Question Number 40699 by Necxx last updated on 26/Jul/18
$${Two}\:{point}\:{charges}\:{q}_{\mathrm{1}} =\mathrm{1}.\mathrm{5}×\mathrm{10}^{−\mathrm{9}} {C} \\ $$$${and}\:{q}_{\mathrm{2}} =\mathrm{3}.\mathrm{0}×\mathrm{10}^{−\mathrm{9}} {C}\:{are}\:{seperated} \\ $$$${by}\:{a}\:{distance}\:{of}\:\mathrm{200}{cm}.{Calculate} \\ $$$${the}\:{point}\:{at}\:{which}\:{the}\:{total}\:{electric} \\ $$$${field}\:{is}\:{zero}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18
$$\frac{\mathrm{1}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} }\frac{{q}_{\mathrm{1}} }{{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} }\frac{{q}_{\mathrm{2}} }{\left({l}−{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{l}−{x}}{{x}}=\sqrt{\frac{{q}_{\mathrm{2}} }{{q}_{\mathrm{1}} }}\: \\ $$$$\frac{{l}}{{x}}−\mathrm{1}=\sqrt{\frac{{q}_{\mathrm{2}} }{{q}_{\mathrm{1}} }}\: \\ $$$$\frac{{l}}{{x}}=\mathrm{1}+\sqrt{\frac{{q}_{\mathrm{2}} }{{q}_{\mathrm{1}} }}\: \\ $$$${x}=\frac{{l}}{\mathrm{1}+\sqrt{\frac{{q}_{\mathrm{2}} }{{q}_{\mathrm{1}} }}}=\frac{\mathrm{2}}{\mathrm{1}+\sqrt{\mathrm{2}}}=\mathrm{2}\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right) \\ $$$${so}\:{null}\:{point}\:{is}\:{at}\:{a}\:{distance}\:\mathrm{2}\left(\mathrm{0}.\mathrm{41}\right)=\mathrm{0}.\mathrm{82} \\ $$$${meter}\:{from}\:{q}_{\mathrm{1}} ... \\ $$
Commented by Necxx last updated on 26/Jul/18
$${Thank}\:{you}\:{so}\:{much}\:{sir}...{Its}\:{really} \\ $$$${accurate}. \\ $$