Question

An infinite number of point charges, each carrying 1 µC charge, are placed along the y-axis at y = 1 m, 2 m, 4 m, 8 m ............ . The total force on a 1 C point charge, placed at the origin, is x \(\times 10^3\) N. The value of x, to the nearest integer, is ________. [Take \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\) Nm²/C²]

Hint:

For infinite charge distributions at doubling distances, the force follows an inverse square GP sum.

Answer 1

Correct Answer: 12

Step 1: Use the principle of superposition. The total force \(F\) is the sum of forces from each charge \(q\). \[ F = \frac{k Q q}{r_1^2} + \frac{k Q q}{r_2^2} + ....... = k Q q \left[ \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} ....... \right] \]
Step 2: Identify the infinite geometric progression (GP). The series is \(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} .......\) First term (\(a\)) = 1, Common ratio (\(r\)) = \(1/4\). Sum \(S_\infty = \frac{a}{1-r} = \frac{1}{1 - 1/4} = \frac{4}{3}\).
Step 3: Calculate the final value. \[ F = (9 \times 10^9) \times (1) \times (1 \times 10^{-6}) \times \frac{4}{3} \] \[ F = 3 \times 10^3 \times 4 = 12 \times 10^3 \text{ N} \] Comparing with \(x \times 10^3\), we get \(x = 12\).