An electric dipole of dipole moment \( \vec{p} = (0.8\,\hat{i} + 0.6\,\hat{j}) \times 10^{-29}\,\text{Cm} \) is placed in an electric field \( \vec{E} = 1.0 \times 10^7\,\hat{k}\,\text{V/m} \). Calculate the magnitude of the torque acting on it and the angle it makes with the x-axis, at this instant.
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Solution and Explanation
Step 1: Torque on an Electric Dipole in Electric Field
Torque on an electric dipole in an electric field is given by: \[ \vec{\tau} = \vec{p} \times \vec{E} \] Given: \[ \vec{p} = (0.8\,\hat{i} + 0.6\,\hat{j}) \times 10^{-29} \text{ Cm}, \quad \vec{E} = 1.0 \times 10^7\, \hat{k} \text{ V/m} \] Compute the cross product \( \vec{p} \times \vec{E} \): \[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.8 \times 10^{-29} & 0.6 \times 10^{-29} & 0 \\ 0 & 0 & 1.0 \times 10^7 \\ \end{vmatrix} \] \[ = \hat{i}(0.6 \times 10^{-29} \cdot 0 - 0 \cdot 10^7) - \hat{j}(0.8 \times 10^{-29} \cdot 0 - 0 \cdot 10^7) + \hat{k}(0.8 \times 10^{-29} \cdot 0 - 0.6 \times 10^{-29} \cdot 0) \] \[ = \hat{i}(0) - \hat{j}(0) + \hat{k}[(0.8 \times 10^{-29})(0) - (0.6 \times 10^{-29})(0)] = \hat{i}(-0.6 \times 10^{-22}) - \hat{j}(0.8 \times 10^{-22}) \] \[ \Rightarrow \vec{\tau} = (-0.6\,\hat{i} - 0.8\,\hat{j}) \times 10^{-22} \text{ Nm} \] Step 2: Magnitude of Torque \[ |\vec{\tau}| = \sqrt{(-0.6)^2 + (-0.8)^2} \times 10^{-22} = \sqrt{0.36 + 0.64} \times 10^{-22} = \sqrt{1} \times 10^{-22} = 1.0 \times 10^{-22} \text{ Nm} \] Step 3: Angle of Dipole with X-axis Let angle \( \theta \) be the angle of \( \vec{p} \) with x-axis: \[ \tan \theta = \frac{0.6}{0.8} = 0.75 \Rightarrow \theta = \tan^{-1}(0.75) \approx 36.87^\circ \]
Answer: - Magnitude of torque: \( 1.0 \times 10^{-22} \, \text{Nm} \)
- Angle made by the dipole with the x-axis: \( \theta \approx 36.87^\circ \)
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