A +ve point charge of magnitude q is located on the y−axis at a point of y=+d and another −ve charge of same magnitude is located on the point y=−d. A third +ve charge of same magnitude is located at some point on x−axis. (1)− what is the magnitude and direction of the force on third charge ? (a)− when its located on the origin. (b)− when its coordinate is x. (2)− show that when x is large compared to distance d, the force in (b) is inversely proportionalto the cube of the distance from origin.


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Question Number 46292 by Umar last updated on 23/Oct/18

$A + v e p o i n t c h a r g e o f m a g n i t u d e q i s l o c a t e d o n t h e y - a x i s$ $a t a p o i n t o f y = + d a n d a n o t h e r - v e c h a r g e o f s a m e m a g n i t u d e$ $i s l o c a t e d o n t h e p o i n t y = - d .$ $A t h i r d + v e c h a r g e o f s a m e m a g n i t u d e i s l o c a t e d$ $a t s o m e p o i n t o n x - a x i s .$ $(1) - w h a t i s t h e m a g n i t u d e a n d d i r e c t i o n o f t h e f o r c e o n$ $t h i r d c h a r g e ?$ $(a) - w h e n i t s l o c a t e d o n t h e o r i g i n .$ $(b) - w h e n i t s c o o r d i n a t e i s x .$ $(2) - s h o w t h a t w h e n x i s l a r g e c o m p a r e d t o d i s t a n c e d,$ $t h e f o r c e i n (b) i s i n v e r s e l y p r o p o r t i o n a l t o t h e$ $c u b e o f t h e d i s t a n c e f r o m o r i g i n .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18
	$a)Force on 3rd charge by charge A i s repulsive force on 3rd charge by charge B is attrctive net force on 3rd charge when it is at origin is F=(1/(4πε_0 ))(q^2 /d^2 )+(1/(4πε_0 ))(q^2 /d^2 )=2×(1/(4πε_0 ))×(q^2 /d^2 ) direction is −ve yaxis vecorically F_1 ^→ =(1/(4πε_0 ))(q^2 /d^2 )(−j^→ ) F_2 ^→ =(1/(4πε_0 ))(q^2 /d^2 )(−j^→ ) net force=F_1 ^→ +F_2 ^→ =2×(1/(4πε_0 ))(q^2 /d^2 )(−j^→ ) b)F_1 ^→ =(1/(4πε_0 ))(q^2 /((d^2 +x^2 )^(3/2) ))(−dj+xi) F_2 ^→ =(1/(4πε_0 ))(q^2 /((d^2 +x^2 )^(3/2) ))(−dj−xi) F_R ^→ =F_1 ^→ +F_2 ^→ =(1/(4πε_0 ))(q^2 /((d^2 +x^2 )^(3/2) ))(−2dj) net force =(1/(4πε_0 ))(q^2 /((d^2 +x^2 )^(3/2) ))×2d direcection −ve y axis F_R =(1/(4πε_0 ))(q^2 /({x^2 (1+(d^2 /x^2 ))}^(3/2) ))×2d≈(1/(4πε_0 ))×((q^2 ×2d)/x^3 ) [(d/x)→0 as x>>d] [$
	$a) F o r c e o n 3 r d c h a r g e b y c h a r g e A i s r e p u l s i v e$ $f o r c e o n 3 r d c h a r g e b y c h a r g e B i s a t t r c t i v e$ $n e t f o r c e o n 3 r d c h a r g e w h e n i t i s a t o r i g i n$ $i s F = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{d^{2}} + \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{d^{2}} = 2 \times \frac{1}{4 π ϵ_{0}} \times \frac{q^{2}}{d^{2}}$ $d i r e c t i o n i s - v e y a x i s$ $v e c o r i c a l l y$ ${\vec{F}}_{1} = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{d^{2}} (- \vec{j})$ ${\vec{F}}_{2} = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{d^{2}} (- \vec{j})$ $n e t f o r c e = {\vec{F}}_{1} + {\vec{F}}_{2} = 2 \times \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{d^{2}} (- \vec{j})$ $b) {\vec{F}}_{1} = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{{(d^{2} + x^{2})}^{\frac{3}{2}}} (- d j + x i)$ ${\vec{F}}_{2} = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{{(d^{2} + x^{2})}^{\frac{3}{2}}} (- d j - x i)$ ${\vec{F}}_{R} = {\vec{F}}_{1} + {\vec{F}}_{2} = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{{(d^{2} + x^{2})}^{\frac{3}{2}}} (- 2 d j)$ $n e t f o r c e = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{{(d^{2} + x^{2})}^{\frac{3}{2}}} \times 2 d d i r e c e c t i o n - v e y a x i s$ $F_{R} = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{{x^{2} (1 + \frac{d^{2}}{x^{2}})}^{\frac{3}{2}}} \times 2 d \approx \frac{1}{4 π ϵ_{0}} \times \frac{q^{2} \times 2 d}{x^{3}} [\frac{d}{x} \to 0 a s x >> d]$ $[$
	Commented by Umar last updated on 23/Oct/18

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