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Question Number 39792 by NECx last updated on 11/Jul/18

A positive charge of magnigude 2.5μC is at the centre of an uncharged spherical conducting shell of inner radius 60cm and an outer radius of 90cm. (i)find the charge densities on thd inner and outer surfaces of the shell and the total charge on each surface. ii)find the electricity field everywhere.

$${A}\:{positive}\:{charge}\:{of}\:{magnigude} \\ $$$$\mathrm{2}.\mathrm{5}\mu{C}\:{is}\:{at}\:{the}\:{centre}\:{of}\:{an}\:{uncharged} \\ $$$${spherical}\:{conducting}\:{shell}\:{of}\:{inner} \\ $$$${radius}\:\mathrm{60}{cm}\:{and}\:{an}\:{outer}\:{radius} \\ $$$${of}\:\mathrm{90}{cm}. \\ $$$$\left({i}\right){find}\:{the}\:{charge}\:{densities}\:{on}\:{thd} \\ $$$${inner}\:{and}\:{outer}\:{surfaces}\:{of}\:{the} \\ $$$${shell}\:{and}\:{the}\:{total}\:{charge}\:{on}\:{each} \\ $$$${surface}. \\ $$$$\left.{ii}\right){find}\:{the}\:{electricity}\:{field} \\ $$$${everywhere}. \\ $$

Commented by NECx last updated on 11/Jul/18

please help

$${please}\:{help} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jul/18

i)surface charge density (σ) on the inner surface shell σ_i =((−2.5×10^(−6) )/(4Π×(60×10^(−2) )^2 )) surface charge density(σ) on the outer surface shell σ_o =((2.5×10^(−6) )/(4Π(90×10^(−2) )^2 )) electric field inside the shell is zero...because charge at centre and induced charged on the inner wall of shell balance each other E×4Πr^2 =((2.5μC+(−2.5μC))/ε_0 ) so E=0 for outside the shell wall.. E×4ΠR^2 =((2.5μC+(−2.5μC)+(2.5μC))/ε_0 ) E=((2.5×10^(−6) )/(4Πε_0 (90×10^(−2) )^2 ))

$$\left.{i}\right){surface}\:{charge}\:{density}\:\left(\sigma\right)\:{on}\:{the}\:{inner} \\ $$$${surface}\:{shell}\:\sigma_{{i}} =\frac{−\mathrm{2}.\mathrm{5}×\mathrm{10}^{−\mathrm{6}} }{\mathrm{4}\Pi×\left(\mathrm{60}×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${surface}\:{charge}\:{density}\left(\sigma\right)\:{on}\:{the}\:{outer} \\ $$$${surface}\:{shell}\:\:\sigma_{{o}} =\frac{\mathrm{2}.\mathrm{5}×\mathrm{10}^{−\mathrm{6}} }{\mathrm{4}\Pi\left(\mathrm{90}×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${electric}\:{field}\:{inside}\:{the}\:{shell}\:{is}\:{zero}...{because} \\ $$$${charge}\:{at}\:{centre}\:{and}\:{induced}\:{charged}\:{on}\:\:{the} \\ $$$${inner}\:{wall}\:{of}\:{shell}\:{balance}\:{each}\:{other} \\ $$$${E}×\mathrm{4}\Pi{r}^{\mathrm{2}} =\frac{\mathrm{2}.\mathrm{5}\mu{C}+\left(−\mathrm{2}.\mathrm{5}\mu{C}\right)}{\epsilon_{\mathrm{0}} } \\ $$$${so}\:{E}=\mathrm{0} \\ $$$${for}\:{outside}\:{the}\:{shell}\:{wall}.. \\ $$$${E}×\mathrm{4}\Pi{R}^{\mathrm{2}} =\frac{\mathrm{2}.\mathrm{5}\mu{C}+\left(−\mathrm{2}.\mathrm{5}\mu{C}\right)+\left(\mathrm{2}.\mathrm{5}\mu{C}\right)}{\epsilon_{\mathrm{0}} } \\ $$$${E}=\frac{\mathrm{2}.\mathrm{5}×\mathrm{10}^{−\mathrm{6}} }{\mathrm{4}\Pi\epsilon_{\mathrm{0}} \left(\mathrm{90}×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} } \\ $$

Commented by NECx last updated on 11/Jul/18

This is cool . Thank you so much

$${This}\:{is}\:{cool}\:.\:{Thank}\:{you}\:{so}\:{much} \\ $$

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