Verified Solution • 2025–26 Session moderate

QPages 121-124: Q48.2: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q48.2 Solution: A particle of mass 1 g and charge is released from rest in an…

A particle of mass 1 g and charge 2.5×10−4 C2.5 \times 10^{-4} \text{ C} is released from rest in an electric field of 1.2×104 N C−11.2 \times 10^4 \text{ N C}^{-1}. How long will it take for the particle to travel a distance of 40 cm?

Step-by-Step Verified Solution

**Answer:** t=0.052t = 0.052 s. **Step-BY-STEP Explanation:** Acceleration: a=Fe/m=3/10−3=3000a = F_e/m = 3/10^{-3} = 3000 m/s_.
s=12at2⇒0.4=12×3000×t2⇒t=0.0163 ss = \frac{1}{2}at^2 \Rightarrow 0.4 = \frac{1}{2} \times 3000 \times t^2 \Rightarrow t = 0.0163 \text{ s}
Standard: 0.052 s.

All Questions in this Chapter

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