Question Number 37045 by ajfour last updated on 08/Jun/18
Commented by ajfour last updated on 08/Jun/18
$${Find}\:{electric}\:{field}\:{at}\:{a}\:{distance} \\ $$$${r}_{\bot} \:{from}\:{a}\:{line}\:{charge}\:{of}\: \\ $$$${uniform}\:{charge}\:{density},\:{net} \\ $$$${charge}\:{q}. \\ $$
Answered by ajfour last updated on 08/Jun/18
$${E}=\int{dE}\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:=\int\frac{\lambda{dy}\mathrm{cos}\:\theta}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}^{\mathrm{2}} } \\ $$$${r}={r}_{\bot} \mathrm{sec}\:\theta \\ $$$${y}={r}_{\bot} \mathrm{tan}\:\theta\:\:\:\Rightarrow\:\:\:{dy}={r}_{\bot} \mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta \\ $$$${E}=\frac{{q}/{l}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\int_{−\alpha} ^{\:\:\alpha} \frac{{r}_{\bot} \mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta\:\mathrm{cos}\:\theta}{{r}_{\bot} ^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta} \\ $$$$\:\:\:=\:\frac{{q}/{l}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}_{\bot} }\left(\mathrm{2sin}\:\alpha\right) \\ $$$$\:\:\:=\frac{{q}/{l}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}_{\bot} }×\frac{\mathrm{2}×{l}/\mathrm{2}}{{r}_{{end}} } \\ $$$${E}\:=\:\frac{{q}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}_{\bot} {r}_{{end}} }\:. \\ $$
Commented by Tinkutara last updated on 08/Jun/18
Thank you very much Sir! I got the answer. ��������