Question

Two thin parallel threads carry a uniform charge with linear densities $$\lambda$$ and $$-\lambda$$. The distance between the threads is equal to $$l$$. Find the potential of the electric field and the magnitude of its strength vector at the distance $$r > > l$$ at the angle $$\theta$$ to the vector $$1$$. (Fig).

Answer 1

Let $$P$$ be a point, at distance $$r > > l$$ and at an angle to $$\theta$$ the vector $$\vec {l}$$ (Fig.).

Thus $$\vec {E}$$ at $$P = \dfrac {\lambda}{2\pi \epsilon_{0}} \dfrac {\vec {r} + \dfrac {\vec {l}}{2}}{\left |\vec {r} + \dfrac {\vec {l}}{2}\right |^{2}} - \dfrac {\lambda}{2\pi \epsilon_{0}} \dfrac {\vec {r} - \dfrac {\vec {l}}{2}}{\left |\vec {r} - \dfrac {\vec {l}}{2}\right |^{2}}$$

$$= \dfrac {\lambda}{2\pi \epsilon_{0}}\left [\dfrac {\vec {r} + \vec {l}/2}{r^{2} + \dfrac {l^{2}}{4} + rl\cos \theta} - \dfrac {\vec {r} - \vec {l}/2}{r^{2} + \dfrac {l^{2}}{4} - rl\cos \theta}\right ]$$

$$= \dfrac {\lambda}{2\pi \epsilon_{0}}\left (\dfrac {\vec {l}}{r^{2}} - \dfrac {2l\vec {r}}{r^{3}} \cos \theta \right )$$

Hence $$E = |\vec {E}| = \dfrac {\lambda l}{2\pi \epsilon_{0} r^{2}}, r > > l$$

Also, $$\varphi = \dfrac {\lambda}{2\pi \epsilon_{0}} ln |\vec {r} + \vec {l}/2| - \dfrac {\lambda}{2\pi \epsilon_{0}} ln |\vec {r} - \vec {l}/2]$$

$$= \dfrac {\lambda}{4\pi \epsilon_{0}} ln \dfrac {r^{2} + rl\cos \theta + l^{2}/ 4}{r^{2} - rl\cos \theta + l^{2}/4} = \dfrac {\lambda l\cos \theta}{2\pi \epsilon_{0}r}, r > > l$$.