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Question Number 37058 by Tinkutara last updated on 08/Jun/18

	Answered by ajfour last updated on 08/Jun/18

	$E_{c u b e} + E_{q} = 0$ $E_{c u b e} = - E_{q} = \frac{- q}{4 π ϵ_{0} d^{2}} .$
	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

	$i t h i n k . . i n s i d e a c o n d u c t o r n e t e l e c t r i c f i e l d$ $i s z e r o . . s o E + E_{q} = 0 . . . e l e c t i c f i e l d a t t h e c e n t r e$ $o f c u b e d u e t o e x t e n a l c h a r g e i s E_{q} = \frac{1}{4 Π ϵ_{0}} \times \frac{q}{d^{2}}$ $E i s t h e v a l u e o f e l e c t r i c f i e l d a t t h e c e n t r e$ $o f c u b e d u e t o I N D U C E c h a r g e p r d u c e d f o r t h e$ $p r e s e n c e o f q . . .$
	Commented by Tinkutara last updated on 10/Jun/18
	Thanks Sir!
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