Two point charges,q_1 =0.4μC and q_2 =−0.3μC are placed at 10cm apart.Calculate (a)the potential at point A which is midway between them,and (b)point B which is 6cm from q_1 and 8cm from q


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Question Number 40711 by Necxx last updated on 26/Jul/18

$T w o p o i n t c h a r g e s, q_{1} = 0.4 μ C a n d$ $q_{2} = - 0.3 μ C a r e p l a c e d a t 10 c m$ $a p a r t . C a l c u l a t e$ $(a) t h e p o t e n t i a l a t p o i n t A w h i c h i s$ $m i d w a y b e t w e e n t h e m, a n d$ $(b) p o i n t B w h i c h i s 6 c m f r o m q_{1}$ $a n d 8 c m f r o m q_{2}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

	$a) p o t e n t i a l = 9 \times 10^{9} (\frac{0.4 \times 10^{- 6}}{0.05} + \frac{- 0.3 \times 10^{- 6}}{0.05})$ $= 9 \times 10^{9} (\frac{0.1 \times 10^{- 6}}{0.05}) = \frac{9 \times 10^{3} \times 1 \times 100}{10 \times 5} = 18 \times 10^{3}$ $b) 9 \times 10^{9} (\frac{0.4 \times 10^{- 6}}{0.06} + \frac{- 0.3 \times 10^{- 6}}{0.08})$
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