Please show the solution. Consider two charges Q placed at fixed positions a hor
ID: 1661772 • Letter: P
Question
Please show the solution.
Consider two charges Q placed at fixed positions a horizontal distance L apart, as shown in the figure. A third particle of mass m and unknown charge q is then placed a vertical distance s above the fixed charges. The position of this third particle is arranged so that it is suspended in mid-air above the other two! This equilibrium condition occurs because the electrical and gravitational forces on the charge precisely cancel each other. r thns third particle is arranged to t d gravitational q, m (a) Calculate the charge q that is required to balance the third particle in mid-air. Express your answer in terms of the given parameters plus physical constants. 8 (b) Does your answer make sense? Find three limiting cases where you know what the answer should be and make sure your expression for q gives the right result. And here's another important check: does the sign of q make sense? (c) Equilibrium situations like the one you have analyzed can be described as either stable or unstable. Stable equilibrium means that the system will return to its equilibrium configuration when disturbed by a small amount (like a swinging pendulum). Unstable equilibrium describes situations like a ball balanced on top of a hill- the slightest disturbance will drastically affect the system. Think about your suspended charge problem for a moment what sort of equilibrium condition is present? Take your best guess before you turn to the next page.Explanation / Answer
due to the symmetry of the arrangement, force between Q and q will be of same magnitude for both left side Q and right side Q.
distance between Q and q=D=sqrt((L/2)^2+s^2)
angle made by this force with the perpendicular from q to line connecting Qs =theta=arccos(s/D)
force between Q and q=F=k*Q*q/D^2
where k=coloumb’s constant
total force in vertical direction=F*cos(theta)+F*cos(theta)
=2*F*s/D
=2*(k*Q*q/D^2)*s/D
=2*k*Q*q*s/D^3
balancing this force with gravitational force,
2*K*q*Q*s/D^3=m*g
==>q=m*g*D^3/(2*k*Q*s)
=m*g*(0.25*L^2+s^2)^1.5/(2*k*Q*s)
part b:
limiting case 1:
when L=0, q=m*g*s^2/(2*k*Q)
limiting case 2:
when L=inifinity,
q=infinity
limiting case 3:
when s=0,
q=infinity
sign of q would be positive .
part c:
this system is in a stable equilibrium.
because if s increases, electrical force will decrease and due to higher weight, it will come back down.
if s decreases, due to higher electrical force, it will go back up to its stable condition.