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Wierzycaz

Answered question

2021-04-27

A charged wire of negligible thickness has length 2L units and has a linear charge density ?. Consider the electric field E? at the point P, a distance d above the midpoint of the wire.

Answer & Explanation

unett

Skilled2021-04-29Added 119 answers

Note that the fields along x cancel because of symmetry.
Thus, as it is positively charged, it points away from the line of charge.
Thus, E points towards the positive y axis.

Jeffrey Jordon

Expert2021-10-01Added 2605 answers

Explanation:

Given that

Length= 2L

Linear charge density= $λ$

Distance=d

$K = \frac{1}{(4 π ϵ)}$

The electric field at point P

$E = 2 K \int_{0}^{L} \frac{λ}{r^{2}} d x \sin θ$

$\sin θ = \frac{d}{\sqrt{d^{2} + x^{2}}}$

$r^{2} = d^{2} + x^{2}$

$E = 2 K λ d \int_{0}^{L} \frac{d x}{(x^{2} + d^{2})^{\frac{3}{2}}}$

Now by integrating above equation

$E = 2 K λ d \frac{L}{d^{2} \sqrt{L^{2} + d^{2}}}$

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