Electric field calculation for 3 point charges

In summary, the conversation discusses using the equation E=FB/qB to determine the magnitude of the electric field at a point where sphere B has been removed from a system of three charged spheres. This is done by bringing a positive test charge to that point and measuring the force on it, which is then divided by the value of the test charge to find the electric field. This process is described using the example of sphere B being the test charge.
  • #1
vf_one
14
0
Hi I'm having difficulty understanding why the solution for the answer to the following problem is worked out this way.

Three spheres are placed at fixed points along the x axis, whose positive direction points towards the right.

Sphere A is at x=47cm with a charge of 5x10-6C
Sphere B is at x=50cm wtih a charge of -5x10-6C
Sphere C is at x=54cm with a charge of 10x10-6C

Sphere B is removed. What would be the magnitude of the electric field at the point where sphere B was located?

I've managed to work out the electrostatic force on sphere B using F= kqAqB/r2 and F= kqBqC/r2.

In the answer it says that we use the equation E=FB/qB. I don't understand why we use the electrostatic force on sphere B and the charge of sphere B to work out the electric field since B is removed and only spheres A and C are left.

Can anyone explain this to me?
Thanks
 
Physics news on Phys.org
  • #2
vf_one said:
In the answer it says that we use the equation E=FB/qB. I don't understand why we use the electrostatic force on sphere B and the charge of sphere B to work out the electric field since B is removed and only spheres A and C are left.

Can anyone explain this to me?
Thanks
To figure out if there is an electric field at a point in space, in what direction it points and how strong it is, you need to bring a positive test charge of known value q at that point and measure the force on that charge. The magnitude of the force divided by the value of the test charge is (by definition) the magnitude of the electric field. The direction of the force is the direction of the electric field. So this question describes exactly this process with sphere B being the test charge.
 
  • #3
We didn't go over this in class so this really helped. Thank you!
 

What is an electric field?

An electric field is a physical quantity that describes the strength and direction of the force exerted on a charged particle by other charged particles in its vicinity.

How is an electric field calculated?

The electric field at a point is calculated by adding the individual electric fields created by each charge at that point. This can be expressed mathematically as E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the particle, and r is the distance between the particle and the point at which the electric field is being calculated.

What are the key factors that affect the electric field calculation for 3 point charges?

The key factors that affect the electric field calculation for 3 point charges are the magnitude and sign of each charge, as well as the distance between each charge and the point at which the electric field is being calculated.

What is the principle of superposition in electric field calculation for 3 point charges?

The principle of superposition states that the total electric field at a point is equal to the vector sum of the individual electric fields created by each charge at that point. This principle allows us to calculate the electric field at a point by considering the individual contributions of each charge separately.

How does the direction of the electric field change in the presence of 3 point charges?

The direction of the electric field at a point is always directed away from positively charged particles and towards negatively charged particles. In the presence of 3 point charges, the direction of the electric field at a point is determined by the vector sum of the individual electric fields created by each charge, taking into account their magnitudes and directions.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
325
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
730
  • Introductory Physics Homework Help
Replies
21
Views
605
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
950
  • Introductory Physics Homework Help
Replies
3
Views
733
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
174
Back
Top