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Question Number 41530 by physics last updated on 09/Aug/18

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

$$E=\frac{\lambda }{2\mathrm{\Pi }{ϵ}_0{r}_{\mathrm{\perp }}}E=electricfieldatpointwhichisag$$$$aperpediculardistance{r}_{\mathrm{\perp }}frominfinitechargerod=$$$$\lambda =chargeperunitlength$$$${ϵ}_0=permitivityofvaccume$$$$forceon-qis{F}_1=\frac{-q\lambda }{2\mathrm{\Pi }{ϵ}_0{r}_{\mathrm{\perp }}}$$$$forceon+qis{F}_2=\frac{q\lambda }{2\mathrm{\Pi }{ϵ}_0(r_{\mathrm{\perp }}+l)}$$$$netforce{F}_1+F_2=\frac{q\lambda }{2\mathrm{\Pi }{ϵ}_0}(\frac{-1}{r_{\mathrm{\perp }}}+\frac{1}{r_{\mathrm{\perp }}+l})$$$$herel=2mm=2\times {10}^{-3}meter$$$$r_{\mathrm{\perp }}=2cm=2\times {10}^{-2}meter$$$$q=2\times {10}^{-5}C$$$$\lambda =4\times {10}^{-4}$$

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