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 rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

      • 80 Wb
      • 16 Wb
      • \(8 \times 10^{-2}\) Wb
      • \(8 \times 10^{-3}\) Wb

    • 2.

      A parallel beam of monochromatic light falls normally on a single slit of width $a$ and a diffraction pattern is observed on a screen placed at distance $D$ from the slit. Explain:

      (I) the formation of maxima and minima in the diffraction pattern, and

      (II) why the maxima go on becoming weaker and weaker with its increasing number $n$.


        • 3.
          A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.


            • 4.

              Consider a cylindrical conductor of length $l$ and area of cross-section $A$. Current $I$ is maintained in the conductor and electrons drift with velocity $\vec{v}_d$ ($|\vec{v}_d| = \frac{e|\vec{E}|}{m}\tau$), (where symbols have their usual meanings). Show that the conductivity $\sigma$ of the material of the conductor is given by

              $$\sigma = \frac{ne^2}{m}\tau.$$


                • 5.
                  Consider the nuclear reaction \( X \to Y + Z \). Let \( M_x \), \( M_y \), and \( M_z \) be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

                    • \( (M_x - M_z)<M_y \)
                    • \( (M_x - M_y)<M_z \)
                    • \( M_x>(M_y + M_z) \)
                    • \( M_x<(M_y + M_z) \)

                  • 6.

                    A circular coil of 100 turns and radius $\left(\frac{10}{\sqrt{\pi}}\right)$ cm carrying current of $5.0\,\text{A}$ is suspended vertically in a uniform horizontal magnetic field of $2.0\,\text{T}$. The field makes an angle $30^\circ$ with the normal to the coil. Calculate:

                    (I) the magnetic dipole moment of the coil, and

                    (II) the magnitude of the counter torque that must be applied to prevent the coil from turning.

                      CBSE CLASS XII Previous Year Papers

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