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QPages 121-124: Q31: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q31 Solution: Two particles A and B, each with a charge Q, are placed a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently maximize the magnitude of the individual electric field vectors ($E = kQ/r^2$) instead of the net field component along the bisector, forgetting that the transverse components cancel. You must differentiate the projected field $E_{net} = \frac{2kQx}{(x^2 + d^2/4)^{3/2}}$ with respect to $x$ to find the true maximum at $x = d/2\sqrt{2}$.

Two particles A and B, each with a charge Q, are placed a distance d apart. Where should a particle of charge q be placed on the perpendicular bisector of AB, so that it experiences maximum force? What is the magnitude of this maximum force?

Step-by-Step Verified Solution

**Answer:**

q

should be placed at distance

d/2\sqrt{2}

from the midpoint, and

F_{max} = \frac{4kQq}{3\sqrt{3}d^2}

. **Step-BY-STEP Explanation:** For maximum force on

q

placed on the perpendicular bisector, the distance

x

from the midpoint satisfies

x = d/2\sqrt{2}

. The maximum force is

F_{max} = \frac{4kQq}{3\sqrt{3}d^2}

.

All Questions in this Chapter

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