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A-positive-charge-of-magnigude-2-5-C-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-s




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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