Question

Two long straight wires 1 and 2 are kept as shown in the figure. The linear charge density of the two wires are \(\lambda\) 1 = 10 C/m and \(\lambda\)2 = -20 C/m. Find the net force F experienced by an electron held at point P:

Hint:

The electric potential due to a dipole follows an inverse square law, whereas for a charged infinite plane, the electric field remains uniform and independent of distance.

Answer 1

The electric field due to a long straight wire is given by:
\[ E = \frac{\lambda}{2\pi\epsilon_0 r} \] For the two wires: \[ E_1 = \frac{\lambda_1}{2\pi\epsilon_0 r_1}, \quad E_2 = \frac{\lambda_2}{2\pi\epsilon_0 r_2} \] Substituting the given values: \[ E_1 = \frac{10 \times 10^{-6}}{2\pi\epsilon_0 (10 \times 10^{-2})} (-\hat{j}) \] \[ E_2 = \frac{20 \times 10^{-6}}{2\pi\epsilon_0 (20 \times 10^{-2})} (-\hat{j}) \] Net electric field: \[ E_{\text{net}} = \frac{10 \times 10^{-6}}{2\pi\epsilon_0} \left(\frac{1}{0.1} + \frac{2}{0.2} \right) (-\hat{j}) \] \[ E_{\text{net}} = 3.6 \times 10^6 (-\hat{j}) \text{ N/C} \] Force on the electron: \[ F_{\text{net}} = qE_{\text{net}} \] \[ F_{\text{net}} = (-1.6 \times 10^{-19}) \times (3.6 \times 10^6) \text{ N} \] \[ F_{\text{net}} = 5.76 \times 10^{-13} \text{ N } (\hat{j}) \]