The x and y components of the E-field produced by the charge distribution Q at points on the positive x axis is; \(E = \frac{\lambda*d\theta}{4\pi \epsilon_{0} r_{0} }(sin\theta i + (cos\theta + 1)j)\)
What is the magnitude of magnetic field?
Let the linear charge density = λ
Let the length of the rod is l and the total charge distributed on the rod is Q. Thus;
\(dE = \frac{dQ}{4\pi \epsilon_{0} r^{2} }\)
Thus;
\(dE = \frac{\lambda*da}{4\pi \epsilon_{0} x^{2} }\)
cos θ = \(\frac{r_{0} }{r}\)
r = r₀ * sec θ
\(dE = \frac{\lambda*da}{4\pi \epsilon_{0} r_{0} ^{2} sec^{2}\theta }\)
taking x component and y component gives;
\(dE = \frac{\lambda*d\theta}{4\pi \epsilon_{0} r_{0} }(cos\theta i + sin\thetaj)\)
Integrating both sides gives;
\(\int\limits^E_0 {E}= \frac{\lambda}{4\pi \epsilon_{0} r_{0} }(\int\limits^\theta _0 { cos\theta d\theta i + \int\limits^\theta _0 sin\theta d\theta j)\)
⇒ \(E = \frac{\lambda*d\theta}{4\pi \epsilon_{0} r_{0} }(sin\theta i + (cos\theta + 1)j)\)
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