Question-37045

Question Number 37045 by ajfour last updated on 08/Jun/18

Commented by ajfour last updated on 08/Jun/18

Find electric field at a distance r_⊥ from a line charge of uniform charge density, net charge q.

$${Find}\:{electric}\:{field}\:{at}\:{a}\:{distance} \\ $$$${r}_{\bot} \:{from}\:{a}\:{line}\:{charge}\:{of}\: \\ $$$${uniform}\:{charge}\:{density},\:{net} \\ $$$${charge}\:{q}. \\ $$

Answered by ajfour last updated on 08/Jun/18

E=∫dEcos θ =∫((λdycos θ)/(4πε_0 r^2 )) r=r_⊥ sec θ y=r_⊥ tan θ ⇒ dy=r_⊥ sec^2 θdθ E=((q/l)/(4πε_0 ))∫_(−α) ^( α) ((r_⊥ sec^2 θdθ cos θ)/(r_⊥ ^2 sec^2 θ)) = ((q/l)/(4πε_0 r_⊥ ))(2sin α) =((q/l)/(4πε_0 r_⊥ ))×((2×l/2)/r_(end) ) E = (q/(4πε_0 r_⊥ r_(end) )) .

$${E}=\int{dE}\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:=\int\frac{\lambda{dy}\mathrm{cos}\:\theta}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}^{\mathrm{2}} } \\ $$$${r}={r}_{\bot} \mathrm{sec}\:\theta \\ $$$${y}={r}_{\bot} \mathrm{tan}\:\theta\:\:\:\Rightarrow\:\:\:{dy}={r}_{\bot} \mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta \\ $$$${E}=\frac{{q}/{l}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\int_{−\alpha} ^{\:\:\alpha} \frac{{r}_{\bot} \mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta\:\mathrm{cos}\:\theta}{{r}_{\bot} ^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta} \\ $$$$\:\:\:=\:\frac{{q}/{l}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}_{\bot} }\left(\mathrm{2sin}\:\alpha\right) \\ $$$$\:\:\:=\frac{{q}/{l}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}_{\bot} }×\frac{\mathrm{2}×{l}/\mathrm{2}}{{r}_{{end}} } \\ $$$${E}\:=\:\frac{{q}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}_{\bot} {r}_{{end}} }\:. \\ $$

Commented by Tinkutara last updated on 08/Jun/18

Thank you very much Sir! I got the answer. ��

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Question-37045

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