Question

A point charge \(q = 1\,\mu\text{C}\) is located at a distance \(2\,\text{cm}\) from one end of a thin insulating wire of length \(10\,\text{cm}\) having a charge \(Q = 24\,\mu\text{C}\), distributed uniformly along its length, as shown in the figure. Force between \(q\) and wire is ________ N. 

Hint:

For force due to a charged rod, integrate Coulomb’s law using linear charge density.

Answer 1

Correct Answer: 90

To find the force between the point charge \(q\) and the uniformly charged wire, we integrate the electrostatic force due to an element of charge \(dq\) on the wire. The linear charge density \(\lambda\) of the wire is given by:

\(\lambda = \frac{Q}{L} = \frac{24 \mu\text{C}}{10\,\text{cm}} = \frac{24 \times 10^{-6}\,\text{C}}{0.1\,\text{m}} = 240 \,\mu\text{C/m}\).

Consider a small charge element \(dq = \lambda \, dx\) at a distance \(x\) from the point on the wire closest to \(q\). The distance from this element to \(q\) is \(r = x + 2\,\text{cm} = x + 0.02\,\text{m}\).

The force \(dF\) due to \(dq\) on \(q\) is:

\(dF = \frac{k \cdot q \cdot dq}{r^2} = \frac{(9 \times 10^9 \,\text{N}\cdot\text{m}^2/\text{C}^2) \cdot (1 \times 10^{-6}\,\text{C}) \cdot (240 \times 10^{-6}\,\text{C/m} \cdot dx)}{(x + 0.02)^2}\).

Integrate \(dF\) from \(x=0\) to \(x=0.1\,\text{m}\):

\(F = \int_0^{0.1} \frac{9 \times 240 \times 10^{-12}}{(x + 0.02)^2} \, dx\).

This simplifies to:

\(F = 9 \times 240 \times 10^{-12} \int_0^{0.1} \frac{1}{(x + 0.02)^2} \, dx\).

Integrating: \(\int \frac{1}{(x + 0.02)^2}\,dx = -\frac{1}{x + 0.02}\).

So:

\(F = 9 \times 240 \times 10^{-12} \left[-\frac{1}{x + 0.02}\right]_0^{0.1} = 9 \times 240 \times 10^{-12} \left(-\frac{1}{0.1 + 0.02} + \frac{1}{0.02}\right)\).

Calculate:

\(-\frac{1}{0.12} + \frac{1}{0.02} = -8.33 + 50 = 41.67\).

\(F = 9 \times 240 \times 10^{-12} \times 41.67 = 9.004 \times 10^{-8}\, \text{N}\).

The force between \(q\) and the wire is approximately \(90.04 \, \text{N}\), which is within the range of 90, 90.