Question Number 39794 by NECx last updated on 11/Jul/18
$${Two}\:{charged}\:{concentric}\:{spheres} \\ $$$${have}\:{radii}\:{of}\:\mathrm{10}{cm}\:{and}\:\mathrm{18}{cm}.{The} \\ $$$${charge}\:{on}\:{the}\:{inner}\:{sphere}\:{is} \\ $$$$\mathrm{6}.\mathrm{0}×\mathrm{10}^{−\mathrm{8}} {C}\:{and}\:{that}\:{on}\:{the}\:{outer} \\ $$$${sphere}\:{is}\:\mathrm{2}.\mathrm{0}×\mathrm{10}^{−\mathrm{8}} {C}.{Find}\:{the} \\ $$$${electric}\:{field}\:\left({i}\right){at}\:{r}=\mathrm{12}{cm}\:{and} \\ $$$$\left({ii}\right){at}\:{r}=\mathrm{25}{cm} \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jul/18
$$\left.{i}\right){E}.\mathrm{4}\Pi{r}^{\mathrm{2}} =\frac{{q}}{\epsilon_{\mathrm{0}} }\:\:\:{r}=\mathrm{12}×\mathrm{10}^{−\mathrm{2}} \:\:{and}\:{q}=\mathrm{6}.\mathrm{0}×\mathrm{10}^{−\mathrm{8}} {C} \\ $$$${E}=\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{6}.\mathrm{0}×\mathrm{10}^{−\mathrm{8}} }{\left(\mathrm{12}×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{0}.\mathrm{375}×\mathrm{10}^{\mathrm{5}} \\ $$$$\left.{ii}\right){E}=\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{8}×\mathrm{10}^{−\mathrm{8}} }{\left(\mathrm{25}×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{0}.\mathrm{1152}×\mathrm{10}^{\mathrm{5}} \\ $$$${pls}\:{check} \\ $$