HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q22 Solution: Two identical balls, each with a charge of and a mass of 100 g, …

QPages 121-124: Q22: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q22 Solution: Two identical balls, each with a charge of and a mass of 100 g, …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume the repulsive force acts entirely perpendicular to the string, setting the along‑string component to zero; however the line joining the balls makes an angle θ (sinθ = d/2l) with the horizontal, so the electric force must be resolved into parallel and perpendicular components, affecting both tension and acceleration.

Two identical balls, each with a charge of

2.00 \times 10^{-7} \text{ C}

and a mass of 100 g, are suspended from a common point by two insulating strings, each 50 cm long. The balls are held at a separation 5.0 cm apart and then released. Find a) the electric force on one of the charged balls b) the components of the resultant force on it along and perpendicular to the string c) the tension in the string d) the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release.

Step-by-Step Verified Solution

**Answer:** (a)

F_e = 0.144

N (b) Along string: 0, perpendicular: 0.144 N (c)

T = 1

N (d)

a = 1.44

m/s_. **Step-BY-STEP Explanation:** Given:

q = 2 \times 10^{-7}

m = 0.1

kg,

l = 0.5

d = 0.05

m. (a)

F_e = \frac{k q^2}{d^2} = \frac{9 \times 10^9 \times (2 \times 10^{-7})^2}{(0.05)^2} = 0.144

N. (b) The resultant force along the string is zero (mg balances T cos _). Perpendicular component =

F_e = 0.144

N. (c)

T\cos\theta = mg \Rightarrow T \approx mg = 0.1 \times 10 = 1

N. (d)

a = F_e/m = 0.144/0.1 = 1.44

m/s_.

QPages 121-124: Q22: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q22 Solution: Two identical balls, each with a charge of and a mass of 100 g, …

Step-by-Step Verified Solution

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