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Question Number 36980 by Tinkutara last updated on 07/Jun/18

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jun/18

$$chargedensigy=\lambda$$$$letlengthofrod=l=10cm$$$$distanceofpointpfromthecentreofrodis$$$$h=\frac{\sqrt{3}}{2}l....asperquestionpointpandtwo$$$$endofrodsmakeaequilateraltriangle$$$$fromcentreofrodalongthelengthlfrod$$$$atadistancexsetakdasmallstrip$$$$dx.$$$$thechargeindxelementis=\lambda dx$$$$electricfielddE=\frac{1}{4\mathrm{\Pi }{ϵ}_0}\times \frac{\lambda dx}{(\sqrt{{h^2+x^2)}^2}}$$$$effectivedE=\frac{1}{4\mathrm{\Pi }{ϵ}_0}\times \frac{\lambda dx}{(\sqrt{{h_{}^2+x^2)}^2}}cos\theta$$$$=\frac{1}{4\mathrm{\Pi }{ϵ}_0}\times (\frac{\lambda hdx}{(h^2+x^2)\frac{3}{2}}whencos\theta =\left(\frac{h}{\sqrt{h^2+x^3}}\right)$$$$Eeffective.=\frac{1}{4\mathrm{\Pi }{ϵ}_0}{\int }_{-\frac{l}{2}}^{\frac{l}{2}}\frac{\lambda hdx}{{(h^2+x^2)}^{\frac{3}{2}}}$$$$=\frac{\lambda h}{4\mathrm{\Pi }{ϵ}_0}{\int }_{-\frac{l}{2}}^{\frac{l}{2}}\frac{dx}{{(h^2+x^2)}^{\frac{3}{2}}}......eqn1$$$$$$$$x=htan\theta dx={hsec}^2\theta d\theta$$$$letI=\int \frac{dx}{{(h^2+x^2)}^{\frac{3}{2}}}$$$$=\int \frac{{hsec}^2\theta d\theta }{h^3{sec}^3\theta }$$$$=\frac{1}{h^2}\int cos\theta d\theta$$$$=\frac{1}{h^2}sin\theta$$$$=\frac{1}{h^2}\frac{x}{\sqrt{x^2+h^2}}$$$$=\frac{\lambda h}{4\mathrm{\Pi }{ϵ}_0}\times \frac{1}{h^2}\mid \frac{x}{\sqrt{x^2+h^2}}{\mid }_{\frac{-l}{2}}^{\frac{l}{2}}$$$$=\frac{\lambda }{4\mathrm{\Pi }{ϵ}_{0}h}\times \{\frac{\frac{l}{2}}{\sqrt{\frac{l^2}{4}+h^2}}-\frac{\frac{-l}{2}}{\sqrt{\frac{l^2}{4}+h^2}}\}$$$$=\frac{\lambda }{4\mathrm{\Pi }{ϵ}_{0}h}\times \left\{\frac{l}{\sqrt{\frac{l^2}{4}+\frac{3l^2}{4}}}\right\}$$$$=\frac{\lambda }{4\mathrm{\Pi }{ϵ}_{0}h}\times 1$$$$=\frac{\frac{q}{l}}{4\mathrm{\Pi }{ϵ}_0\left(\frac{\sqrt{3}}{2}l\right)}=\frac{q\times 2}{4\mathrm{\Pi }{ϵ}_0\times \sqrt{3}\times l^2}$$$$=\frac{50\times {10}^{-6}\times 9\times {10}^9\times 2}{\sqrt{3}\times {(0.1)}^2}$$$$=\frac{450}{\sqrt{3}}\times 2\times {10}^{-6+9+2}$$$$=150\sqrt{3}\times 2\times {10}^5$$$$=300\sqrt{3}\times {10}^5$$$$=3\sqrt{3}\times {10}^7$$$$$$$$$$

Commented by ajfour last updated on 08/Jun/18

see Q.37045

Commented by tanmay.chaudhury50@gmail.com last updated on 08/Jun/18

$$ihavedoneitbutyourmethodisshort...$$

Commented by Tinkutara last updated on 08/Jun/18

Thanks Sir!

Answered by ajfour last updated on 08/Jun/18

$$E=\frac{q}{4\pi {ϵ}_0{r}_{\mathrm{\perp }}{r}_{end}}$$$$r_{\mathrm{\perp }}=\sqrt{100-25}=5\sqrt{3}cm$$$$r_{end}=10cm$$$$E=\frac{50\times {10}^{-6}\times 9\times {10}^9}{50\sqrt{3}\times {10}^{-4}}$$$$=3\sqrt{3}\times {10}^7\frac{N}{C}.$$

Commented by Tinkutara last updated on 08/Jun/18

Thank you Sir.

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