Electric Field Halfway Between Two Charges

In summary: Well, if it's a dumb question to ask, I won't ask any questions anymore. In summary, the conversation discusses the calculation of the E field value at the half-way point between two charges. The equation for calculating the electric field at a point is mentioned, along with the use of the constant 8.99E9 Nmm/(CC). The questioner makes an attempt at solving the problem, but gets it wrong and is unsure of how to proceed. They mention using a distance of .135m for the R value, but the responder suggests that without showing their work, it is difficult to determine where the error occurred. The questioner then seems to become discouraged and states that they will no longer ask questions.
  • #1
Xet
3
0

Homework Statement


q1 (-6.5E-7 C) @ 6 cm to the right of origin (X,Y coordinate is 6,0)
q2 (6.5E-7 C) @ 21 cm to the left of origin (X,Y coordinate is -21,0)

At the half-way point of these two charges, what is the E field value...?

Homework Equations


sum of electric fields
electric field at point is equal to constant times the charge, all divided by radius squared
constant is 8.99E9 Nmm/(CC)

The Attempt at a Solution


I found the two E's from the point to the half-way mark...I'm suppose to add the two E's because it's going from negative to positive...right? Well...I got it wrong...
 
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  • #2
Xet said:

Homework Statement


q1 (-6.5E-7 C) @ 6 cm to the right of origin (X,Y coordinate is 6,0)
q2 (6.5E-7 C) @ 21 cm to the left of origin (X,Y coordinate is -21,0)

At the half-way point of these two charges, what is the E field value...?

Homework Equations


sum of electric fields
electric field at point is equal to constant times the charge, all divided by radius squared
constant is 8.99E9 Nmm/(CC)


The Attempt at a Solution


I found the two E's from the point to the half-way mark...I'm suppose to add the two E's because it's going from negative to positive...right? Well...I got it wrong...


What did you use for the distance when you calculated the E field?
 
  • #3
kdv said:
What did you use for the distance when you calculated the E field?

Rather dumb question to ask...but I used .135 m as R.
 
  • #4
Xet said:
Rather dumb question to ask...but I used .135 m as R.

It is not a dumb question at all. You didn't show your work. How can we tell you what you did wrong if you don't show any calculations?
 
  • #5
Xet said:
Rather dumb question to ask...but I used .135 m as R.

well, if it's a dumb question to ask I won't ask any question anymore.

Best luck
 

1. What is an electric field?

An electric field is a physical quantity that describes the influence of electrically charged objects on each other. It is a vector quantity, meaning it has both magnitude and direction, and is measured in units of volts per meter (V/m).

2. How do you calculate the electric field halfway between two charges?

To calculate the electric field halfway between two charges, you can use the formula E = kQ/r^2, where E is the electric field, k is the proportionality constant (9 x 10^9 Nm^2/C^2), Q is the magnitude of the charge, and r is the distance between the two charges.

3. What is the direction of the electric field halfway between two charges?

The direction of the electric field halfway between two charges depends on the sign of the charges. If the charges are of the same sign, the electric field will point away from the charges. If the charges are of opposite signs, the electric field will point towards the charges.

4. Can the electric field be zero halfway between two charges?

Yes, the electric field can be zero halfway between two charges if the charges are of equal magnitude and opposite signs. In this case, the electric field vectors cancel each other out, resulting in a net electric field of zero.

5. How does the distance between two charges affect the electric field halfway between them?

The electric field halfway between two charges is inversely proportional to the square of the distance between the charges. This means that as the distance between the charges increases, the electric field decreases, and vice versa.

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