Find the magnitude of the electrostatic force

Click For Summary
SUMMARY

The discussion focuses on calculating the electrostatic force between two charged particles, specifically finding the placement of a third charge, q3 = +5.0 µC, such that the net electrostatic force on it is zero. The charges are q1 = +3.5 µC and q2 = -4.0 µC, located at specified coordinates in the xy plane. The magnitude of the force on q2 is calculated as 34.05 N, with a direction of 350.54° from the +x axis. The solution to part (c) involves determining the coordinates for q3 by balancing the forces exerted by q1 and q2.

PREREQUISITES
  • Understanding of Coulomb's Law for electrostatic forces
  • Familiarity with vector components in the xy coordinate system
  • Knowledge of charge interactions and net force calculations
  • Ability to sketch and interpret diagrams in physics
NEXT STEPS
  • Study Coulomb's Law and its applications in electrostatics
  • Learn how to resolve forces into their x and y components
  • Explore the concept of equilibrium in electrostatic systems
  • Practice problems involving multiple charges and net force calculations
USEFUL FOR

Students studying physics, particularly those focusing on electrostatics, as well as educators and anyone involved in solving problems related to electric forces and charge interactions.

DeadxBunny
Messages
29
Reaction score
0
I am having difficulties with the following problem. I have gotten parts (a) and (b) (my correct answers are shown below), but I cannot figure out how to do part (c). Any help with it would be greatly appreciated. Thanks!
------------
The charges and coordinates of two charged particles held fixed in an xy plane are q1 = +3.5 µC, x1 = 4.0 cm, y1 = 0.50 cm, and q2 = -4.0 µC, x2 = -2.0 cm, y2 = 1.5 cm.
(a) Find the magnitude of the electrostatic force on q2.
Answer: 34.05N
(b) Find the direction of this force.
Answer: 350.54° (counterclockwise from the +x axis)
(c) At what coordinates should a third charge q3 = +5.0 µC be placed such that the net electrostatic force on particle 3 due to particles 1 and 2 is zero?
 
Physics news on Phys.org
first of all it will help if you drew a diagram
(a rough sketch of the xy plane and plot these points)

now the force of Q1 on Q3 shouldbe equal to the force of Q1 on Q2.
(there's many ways to do this, here's on of them)

consider the X coordinate system ONLY
the distance between Q1 and Q2 is 6 units. If a charge Q3 were to be placed such taht F (of Q1 on Q3) = F(of Q3 on Q2) then it would be placed along the line joining Q1 and A3. If Q3 were placed x units from Q1 how was is placed from Q2? Now using F (of Q1 on Q3) = F(of Q3 on Q2) and you know how far Q2 is from Q1 and Q3, you can solve for x.
DO the similar thing for the Y direction. Thus you have your answer
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
17
Views
3K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
1
Views
1K