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QPages 121-124: Q71.3: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q71.3 Solution: Two particles A and B, of opposite charges and , are placed at…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the electric field of a dipole with that of a point charge, using an inverse square law ($1/r^2$) instead of the correct inverse cube dependence ($1/r^3$) for the equatorial line. Always verify the power of the distance term in the dipole field formula $E = \frac{kp}{r^3}$ to avoid order-of-magnitude errors.

Two particles A and B, of opposite charges 2.0×10−6 C2.0 \times 10^{-6} \text{ C} and −2.0×10−6 C-2.0 \times 10^{-6} \text{ C}, are placed at a separation of 1.0 cm. Calculate the electric field at a point on the perpendicular bisector of the dipole and 1.0 m away from the centre.

Step-by-Step Verified Solution

**Answer:** E=180E = 180 N/C. **STEP-BY-STEP Explanation:** For an equatorial point at distance x=1x = 1 m (with x≫l/2x \gg l/2):
E=kpx3=9×109×2.0×10−813=180 N/CE = \frac{kp}{x^3} = \frac{9 \times 10^9 \times 2.0 \times 10^{-8}}{1^3} = 180 \text{ N/C}

All Questions in this Chapter

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