A charge q_1 =2.0×10^(−9) C is placed at the point,(x=0,y=4cm) and another charge,q_2 =−3.0×10^(−9) C is located at the point,(x=3cm,y=4cm). If a third charge,q_3 =4.0×10^(−9) C is placed at the origin, a)obtain the x and y component of the total force on q_3 b)Calculate the magnitude and direction of the total force on q


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Question Number 40700 by Necxx last updated on 26/Jul/18

$A c h a r g e q_{1} = 2.0 \times 10^{- 9} C i s p l a c e d$ $a t t h e p o i n t, (x = 0, y = 4 c m) a n d$ $a n o t h e r c h a r g e, q_{2} = - 3.0 \times 10^{- 9} C$ $i s l o c a t e d a t t h e p o i n t, (x = 3 c m, y = 4 c m) .$ $I f a t h i r d c h a r g e, q_{3} = 4.0 \times 10^{- 9} C$ $i s p l a c e d a t t h e o r i g i n,$ $a) o b t a i n t h e x a n d y c o m p o n e n t o f$ $t h e t o t a l f o r c e o n q_{3}$ $b) C a l c u l a t e t h e m a g n i t u d e a n d$ $d i r e c t i o n o f t h e t o t a l f o r c e o n q_{3 .}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

	$q_{1} = + v e q_{2} = - v e q_{3} = + v e$ $s o f o r c e o n q_{3} b y q_{1} i s r e p u l s i v e d i r e c t i o n$ $- v e y a x i s$ ${\vec{F}}_{1} = 9 \times 10^{9} \times \frac{2.0 \times 10^{- 9} \times 4.0 \times 10^{- 9}}{{(0.04)}^{2}} \times (- \vec{j})$ $f o r c e o n q_{3} b y q_{2} i s a t t r a c t i v e$ ${\vec{F}}_{2} = 9 \times 10^{9} . \frac{3.0 \times 10^{- 9} \times 4.0 \times 10^{- 9}}{{(0.05)}^{2}} (\frac{0.03 \vec{i} + 0.04 \vec{j}}{0.05})$ $n e t f o r c e \vec{F} = {\vec{F}}_{1} + {\vec{F}}_{2}$ ${\vec{F}}_{1} = \frac{72}{16} \times 10^{9 + 4 - 18} (- \vec{j}) = 4.5 \times 10^{- 5} (- \vec{j})$ ${\vec{F}}_{2} = \frac{108}{25} \times 10^{9 + 4 - 18} (\frac{0.03 \vec{i} + 0.04 \vec{j}}{0.05})$ ${\vec{F}}_{2} = 4.32 \times 10^{- 5} (\frac{3}{5} i + \frac{4}{5} j)$ $= 2.592 \times 10^{- 5} i + 3.456 \times 10^{- 5} j$ $n e t f o r c e$ $\vec{F} = 2.592 \times 10^{- 5} i + 7.956 \times 10^{- 5} j$ $m i a g n e t u d e \approx (\sqrt{9 + 64}) \times 10^{- 5} = 8.54 \times 10^{- 5} N$ $d i r e c t i o n {t a n}^{- 1} (\frac{7.956}{2.592}) = {t a n}^{- 1} (3.07) . . .$
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