Short webpage title: Calculating Electric Field and Force from Two Charges

  • Thread starter dgl7
  • Start date
In summary, the conversation discusses two charges located on a horizontal axis and determining the electric field at a given point on a vertical axis. The Coulomb constant is used in calculating the electric field, and the vertical component of the electric force is also calculated. The electric field is a vector, so the direction must be taken into account when adding the magnitudes of the electric fields from each charge.
  • #1
dgl7
8
0
Two charges create an electric field--electric field strenght at a point above h.fiel

Homework Statement


Two charges are located on a horizontal axis. The Coulomb constant is 8.98755x10^9 Nm^2/C^2.

a) Determine the electric field at p on a vertical axis as shown in the attachment. Up is the positive direction. Answer in units of V/m.

b) Calculate the vertical component of the electric force on a -3.1e-6C charge placed at point p. Answer in units of N


Homework Equations



a) E=kQ/r^2
b) F=Eq


The Attempt at a Solution



a) I'm fairly certain I know how to find the field strength from the two charged particles, if point p was directly on the field and in the center.

E=E1+E2
E=k/r^2(Q+q)
E=8.98755e9/3^2(2.2e-6+2.2e-6)
E=4393 V/m

But I think I also need to take into account that p is above the two particles, but now I'm stuck...

b) I think this part would be easier, I just can't do it since I haven't gotten the answer to part a.

F=Eq
F=E(that would be found in part a)(-3.1e-6)
 

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  • #2
Hi dgl7! :smile:
dgl7 said:
E=k/r^2(Q+q)
E=8.98755e9/3^2(2.2e-6+2.2e-6)

That's correct, if p was on the x-axis, you would use r = 3 for each charge, and add.

Instead, use r = the actual distance between p and each charge.

That will give you the force, so remember force is a vector, and add the two vectors.
 
  • #3


AHHH that makes sense. Thanks very much!
 
  • #4


Ok nope. Nevermind, I'm still confused.
This is what I've been doing and attempting:
E=Eleft+Eright
E=kQleft/r^2+kQright/r^2
(r=r and Qleft=Qright)
E=2kQ/r^2
E=2*8.98755e9*2.2e-6/(1.8^2+3^2)
E=3230.818627 V/m

Not sure what I'm doing wrong.
 
  • #5
The electric field is a vector,

so the electric field from each charge has the magnitude kQ/r2, but it also has a direction.

The direction is different for each charge, so you can't just add the magnitudes, can you? :smile:
 

What is the meaning of the equation F=(4393)(-3.1e-6)F=-13.5 N?

The equation F=(4393)(-3.1e-6)F=-13.5 N represents the force (F) of an object with a charge of 4393 Coulombs in an electric field with a strength of -3.1 micro Newtons, resulting in a force of -13.5 Newtons.

What do the values 4393 and -3.1e-6 represent in the equation?

The value 4393 represents the charge (in Coulombs) of the object in the electric field, while -3.1e-6 represents the strength (in micro Newtons) of the electric field.

What is the unit of measurement for the force in the equation?

The unit of measurement for the force in the equation is Newtons (N).

What does the negative sign indicate in front of the force value?

The negative sign in front of the force value indicates that the force is acting in the opposite direction of the electric field. In this case, the force is acting in the opposite direction as the electric field's direction.

How is this equation applicable in real-life scenarios?

This equation can be applicable in various scenarios, such as calculating the force on a charged particle in an electric field, determining the strength of the electric field based on the force experienced by an object, or predicting the motion of an object in an electric field. It is commonly used in the field of physics and engineering to understand the behavior of electrically charged objects in different situations.

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