8 \times 10^4\, \text{N}\,\text{C}^{-1}\,\text{m}^2, \] calculate the electrostatic potential energy of the system.


          • 4.
            A current of 5 A is passing along +X direction through a wire lying along X-axis. Find the magnetic field \( \vec{B} \) at a point \( \vec{r} = (3 \hat{i} + 4 \hat{j}) \, \text{m} \) due to 1 cm element of the wire, centered at the origin.


              • 5.
                What is the source of force acting on a current-carrying conductor placed in a magnetic field? Obtain the expression for the force acting between two long straight parallel conductors carrying steady currents and hence define Ampère’s law.


                  • 6.
                    A coil has 100 turns, each of area \( 0.05 \, \text{m}^2 \) and total resistance \( 1.5 \, \Omega \). It is inserted at an instant in a magnetic field of \( 90 \, \text{mT} \), with its axis parallel to the field. The charge induced in the coil at that instant is:

                      • \( 3.0 \, \text{mC} \)
                      • \( 0.30 \, \text{C} \)
                      • \( 0.45 \, \text{C} \)
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                    CBSE CLASS XII Previous Year Papers

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