Pleas help electric field

1. Sep 14, 2005

stunner5000pt

pleas help!! electric field

a semi infinite insulting rod carries a constant charge per unit length of lambda. SHow that he electric field at the point P makes an angle of 45 with the rod and that this result is independant of R

$$q = \lambda x$$
$$dq = \lambda dx$$

for the Y direction poiting downward
$$dE_{y} = \int_{0}^{\infty} \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{R^2 + x^2} \sin{\theta}$$
$$\sin{\theta} = \frac{x}{R^2 + x^2}$$
$$dE_{y} = \int_{0}^{\infty} \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda R dx}{(R^2 + x^2)^\frac{3}{2}}$$
$$dE_{y} = \frac{\lambda R}{4 \pi \epsilon_{0}} \int_{0}^{\infty} \frac{dx}{(R^2 + x^2)^\frac{3}{2}}$$
$$dE_{y} = \frac{\lambda R}{4 \pi \epsilon_{0}} [\frac{x}{R^2 \sqrt{R^2+x^2}}]_{0}^{\infty}$$

for the horizontal pointing left
$$\frac{\lambda R}{4 \pi \epsilon_{0}} \int_{0}^{\infty} \frac{xdx}{(R^2 + x^2)^\frac{3}{2}}$$
same problem here

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Last edited: Sep 14, 2005
2. Sep 15, 2005

quasar987

The integral is right and the limit to evaluate is almost trivial: pull x² out of the square root.

3. Sep 15, 2005

stunner5000pt

ok for the Y part
$$E_{y} = \frac{\lambda}{4 \pi \epsilon_{0}R}$$
i swhat i got the Y part. Am i right??

for the X part
i got the same thing.
However the resultant for these two given a 45 degree angle... however it isn ot independat of R as in
$$E = \sqrt{2} \frac{\lambda}{4 \pi \epsilon_{0}R}$$