Electrical fields between two objects

  • Thread starter Thread starter cougarsoccer
  • Start date Start date
  • Tags Tags
    Electrical Fields
Click For Summary
To determine the point where the electric field is zero between two charges of -2.5 microC and 6 microC, a coordinate system should be established. The electric field equation E=(1/4PI(8.854*10^(-12)))*(q/r^2) is relevant, but the key is to find the distance from each charge to the point where the fields cancel. By using symmetry, if a point is 0.2 m from the -2.5 microC charge, the distance to the 6 microC charge can be calculated accordingly. This approach simplifies the problem and helps identify the exact location of the zero electric field.
cougarsoccer
Messages
12
Reaction score
0

Homework Statement


Two charges are 1m apart, the first charge is -2.5microC and the second on is 6 microC, determine the point(other than infinity) at which the electric field is zero.


Homework Equations


E=(1/4PI(8.854*10^(-12)))*(q/r^2)


The Attempt at a Solution


ok maybe i am just really overthinking the problem at hand. but for my first attempt, i simply let E=0 and set it equal to the rest of the equation(with r=1) the thing is, i do not know how to find the point. (i'm assuming i am solving for q but i don't think that is right) I just do not know how to solve for the point
 
Physics news on Phys.org
You started off well enough. You are not solving for q; both of those are given. You are solving for something like r, except that is given, too, so that's probably what's confusing you. You have to put these charges in a coordinate system, and then give it some thought. Try to use symmetry to make this problem as simple as possible. Hint: if a point in space is 0.2 m from -2.5~\mu{}C, then how far away is it from +6.0~\mu{}C?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 1 ·
Replies
1
Views
930
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K