Two small plastic balls hang from threads of negligible mass. Each ball has a ma

Question

Two small plastic balls hang from threads of negligible mass. Each ball has a mass of 0.140g and a charge of magnitude q. The balls are attracted to each other, and the threads attached to the balls make an angle of 20.0 degrees with the vertical, as shown in the figure.(Figure 1)

same picture as here http://www.chegg.com/homework-help/questions-and-answers/small-plastic-balls-hang-threads-negligible-mass-eachball-mass-014-g-chargeof-magnitude-q--q254221

Find the magnitude of the charge on the balls in (nC.)

Explanation / Answer

Force diagram time!
Up and right are positive, and I'll call the left one q1, the right one q2. Also, since the angle given is with the vertical and we'll want to use the angle the threads makes with the horizontal, I'll say =90-, where is the given angle.

q1:

x: Fq2-Tcos()=0

y: Tsin()-mg=0

q2:

x: Tcos()-Fq1=0

y: same as q1, since the masses are the same.

Now, Fq1 and Fq2 are equal in magnitude, but opposite in direction. Since signs are taken care of from the force diagram, we'll just deal with magnitudes.

F=kq2/r2

Since r is given, this is the expression we'll need, no need to change it. All we need now is to get rid of T, and solve for q.

T=mg/sin()

Fq2=kq2/r2=Tcos()=mg*cos()/sin()

q2=r2mg*cot()/k

Since =90-, and cot(90-)=tan(), this simplifies to:

q=( r2mg*tan()/k )^.5

Just be sure to keep track of your units, and that will give you the answer to parts a and c. It's pretty easy to see how to solve for T, so I'll let you go ahead and do that.

Plug and chug!

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.