Q56P - Fundamentals Of Physics

Question

An electric dipole consists of charges +2e and -2e separated by 0.78 nm. It is in an electric field of strength$3.4\times {10}^6 N/C$. Calculate the magnitude of the torque on the dipole when the dipole moment is (a) parallel to, (b) perpendicular to, and (c) antiparallel to the electric field.

Step-by-Step Solution

Answer

• a) The magnitude of the torque on the dipole when the dipole moment is parallel to the electric field is 0.
• b) The magnitude of the torque on the dipole when the dipole moment is perpendicular to the electric field is$8.5\times {10}^{-22}N.m$.
• c) The magnitude of the torque on the dipole when the dipole moment is anti-parallel to the electric field is 0.

Step 1: The given data

• a) Charges of values, +2e and-2e.
• b) Separation between the charges, r =0.78 nm.
• c) Magnitude of the electric field, $E= 3.4\times {10}^6 N/C$

Step 2: Understanding the concept of torque

Using the concept of the torque acting on a dipole, the torque for the given cases is given by considering the given situations of the angle made by the dipole to the electric field.

 

Formula:

The magnitude of torque $\left(\zeta \right)$in terms of the magnitudes of the electric field E and the dipole moment  p = qd is given by:

$\zeta =pE \sin \theta$   

Step 3: a) Calculation of the torque when dipole is parallel

From equation (i), we can see that the torque for $\theta =0°$ is given as:

$$\begin{gathered} \zeta =pE \sin 0° \\ =0. \end{gathered}$$

Hence, the value of the torque is 0.

Step 4: b) Calculation of the torque when dipole is perpendicular

From equation (i), we can see that the torque for $\theta =90°$is given as:

$$\begin{gathered} \zeta =pE \\ =\left(2.16\times {10}^{-19}C\right)\left(0.78\times {10}^{-9}m\right)\left(3.4\times {10}^{6}N/C\right) \\ =8.5\times {10}^{-22}N.m \end{gathered}$$

Hence, the value of the torque is $8.5\times {10}^{-22}N.m$.

Step 5: c) Calculation of the torque when dipole is anti-parallel

From equation (i), we can see that the torque for $\theta =180°$ is given as:

$$\begin{gathered} \zeta =pE \sin 180° \\ =0. \end{gathered}$$

Hence, the value of the torque is 0.