Graded Homework Assignment 3 PHYS-1610 Fall 2018 Prof. Comes Problem m l Two sma
ID: 1879215 • Letter: G
Question
Graded Homework Assignment 3 PHYS-1610 Fall 2018 Prof. Comes Problem m l Two small insulating balls with charge +q and mass m are placed in a circular bowl with radius R as shown above and are free to move without friction. You may assume that the radius of the balls is negligible compared to the radius of the bowl If we begin with the balls at the top of the bowl (i.e. gravitational potential energy of both balls? You may assume that the potential energy is zero if the ball is at the bottom of the bowl. (2 points) a) 90°), what is the total b) Write an equation for the total gravitational potential energy of the two balls, Ug, in terms c) If the balls are initially at the top of the bowl, what is the electrostatic potential energy of d) Write an equation for the total electrostatic potential energy, Ue, as a function of R, 0, and e) Combine the equations for Uc and Ug to get the total potential energy of the system, Uro of R, 6, and m, and g. (2 points) the two charges? (2 points) q. (2 points) Find an equation for the equilibrium value of that minimizes the potential energy in terms of the parameters that were given in the problem (R, q, m, and fundamental constants). Your answer will look similar to the solution for problem 5.68. (6 points) Solve for the value of q that makes the equilibrium value of = 45°. (3 points) If the value of 0 is changed so that it is slightly less than 45, will the balls move towards each other or away from each other? What about if the value of 0 is slightly greater than 45°? What does this imply about the motion of the charges about the equilibrium value of f) g) ? (3 points)Explanation / Answer
given two small insulating balls of charge q each
a. at theta = 90 deg
gravitational PE of both balls, PE = 2mgR
b. at some angle theta
gPE = 2mgR(1 - cos(theta))
c. at the top of the bowl
electrostatic PE = kq^2/2R = kq^2/2R
d. at some angle theta
electrostatic PE = kq^2/d
sin(theta) = d/2*R
hence
ePE= kq^2/2Rsin(theta)
e. Ut = Ug + Ue = 2mgR(1 - cos(theta)) + kq^2/2Rsin(theta)
for minimising Ut
dUt/d(tehta) = 0
4mgR^2(sin^3(theta)) = kq^2*cos(tehta)
solving this we get value of theta for equilibrium
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