Question

18. (III) Two charges, -Q and -3Q, are a distance l apart. These two charges are free to move but do not because there is a third (fixed) charge nearby. What must be the magnitude of the third charge and its placement in order for the first two to be in equilibrium?

Answer 1

Solution let us start by placing charge oa; at different position When q, is b negative qu=placed charge of 70 -qi from same changes repel each other opposite changes attract each other. At positin & net force is toward (-x) so it can't be Equilbrium position At position , net force is toward C+²) so we can't place charge at this position. At position , net force are is zero cor they are balancing each othen). As whole system is in equilbrium when we check force at change tQ), .net force on this charge is toward (*) . Similarly, net force on charge (-3Q) is towards (+x). so we can't place negative change between these two charge ☺ when 92 is positive qi At basition , net force is to ward (+x), We can't place Charge at this location. At position @ net force is toward (-x) we can't place charge at this location

At position net force on all the three charges are zeno las forces are batanung each other) This can be position of charge. position of change = blw the two charges sign of charge - positive change. - 3Q clue fl- lon- from (-Q) let us place charge at a distance fonce on charge a As system is in Equil buíum, net EFq, = 0 Fi2 = Force on q, "to-Q=1 .q IQ (0) - Aato x2 Fara Force on q, due to -3Q=- 2, 3Q (+7) 40to (e-x7² Fiat Fl=0 Fize - Fai = -Kol At R. Fizl=Fril = kVQ _ kq,30 x² [e-oc)2 = (1-x)² = 3x² = e²+x²_ 20x= 3x² = 300² - x ² a 2 + 2lx=0 = 200² a ² + 2 lx=0 = 200² +2 ex-e²=0

Using x=- b ± √b²4aC (ant but c =o) za b=2l, a=2 , c=-1² x=-26 + √ 40²7462) (12) 2(2) X=-26 IV1212 = -26+43 e x = -l Ibe As charge will lie blo two charge, 80 taking only x=-1+30 = +0.732 e |x=0.367 et position of charge will be 0.367 e from -Q. As system is in Equibuium, force on each change will be Zero..., Using this, baštinut Whiting total force on Q? and balacing them Force on-Q due qi= 1.q1Q ) 47to x2 e Force on-a due to -3Q= I (@130) (-x²) 4nto e These two force must balance each other. qila Anto ocz (4atol 12 - 9,= (doc)COL3 @) 12 qı = 3x 0-134 42 Q. 19,- +0.404 Q | magnitude of charge