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: \[ F_{\text{net}} = qE_{\text{net}} \] \[ F_{\text{net}} = (-1.6 \times 10^{-19}) \times (3.6 \times 10^6) \text{ N} \] \[ F_{\text{net}} = 5.76 \times 10^{-13} \text{ N } (\hat{j}) \]

Answer 2

Two infinitely long, straight, parallel wires. Linear charge densities: \(\lambda_1=+10\,\mu\text{C/m}\), \(\lambda_2=-20\,\mu\text{C/m}\). The electron at \(P\) is at the same perpendicular distance \(r=15\text{ cm}=0.15\text{ m}\) from each wire, and the two field vectors at \(P\) are colinear so their magnitudes add. (Electron charge: \(q_e=-e=-1.602\times10^{-19}\,\text{C}\).)

1) Electric field of an infinite line charge

\[ E=\frac{|\lambda|}{2\pi\varepsilon_0\,r},\qquad \frac{1}{2\pi\varepsilon_0}\approx 1.7975\times10^{10}\ \frac{\text{N}\cdot\text{m}}{\text{C}^2}. \] For \(\lambda_1=+10\,\mu\text{C/m}\) and \(\lambda_2=-20\,\mu\text{C/m}\) at the same distance \(r\), the net field magnitude at \(P\) (additive directions) is \[ E_{\text{net}}=\frac{|\lambda_1|+|\lambda_2|}{2\pi\varepsilon_0\,r} =\frac{30\times10^{-6}}{2\pi\varepsilon_0\,\cdot\,0.15} \approx 3.595\times10^{6}\ \text{N/C}. \]

2) Force on the electron

\[ |\vec F|=|q_e|\,E_{\text{net}}=e\,E_{\text{net}} \approx (1.602\times10^{-19})\times(3.595\times10^{6}) \approx 5.76\times10^{-13}\ \text{N}. \] The direction of \(\vec F\) is opposite to \(\vec E_{\text{net}}\) (since the electron is negatively charged).

Final Answer: \[ \boxed{\,|\vec F|\ \approx\ 5.75\times10^{-13}\ \text{N}\,}. \]

Notes:

• If the figure places \(P\) at equal distances and the two field vectors at \(P\) are colinear (one toward and one away so both point the same way), magnitudes add as shown. If instead they were perpendicular, you would combine them by the Pythagorean sum.
• Using \(\lambda\) in \(\mu\text{C/m}\) is the standard convention in such problems; the given numerical answer matches this unit choice with \(r=15\text{ cm}\).