MP Board · Class 12 · Physics · Electric Charges and FieldsQuestion 3:\nSolve the following numerical problems based on electrostatics: (a) Two point charges $q1 = +4\,\mu\text{C}$ and $q2 = -4\,\mu\text{C}$ are placed $10\,\text{cm}$ apart in vacuum, forming an electric dipole. (i) Calculate the electric dipole moment of the system. (ii) Calculate the exact magnitude of the electric field intensity at a point situated on its axial line at a distance of $20\,\text{cm}$ from the center of the dipole. (b) The same dipole is now placed in a uniform electric field of $2.5 \times 10^5\,\text{N/C}$ such that its dipole moment vector makes an angle of $30^\circ$ with the field lines. (i) Calculate the magnitude of the torque acting on the dipole. (ii) Calculate the work done in rotating the dipole from $\theta1 = 0^\circ$ to $\theta2 = 180^\circ$ in this field.

Step-by-Step Solution

Part (a): Dipole Moment and Axial Electric Field

1. Given Data:

• Charge magnitude, $q = 4,\mu\text{C} = 4 \times 10^{-6},\text{C}$
• Distance between charges (dipole length), $2l = 10,\text{cm} = 0.10,\text{m} \implies l = 0.05,\text{m}$
• Distance of point from dipole center, $r = 20,\text{cm} = 0.20,\text{m}$
• Permittivity constant, $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9,\text{N}\cdot\text{m}^2/\text{C}^2$

2. (i) Calculation of Electric Dipole Moment ($p$): $$\text{Formula: } p = q \times 2l$$ $$p = (4 \times 10^{-6},\text{C}) \times (0.10,\text{m})$$ $$\mathbf{p = 4 \times 10^{-7},\text{C}\cdot\text{m}}$$

3. (ii) Calculation of Electric Field on Axial Line ($E_{\text{axial}}$): $$\text{Formula: } E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2 p r}{(r^2 - l^2)^2}$$

• Intermediate Calculations:

  • $r^2 = (0.20)^2 = 0.04,\text{m}^2$
  • $l^2 = (0.05)^2 = 0.0025,\text{m}^2$
  • $(r^2 - l^2) = 0.04 - 0.0025 = 0.0375,\text{m}^2$
  • $(r^2 - l^2)^2 = (0.0375)^2 = 0.00140625,\text{m}^4 = 1.40625 \times 10^{-3},\text{m}^4$

• Numerator Value: $$\text{Numerator} = 9 \times 10^9 \times 2 \times (4 \times 10^{-7}) \times 0.20$$ $$\text{Numerator} = 9 \times 10^9 \times 1.6 \times 10^{-7} = 1440,\text{N}\cdot\text{m}^2/\text{C}$$

• Substituting values into $E_{\text{axial}}$: $$E_{\text{axial}} = \frac{1440}{0.00140625} = 1,024,000,\text{N/C}$$ $$\mathbf{E_{\text{axial}} = 1.024 \times 10^6,\text{N/C}}$$

Part (b): Torque and Work Done in External Electric Field

1. Given Data:

• Electric Field, $E = 2.5 \times 10^5,\text{N/C}$
• Dipole moment, $p = 4 \times 10^{-7},\text{C}\cdot\text{m}$
• Angle, $\theta = 30^\circ$

2. (i) Torque Calculation ($\tau$): $$\text{Formula: } \tau = p E \sin\theta$$ $$\tau = (4 \times 10^{-7},\text{C}\cdot\text{m}) \times (2.5 \times 10^5,\text{N/C}) \times \sin(30^\circ)$$ $$\text{Since } \sin(30^\circ) = 0.5:$$ $$\tau = (10^{-1}) \times 0.5 = 0.05,\text{N}\cdot\text{m}$$ $$\mathbf{\tau = 5 \times 10^{-2},\text{N}\cdot\text{m}}$$

3. (ii) Work Done Calculation ($W$): $$\text{Formula: } W = pE (\cos\theta_1 - \cos\theta_2)$$

• Here, $\theta_1 = 0^\circ \implies \cos(0^\circ) = 1$
• $\theta_2 = 180^\circ \implies \cos(180^\circ) = -1$

$$W = pE [1 - (-1)] = 2 p E$$ $$W = 2 \times (4 \times 10^{-7}) \times (2.5 \times 10^5)$$ $$W = 2 \times 10^{-1} = 0.2,\text{J}$$ $$\mathbf{W = 0.2,\text{J}}$$