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

Step 1: Find linear charge density of the wire.
\[ \lambda = \frac{Q}{L} = \frac{24 \times 10^{-6}}{0.10} = 2.4 \times 10^{-4}\,\text{C/m} \]
Step 2: Consider an element of wire.
Let an element \(dx\) be at a distance \(x\) from the point charge.
Charge on element, \[ dq = \lambda dx \]
Step 3: Write expression for force due to element.
\[ dF = \frac{1}{4\pi\varepsilon_0} \frac{q\,dq}{x^2} = 9 \times 10^9 \frac{q\lambda}{x^2} dx \]
Step 4: Set limits of integration.
Nearest end is at \(x = 0.02\,\text{m}\), far end at \(x = 0.12\,\text{m}\).
Step 5: Integrate to find total force.
\[ F = 9 \times 10^9 \, q \lambda \int_{0.02}^{0.12} \frac{dx}{x^2} \] \[ F = 9 \times 10^9 \times (1 \times 10^{-6}) \times (2.4 \times 10^{-4}) \left[ -\frac{1}{x} \right]_{0.02}^{0.12} \] \[ F = 9 \times 10^9 \times 2.4 \times 10^{-10} \left( \frac{1}{0.02} - \frac{1}{0.12} \right) \] \[ F = 2.16 \times (50 - 8.33) \approx 90\,\text{N} \]