Two point charges q_1 =1.5×10^(−9) C and q_2 =3.0×10^(−9) C are seperated by a distance of 200cm.Calculate the point at which the total electric field is zero.


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Question Number 40699 by Necxx last updated on 26/Jul/18

$T w o p o i n t c h a r g e s q_{1} = 1.5 \times 10^{- 9} C$ $a n d q_{2} = 3.0 \times 10^{- 9} C a r e s e p e r a t e d$ $b y a d i s t a n c e o f 200 c m . C a l c u l a t e$ $t h e p o i n t a t w h i c h t h e t o t a l e l e c t r i c$ $f i e l d i s z e r o .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

	$\frac{1}{4 Π ϵ_{0}} \frac{q_{1}}{x^{2}} = \frac{1}{4 Π ϵ_{0}} \frac{q_{2}}{{(l - x)}^{2}}$ $\frac{l - x}{x} = \sqrt{\frac{q_{2}}{q_{1}}}$ $\frac{l}{x} - 1 = \sqrt{\frac{q_{2}}{q_{1}}}$ $\frac{l}{x} = 1 + \sqrt{\frac{q_{2}}{q_{1}}}$ $x = \frac{l}{1 + \sqrt{\frac{q_{2}}{q_{1}}}} = \frac{2}{1 + \sqrt{2}} = 2 (\sqrt{2} - 1)$ $s o n u l l p o i n t i s a t a d i s t a n c e 2 (0.41) = 0.82$ $m e t e r f r o m q_{1} . . .$
	Commented by Necxx last updated on 26/Jul/18

	$T h a n k y o u s o m u c h s i r . . . I t s r e a l l y$ $a c c u r a t e .$
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