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Class 12 › Physics › Electric Charge and Field
CBSE2023Class Class 12 · Physics3 Marks · Short✅ Verified
Two charges of +25 × 10⁻⁹ C and -25 × 10⁻⁹ C are placed 6 m apart. Find the electric field at a point 4 m from the centre of the electric dipole (a) On axial line (b) On equatorial line.
✅ Answer & Solution
<b>Given:</b><br>• q = 25 × 10⁻⁹ C<br>• Distance between charges 2l = 6 m, so l = 3 m<br>• Distance of point from centre r = 4 m<br>• k = 9 × 10⁹ N·m²/C²<br><br><b>Step 1: Calculate dipole moment</b><br>p = q × 2l = (25 × 10⁻⁹) × 6<br><b>p = 1.5 × 10⁻⁷ C·m</b><br><br><b>(a) Field on Axial Line</b><br><br><b>Formula:</b><br>E<sub>axial</sub> = k·2pr/(r² - l²)²<br><br><b>Calculation:</b><br>E<sub>axial</sub> = (9 × 10⁹) × 2 × (1.5 × 10⁻⁷) × 4 / (16 - 9)²<br>E<sub>axial</sub> = (9 × 10⁹) × (12 × 10⁻⁷) / 49<br>E<sub>axial</sub> = 1.08 × 10⁴/49<br><b>E<sub>axial</sub> ≈ 220.4 N/C</b><br>Direction: along p (from -q to +q)<br><br><b>(b) Field on Equatorial Line</b><br><br><b>Formula:</b><br>E<sub>eq</sub> = kp/(r² + l²)<sup>3/2</sup><br><br><b>Calculation:</b><br>E<sub>eq</sub> = (9 × 10⁹) × (1.5 × 10⁻⁷) / (16 + 9)<sup>3/2</sup><br>E<sub>eq</sub> = 1350/(25)<sup>3/2</sup><br>E<sub>eq</sub> = 1350/125<br><b>E<sub>eq</sub> = 10.8 N/C</b><br>Direction: antiparallel to p<br><br><b>Note:</b> E<sub>axial</sub> ≈ 2 × E<sub>eq</sub> (consistent with dipole field properties)
