Dear student,let the cordinate of 3 charge be r⇀ from 2 μC chargehence the force between 2,8 μC is equal to -4 and 8 μCusing coulombs force in vector formF⇀2,8=F⇀-4,8k×2×8×(r)2=k×4×8(5+r)2r=12 19 now slope of line tanθ =y2-y1x2-x1=34hence x cordinate will be (1+rcosθ) =(1+12 19×45) =10 75≈11hence y cordinate will be (1+rsinθ) =(1+12 19×35) =8 314≈8Regards

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solve this Coulomb's Law Q 5) The charges qi 2 pc and are located at ccurdinates 31.-11 and (3, 2) respeczivdy lfwe have put one more charge q, : 8 PC in the line joining two charges such that the charges an equilibrium. then find the coordinate ot the location ct A) 111.8) Correct answer Solution:- Let A1 and be at A and 3 slnvn in figure . O Type hereto search Strength-o.meter /100 Attempted 4/10

D e a r s t u d e n t, l e t t h e c o r d i n a t e o f 3 c h a r g e b e \overset{⇀}{r} f r o m 2 μ C c h a r g e h e n c e t h e f o r c e b e t w e e n 2, 8 μ C i s e q u a l t o - 4 a n d 8 μ C u \sin g c o u l o m b s f o r c e i n v e c t o r f o r m {\overset{⇀}{F}}_{2, 8} = {\overset{⇀}{F}}_{- 4, 8} \frac{k \times 2 \times 8 \times}{(r)^{2}} = \frac{k \times 4 \times 8}{(5 + r)^{2}} r = 12.19 n o w s l o p e o f l i n e \tan θ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{3}{4} h e n c e x c o r d i n a t e w i l l b e (1 + r \cos θ) = (1 + 12.19 \times \frac{4}{5}) = 10.75 \approx 11 h e n c e y c o r d i n a t e w i l l b e (1 + r \sin θ) = (1 + 12.19 \times \frac{3}{5}) = 8.314 \approx 8 R e g a r d s

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