Question Number 37058 by Tinkutara last updated on 08/Jun/18
Answered by ajfour last updated on 08/Jun/18
$${E}_{{cube}} +{E}_{{q}} =\mathrm{0} \\ $$$${E}_{{cube}} =−{E}_{{q}} =\frac{−{q}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {d}^{\:\mathrm{2}} }\:. \\ $$
Commented by Tinkutara last updated on 08/Jun/18
Sir can you please explain first line.
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18
$${i}\:{think}..{inside}\:{a}\:{conductor}\:{net}\:{electric}\:{field} \\ $$$${is}\:{zero}..{so}\:{E}+{E}_{{q}} =\mathrm{0}…{electic}\:{field}\:{at}\:{the}\:{centre} \\ $$$${of}\:{cube}\:{due}\:{to}\:{extenal}\:{charge}\:{is}\:{E}_{{q}} =\frac{\mathrm{1}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} }×\frac{{q}}{{d}^{\mathrm{2}} } \\ $$$${E}\:{is}\:{the}\:{value}\:{of}\:{electric}\:{field}\:{at}\:{the}\:{centre} \\ $$$${of}\:{cube}\:{due}\:{to}\:{INDUCE}\:{charge}\:{prduced}\:{for}\:{the} \\ $$$${presence}\:{of}\:{q}… \\ $$
Commented by Tinkutara last updated on 10/Jun/18
Thanks Sir!