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.

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A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

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

CBSE CLASS XII Related Questions

3.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

4.
In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

6.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

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

CBSE CLASS XII Related Questions

3.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

4.In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

6.A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

3.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

4.
In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

6.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.