Find-out-electric-field-on-an-axial-position-due-to-a-ring-having-linear-charge-density-0-cos-

Question Number 33400 by rahul 19 last updated on 15/Apr/18

\boldsymbol F i n d o u t e l e c t r i c f i e l d o n a n a x i a l

p o s i t i o n d u e t o a r i n g h a v i n g l i n e a r

c h a r g e d e n s i t y \boldsymbol λ = λ_{0} \cos θ .

Commented by ajfour last updated on 16/Apr/18

Commented by ajfour last updated on 16/Apr/18

Commented by ajfour last updated on 16/Apr/18

{\bar{E}}_{⊥} = \frac{d Q}{4 π ϵ_{0} r^{2}} (\frac{R}{r}) (- \cos θ \hat{j} - \sin θ \hat{k})

[i n d i a g r a m a b o v e i f o r g o t t o

i n c l u d e t h e (\frac{R}{r}) f a c t o r]

= \frac{λ_{0} R^{2} \cos θ d θ}{4 π ϵ_{0} r^{3}} (- \cos θ \hat{j} - \sin θ \hat{k})

E_{x} = 0, E_{z} = 0

E_{y} = - \frac{λ_{0} R^{2}}{4 π ϵ_{0} {(x^{2} + R^{2})}^{3 / 2}} \int_{0}^{2 π} \cos^{2} θ d θ

= - \frac{λ_{0} R^{2}}{4 π ϵ_{0} {(x^{2} + R^{2})}^{3 / 2}} \times \frac{1}{2} \int_{0}^{2 π} (1 + \cos 2 θ) d θ

= - \frac{λ_{0} R^{2}}{4 π ϵ_{0} {(x^{2} + R^{2})}^{3 / 2}} \times \frac{1}{2} (2 π)

E = ∣ E_{y} ∣= \frac{\boldsymbol λ_{0} R^{2}}{4 \boldsymbol ϵ_{0} {(\boldsymbol x^{2} + \boldsymbol R^{2})}^{3 / 2}} .

Commented by rahul 19 last updated on 16/Apr/18

t h a n k u s i r .

t h a n k u s i r .