Point charges 88μC,−55μC and 70μC are placed in a straight line. The central one is 0.75m from each of the others.Calculate the net force on each due to the other two.


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Question Number 39573 by NECx last updated on 08/Jul/18

$P o i n t c h a r g e s 88 μ C, - 55 μ C a n d$ $70 μ C a r e p l a c e d i n a s t r a i g h t l i n e .$ $T h e c e n t r a l o n e i s 0.75 m f r o m$ $e a c h o f t h e o t h e r s . C a l c u l a t e t h e$ $n e t f o r c e o n e a c h d u e t o t h e o t h e r$ $t w o .$
Commented by NECx last updated on 08/Jul/18

$P l e a s e g i v e a d i a g r a m i f n e c e s s a r y .$ $T h a n k s i n a d v a n c e .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jul/18

	${(F)}_{- 55}^{88} = 9 \times 10^{9} \times \frac{88 \times 10^{- 6} \times (- 55) \times 10^{- 6}}{{(^{} 0.75)}^{2}} f o r c o n$ $Q = 88 μ C b y (- 55 μ C) a t t r a c t i v e$ ${(F)}_{77}^{88} = 9 \times 10^{9} \times \frac{88 \times 10^{- 6} \times 70 \times 10^{- 6}}{{(1.5)}^{2}} r e p u l s i$ $s o n e t f i r c e o n 88 C b y o t h d r s t w o c h a r g e i s$ ${(F)}_{- 55}^{88} \sim {(F)}_{70}^{88} s i m i l a r w a y o t h e r c a n b e c a l c u l a t e d$
	Commented by NECx last updated on 08/Jul/18

	$B u t t h e t e x t b o o k s a y s 53 N, 16 N a n d$ $37 . . . . . P l e a s e c a n y o u e n l i g h t e n m e$ $o n h o w t h e y c a m e a b o u t t h e s e r e s u l t s .$
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