Calculating Electrostatic Force of 3 Point Charges

Click For Summary
SUMMARY

The electrostatic force acting on a charge at the origin due to two identical point charges of 2 micro C located at 50 cm and 100 cm on the x-axis is calculated to be 0.18 N directed to the left. This force is derived by summing the individual forces exerted by the other two charges, specifically F_12 (0.144 N) and F_13 (0.036 N). The principle of superposition confirms that the net force is the vector sum of the individual forces, resulting in a leftward direction due to the repulsive nature of like charges.

PREREQUISITES
  • Understanding of electrostatic forces and Coulomb's Law
  • Familiarity with vector addition in physics
  • Knowledge of the principle of superposition in electrostatics
  • Basic concepts of electric fields and charge interactions
NEXT STEPS
  • Study Coulomb's Law and its applications in electrostatics
  • Learn about vector addition and its role in calculating net forces
  • Explore the principle of superposition in more complex charge configurations
  • Investigate electric field calculations for multiple point charges
USEFUL FOR

Students of physics, educators teaching electrostatics, and anyone interested in understanding the interactions between point charges in electric fields.

Soaring Crane
Messages
461
Reaction score
0
Three identical point charges of 2 micro C are placed on the x-axis. The first charge is at the origin, the second to the right at x = 50 cm (.5 m), and the third at the 100 cm (1 m) mark. What are the mag. and direction of the electrostatic force which acts on the charge on the origin?

The answer is 0.18 to the left.

I calculated the magnitude and got 0.18 by:

finding F_12 and F_13, which is .144 N and .036 N, respectively.

F_1 = F_12 + F_13 = .18 N,

where 1 is charge at origin, 2 is charge at .5 m, and 3 is charge at 1m.

Is this even correct at all? Why must F_12 and F_13 be added?
How is the direction to the left?

Thanks.
 
Physics news on Phys.org
This is correct. The forces must be added because the electric field adds. You could also calculate the electric field from the 2 and 3 charges and then find the force. The force is to the left because 2 and 3 are to the right of 1 and like charges repell each other.
 


Your calculation is correct. The electrostatic force acting on the charge at the origin is the vector sum of the forces exerted by the other two charges. This is why you add F_12 and F_13 to get F_1.

The direction of the electrostatic force can be determined by using the principle of superposition, which states that the net force on a charge is the vector sum of all individual forces acting on it. In this case, since both F_12 and F_13 are acting in the same direction (to the right), the resulting force F_1 will be in the opposite direction, to the left.

To better visualize this, imagine a tug-of-war between the two charges at .5 m and 1 m, with the charge at the origin caught in the middle. The resulting force on the charge at the origin will be in the direction opposite to the stronger force. In this case, F_12 is stronger than F_13, so the resulting force F_1 will be to the left.

I hope this helps clarify the direction of the electrostatic force and why it must be added. Keep up the good work with your calculations!
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
21
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K