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Question-36096

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Question Number 36096 by rahul 19 last updated on 28/May/18
Commented by rahul 19 last updated on 30/May/18
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Commented by tanmay.chaudhury50@gmail.com last updated on 30/May/18
i am trying to solve by the following method 1)calculate potential for infinite −ve charge sheet at p then substruct it by the potential by disc of radius R ...same for +ve sheet and then add 2)same method for electric field... pls wait
$${i}\:{am}\:{trying}\:{to}\:{solve}\:{by}\:{the}\:{following}\:{method} \\ $$$$\left.\mathrm{1}\right){calculate}\:{potential}\:{for}\:{infinite}\:−{ve}\:{charge} \\ $$$${sheet}\:{at}\:{p}\:{then}\:{substruct}\:{it}\:{by}\:{the}\:{potential} \\ $$$${by}\:{disc}\:{of}\:{radius}\:{R}\:…{same}\:{for}\:+{ve}\:{sheet}\:{and} \\ $$$${then}\:{add} \\ $$$$\left.\mathrm{2}\right){same}\:{method}\:{for}\:{electric}\:{field}… \\ $$$${pls}\:{wait} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 30/May/18
Commented by tanmay.chaudhury50@gmail.com last updated on 30/May/18
Answered by tanmay.chaudhury50@gmail.com last updated on 31/May/18
formula E=(σ/(2ε_0 ))(1−(z/( (√(z^2 +R^2 )) ))) for charge disc E=(σ/(2ε_0 )) for infinite sheet value of z=x+(l/2) for −ve charged sheet vslue of z=x−(l/2) for +ve charged sheet now E for −ve charged sheet =(((−σ))/(2ε_0 ))−(((−σ))/(2ε_0 )){1−(((x+(l/2)))/( (√((x+(l/2))^2 +R^2 ))))} =(((−σ))/(2ε_0 )){1−1+(((x+(l/2)))/( (√((x+(l/2))^2 +R^2 ))))} =(((−σ))/(2ε_0 )){(((x+(l/2)))/( (√((x+(l/2))^2 +R^2 ))))} similarly E for +ve charged sheet =(((+σ))/(2ε_0 )){(((x−(l/2)))/( (√((x−(l/2))^2 +R^2 ))))} so net E at p= (σ/(2ε_0 )){(((x−(l/2)))/( (√((x−(l/2))^2 +R^2 ))))−(((x+(l/2)))/( (√((x+(l/2))^2 +R^2 ))))} now approximatation...l≫R =(((x−(l/2)))/((x−(l/2))(√(1_ +((R/(x−(l/2))))^2 )))) ={1+((R/(x−(l/2))))^2 }^(−(1/2)) ≈1−(1/2)((R/(x−(l/2))))^2 (((x+(l/2)))/((x+(l/2))(√(1+((R/(x+(l/2))))^2 )))) =(({1+((R/(x+(l/2))))^2 }^((−1)/2) )/) ≈1−(1/2)((R/(x+(l/2))))^2 =(σ/(2ε_0 )){1−(1/2)((R/(x−(l/2))))^2 −1+(1/2)((R/(x+(l/2))))^2 } =((σR^2 )/(4ε_0 )){(((x+(l/2))^2 −(x−(l/2))^2 )/((x^2 −(l^2 /4))^2 ))} =((σR^2 )/(4ε_0 )){((4.x.(l/2))/((x^2 −(l^2 /4))^2 ))} =((σR^2 xl)/(2ε_0 (x^2 −(l^2 /4))^2 )) ≈((σR^2 xl)/(2ε_0 x^4 )) considering (x≫(l/2)) =((σR^2 l)/(2ε_0 x^3 )) pls check and suggest
$${formula} \\ $$$${E}=\frac{\sigma}{\mathrm{2}\epsilon_{\mathrm{0}} }\left(\mathrm{1}−\frac{{z}}{\:\sqrt{{z}^{\mathrm{2}} +{R}^{\mathrm{2}} }\:}\right)\:{for}\:{charge}\:{disc} \\ $$$${E}=\frac{\sigma}{\mathrm{2}\epsilon_{\mathrm{0}} }\:{for}\:{infinite}\:{sheet} \\ $$$${value}\:{of}\:{z}={x}+\frac{{l}}{\mathrm{2}}\:{for}\:−{ve}\:{charged}\:{sheet} \\ $$$${vslue}\:{of}\:{z}={x}−\frac{{l}}{\mathrm{2}}\:{for}\:+{ve}\:{charged}\:{sheet} \\ $$$${now}\:{E}\:{for}\:−{ve}\:{charged}\:{sheet} \\ $$$$=\frac{\left(−\sigma\right)}{\mathrm{2}\epsilon_{\mathrm{0}} }−\frac{\left(−\sigma\right)}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{\mathrm{1}−\frac{\left({x}+\frac{{l}}{\mathrm{2}}\right)}{\:\sqrt{\left({x}+\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} }}\right\} \\ $$$$=\frac{\left(−\sigma\right)}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{\mathrm{1}−\mathrm{1}+\frac{\left({x}+\frac{{l}}{\mathrm{2}}\right)}{\:\sqrt{\left({x}+\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} \:}}\right\} \\ $$$$=\frac{\left(−\sigma\right)}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{\frac{\left({x}+\frac{{l}}{\mathrm{2}}\right)}{\:\sqrt{\left({x}+\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} }}\right\} \\ $$$${similarly}\:{E}\:{for}\:+{ve}\:{charged}\:{sheet} \\ $$$$=\frac{\left(+\sigma\right)}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{\frac{\left({x}−\frac{{l}}{\mathrm{2}}\right)}{\:\sqrt{\left({x}−\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} }}\right\} \\ $$$${so}\:{net}\:{E}\:{at}\:{p}= \\ $$$$\frac{\sigma}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{\frac{\left({x}−\frac{{l}}{\mathrm{2}}\right)}{\:\sqrt{\left({x}−\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} }}−\frac{\left({x}+\frac{{l}}{\mathrm{2}}\right)}{\:\sqrt{\left({x}+\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} }}\right\} \\ $$$${now}\:{approximatation}…{l}\gg{R} \\ $$$$=\frac{\left({x}−\frac{{l}}{\mathrm{2}}\right)}{\left({x}−\frac{{l}}{\mathrm{2}}\right)\sqrt{\mathrm{1}_{} +\left(\frac{{R}}{{x}−\frac{{l}}{\mathrm{2}}}\right)^{\mathrm{2}} }} \\ $$$$=\left\{\mathrm{1}+\left(\frac{{R}}{{x}−\frac{{l}}{\mathrm{2}}}\right)^{\mathrm{2}} \right\}^{−\frac{\mathrm{1}}{\mathrm{2}}} \approx\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{R}}{{x}−\frac{{l}}{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\frac{\left({x}+\frac{{l}}{\mathrm{2}}\right)}{\left({x}+\frac{{l}}{\mathrm{2}}\right)\sqrt{\mathrm{1}+\left(\frac{{R}}{{x}+\frac{{l}}{\mathrm{2}}}\right)^{\mathrm{2}} }}\: \\ $$$$=\frac{\left\{\mathrm{1}+\left(\frac{{R}}{{x}+\frac{{l}}{\mathrm{2}}}\right)^{\mathrm{2}} \right\}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{} \\ $$$$\approx\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{R}}{{x}+\frac{{l}}{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$=\frac{\sigma}{\mathrm{2}\epsilon_{\mathrm{0}} }\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{R}}{{x}−\frac{{l}}{\mathrm{2}}}\right)^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{R}}{{x}+\frac{{l}}{\mathrm{2}}}\right)^{\mathrm{2}} \right\} \\ $$$$=\frac{\sigma{R}^{\mathrm{2}} }{\mathrm{4}\epsilon_{\mathrm{0}} }\left\{\frac{\left({x}+\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({x}−\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\frac{{l}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\sigma{R}^{\mathrm{2}} }{\mathrm{4}\epsilon_{\mathrm{0}} }\left\{\frac{\mathrm{4}.{x}.\frac{{l}}{\mathrm{2}}}{\left({x}^{\mathrm{2}} −\frac{{l}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\sigma{R}^{\mathrm{2}} {xl}}{\mathrm{2}\epsilon_{\mathrm{0}} \left({x}^{\mathrm{2}} −\frac{{l}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$$\approx\frac{\sigma{R}^{\mathrm{2}} {xl}}{\mathrm{2}\epsilon_{\mathrm{0}} {x}^{\mathrm{4}} }\:\:\:\:\:\:\:\:\:{considering}\:\left({x}\gg\frac{{l}}{\mathrm{2}}\right) \\ $$$$=\frac{\sigma{R}^{\mathrm{2}} {l}}{\mathrm{2}\epsilon_{\mathrm{0}} {x}^{\mathrm{3}} }\:\:{pls}\:{check}\:{and}\:{suggest} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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