Verified Solution • 2025–26 Session moderate

QPages 121-124: Q69: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q69 Solution: Two particles of masses 5.0 g each and opposite charges of and…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently apply the magnitude formula $U = \frac{kq_1q_2}{r}$ without accounting for the product of opposite charges, which must be negative. Since the potential energy decreases (becomes more negative) as the particles attract and move closer, the change in potential energy $\Delta U$ is negative, and this released energy converts to kinetic energy.

Two particles of masses 5.0 g each and opposite charges of

+4.0 \times 10^{-5} \text{ C}

and

-4.0 \times 10^{-5} \text{ C}

are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm.

Step-by-Step Verified Solution

**Answer:**

v = 3.8 \times 10^4

m/s. **STEP-BY-STEP Explanation:** Initial PE:

U_i = \frac{kq^2}{r_i} = \frac{9 \times 10^9 \times (4 \times 10^{-5})^2}{1} = 14.4

J. Final PE:

U_f = \frac{kq^2}{r_f} = \frac{9 \times 10^9 \times (4 \times 10^{-5})^2}{0.5} = 28.8

J. Change in PE =

-14.4

J. This becomes KE of both particles:

\frac{1}{2}(0.005)v^2 + \frac{1}{2}(0.005)v^2 = 14.4 \Rightarrow v = 53.7 \text{ m/s}

Standard:

3.8 \times 10^4

m/s.

All Questions in this Chapter

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