Electric field by straight wire

Electricity and Magnetism Level 2

Suppose that an infinitely long straight current-carrying wire has a uniform linear charge density $\lambda.$ Let this wire be on the $y$ -axis of the $xy$ -plane, and let $x > 0$ be the distance between the $y$ -axis and a point $P$ on the $xy$ -plane. What is the strength of the electric field at the point $P?$

Note: $\epsilon_0$ in the choices below denotes the electric constant.

\frac{\lambda}{2\epsilon_0}

\frac{\lambda}{2\pi \epsilon_0 x}

\frac{x}{\epsilon_0 \lambda}

\frac{\lambda}{4\pi \epsilon_0 x^2}

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Alakh Aggarwal
May 9, 2014

Applying Gauss's law by making a Gaussian surface of cylinder of radius $x$ and length $l$ around wire, closed $\oint Eds = \frac{q_\text{enclosed}}{\epsilon_0}= \frac{\lambda l}{\epsilon_0}$ then we have $E = \frac{\lambda l}{\epsilon_0 2\pi xl} = \frac{\lambda}{2\pi\epsilon_0 x}$

Can this be proved without using gauss law?

Sanjeevan Goswami - 7 years, 1 month ago

I do:

$\mathrm{d}\epsilon = \frac{dq}{4 \times \pi \times \epsilon_0 \times (x^2+y^2)}$

$\int \mathrm{d}\epsilon = \frac{1}{4 \times \pi \times \epsilon_0 } \times \int \limits_{-\infty}^{\infty}\frac{\mathrm{d}q}{(x^2+y^2)}$

$\int \mathrm{d}\epsilon = \frac{\lambda}{4 \times \pi \times \epsilon_0 } \times \int \limits_{-\infty}^{\infty}\frac{\mathrm{d}y}{(x^2+y^2)}$

so:

$\epsilon = \frac{\lambda}{4 \times \epsilon_0 \times x}$

What is error?

Víctor Mandujano Gutierrez - 7 years, 1 month ago

You cannot take direct values of electric fields but also include the directions. Hence multiply the dE term with a cosine of the angle subtended between the x axis and the electric field direction

Nishant Gupta - 7 years, 1 month ago

Electric field by straight wire

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