Verified Solution • 2025–26 Session moderate

QPages 121-124: Q49: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q49 Solution: A ball of mass 100 g and with a charge of is released from rest …

A ball of mass 100 g and with a charge of 4.9×10−5 C4.9 \times 10^{-5} \text{ C} is released from rest in a region where a horizontal electric field of 2.0×104 N C−12.0 \times 10^4 \text{ N C}^{-1} exists. a) Find the resultant force acting on the ball. b) What will be the path of the ball? c) Where will the ball be at the end of 2 s?

Step-by-Step Verified Solution

**Answer:** (a) F=2.2F = 2.2 N (b) Straight line (c) x=10x = 10 m, y=20y = 20 m. **Step-BY-STEP Explanation:** Given: m=0.1m = 0.1 kg, q=4.9×10−5q = 4.9 \times 10^{-5} C, E=2.0×104E = 2.0 \times 10^4 N/C. (a) Fe=qE=4.9×10−5×2×104=0.98F_e = qE = 4.9 \times 10^{-5} \times 2 \times 10^4 = 0.98 N. Fg=1F_g = 1 N.
F=0.982+12=1.4 NF = \sqrt{0.98^2 + 1^2} = 1.4 \text{ N}
(b) The path is a straight line (constant acceleration). (c) ax=Fe/m=9.8a_x = F_e/m = 9.8 m/s_, ay=g=10a_y = g = 10 m/s_. x=12×9.8×4=19.6x = \frac{1}{2} \times 9.8 \times 4 = 19.6 m, y=12×10×4=20y = \frac{1}{2} \times 10 \times 4 = 20 m .

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