Electric Fields and Styrofoam balls

Click For Summary
SUMMARY

The discussion centers on calculating the common charge carried by three identical Styrofoam balls suspended in an equilateral triangle formation. Each ball has a mass of 1.93 g and is suspended by nonconducting threads of 49.0 cm. The initial calculation for the charge, using the equations Fe = mgtan(theta) and q^2 = Fe(r^2)/ke, yields a charge of approximately 8.108e-7 C for two balls. The challenge arises in incorporating the third ball's influence on the system, as it introduces additional repulsive forces that affect the equilibrium distance between the balls.

PREREQUISITES
  • Understanding of electrostatics, specifically Coulomb's Law
  • Knowledge of vector components in physics
  • Familiarity with trigonometric functions and their applications in physics
  • Basic principles of equilibrium in mechanical systems
NEXT STEPS
  • Study Coulomb's Law and its application to multiple charges
  • Learn about vector resolution and force diagrams in equilibrium problems
  • Explore the concept of electric field strength and its relation to charge distribution
  • Investigate the effects of additional charges on the forces in a system
USEFUL FOR

Students and educators in physics, particularly those focusing on electrostatics and mechanics, as well as anyone interested in solving complex equilibrium problems involving multiple forces.

glid02
Messages
54
Reaction score
0
Here is the question:
Three identical small Styrofoam balls (m = 1.93 g) are suspended from a fixed point by three nonconducting threads, each with a length of 49.0 cm and with negligible mass. At equilibrium the three balls form an equilateral triangle with sides of 29.2 cm. What is the common charge q carried by each ball?

I set it up by taking two of the balls and making them into two back-to-back right triangles of the same size. I used the equation Fe=mgtan(theta) where theta is sin^-1(.146/.49)=17.335 deg. For this I got 5.909787*e-3.

Then I used the equation q^2=Fe(r^2)/ke, where r is .292 m.
For this i got q=8.108e-7.

I've triple checked everything and I'm fairly confident that this is the right answer for two balls, but I'm not sure how the third ball plays into the answer. It seems like it would push the two balls apart from each other a little more, which would diminish the charge I got when considering just the two of them, but I'm not sure how to go about finding the amount that the third ball would change the distance between ball 1 and ball 2. Any help would be awesome. Thanks a lot.
 
Physics news on Phys.org
Each of two balls repels the third. If one draws a line between two charges, the third ball is repelled with a net force perpendicular to that line. Each ball is repelled from the equilibrium distance from the center of mass which would coincide with the center of the equilateral triangle.

Each ball sits at the end of thread at some angle. The lateral component of tension in the string (related to the weight of each ball) is equal and opposite the repulsive electric force. You seem to have done that correctly.

Treat the forces as vectors along the line of action, then determine the component acting opposite the tension in the thread.
 
Last edited:

Similar threads

  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
8K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
6
Views
9K