Question

Two identical, small insulating balls are suspended by separate 0.21-m threads that are attached to a common point on the ceiling. Each ball has a mass of 7.4 x 10-4 kg. Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of 50 degrees between the threads. Determine (a) the charge on each ball and (b) the tension in the threads.

Answer 1

draw a free body diagram for this. First, we are given the fact that two threads of equal length make an angle of 50. The two threads and the distance between the two insulating balls make an isosceles triangle. In order to find the distance between the two balls, we need to use the law of sines.

Sin(A)/x = Sin(B)/l

But we only know 1 angle. We do know that the other two angles we don't know are equal to each other due to there opposite lengths being equal. Also, the sum of the angles in any triangle is 180o.

180-50-2T= 0

T=65 degree

Now we can find the distance between the balls.

Sin(53)/x = Sin(63.5)/l

x= l* (Sin(50)/Sin(65)) (l is the length of the string).

x= (.21m)* (Sin(50)/Sin(65))

x=.177499 m (Now we know the distance for columns law).

Now to split the force into x and y components (I am using the left ball to describe the forces using right as +x and up as +y).

If we split the force of tension into x and y components, we find that:

FTx= FT sin(25)
FTy= FT cos(25)

Now for the algebra:

Fx= FTx-Fc=0 (Fc is force of charge)
Fy= FTy-m*g=0

FT*cos(25)= m*g

FT= (m*g)/cos(25) = .008009868 N

Fc = FTx = FT sin(25)
Fc= k*(Q1*Q2)/x^2

(Q1=Q2=Q) (k=9*10^9 N*m^2/C^2) (x=.177499 m)

kQ^2/x^2= FT sin(25)

Q= (x^2*FT*sin(25)/(k)) = 1.185*10^-14

= 1.0885 * 10^-7 = 0.10885 micro C