Electric Field due to Point Charges

In summary, the conversation discusses a practice exam question involving two point charges and their electric potential and field. The solution to part A is found to be 0.36m, but there is confusion with part B. The correct equation for finding the magnitude of the electric field is given as E=k[((5x10-6)/0.362)+((2x10-6)/0.142)], regardless of the plus and minus signs, and the explanation for why the electric field is non-zero is that the potential is changing with distance.
  • #1
Caiti
9
0

Homework Statement

[/B]

The following is on a practice exam I have been completing. In advance, it is part b I am struggling with.

Two point charges Q1 = +5*10^(-6)C and -2*10^(-6)C are 50cm apart.
a) Where along a line that passes through the two charges is the electric potential zero (apart
from at r=infinity)?

b) Determine the magnitude of the electric field at that point. Explain why the electric field is non-zero.

Homework Equations


E=kQ/r^2

V=kQ/r

E=dV/dr

The Attempt at a Solution


I got part A immediately using V=kQ/r and superposition of electric potential, finding it to be 0.36m (which matched the solution). It is part b that I am struggling with. I have as follows:

E=kQ/r^2
Using superposition of electric fields, this gives
E = k(5/(0.36^2) + -2/(0.5-0.36)^2), where k=1/(4*pi*E_0). This gave me an electric field of -570367N/C; however, the solution said it was 1.2*10^6 N/C. I believe the issue was in my implementation of the -2C in the above equation; however, am not entirely sure why this is incorrect or what to do to fix it.

As for the explanation, my belief is that because E is changing in potential with respective to distance, as long as the potential is not set at a constant 0 - that is, the potential is changing as distance changes - the electric field could be non-zero. I think I'm okay with this, but please do not hesitate to correct me if I'm wrong!

I'd really appreciate some help in figuring this out - or even just in where I went wrong. Please let me know if I've forgotten anything or my working isn't clear.

Thank you so much
 
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  • #2
You are right there was something wrong with the use of the -2 x 10-6C. The equation should be E=k[((5x10-6)/0.362)+((2x10-6)/0.142)], because we are finding the magnitude of the electric field, so the plus and minus signs doesn't matter.
 
Last edited:
  • #3
robotpie3000 said:
You are right there was something wrong with the use of the -2 x 10-6C. The equation should be E=k[((5x10-6)/0.362)+((2x10-6)/0.142)], because we are finding the magnitude of the electric field, so the plus and minus signs doesn't matter.

So does that mean that if we were finding the electric field, not magnitude, I would be correct? Or would I still use E=k[((5x10-6)/0.362)+((2x10-6)/0.142)] , then choose direction afterwards?
 

Related to Electric Field due to Point Charges

1. What is an electric field due to a point charge?

The electric field due to a point charge is a force field that surrounds a point charge and exerts a force on other charged particles in its vicinity. It is a measure of the strength and direction of the electric force experienced by a test charge placed in the field.

2. How is the electric field calculated for a point charge?

The electric field at a point due to a point charge can be calculated using the equation E = kQ/r², where E is the electric field, k is the Coulomb's constant (9x10^9 Nm²/C²), Q is the magnitude of the point charge, and r is the distance between the point charge and the point where the electric field is being calculated.

3. What is the direction of the electric field due to a positive point charge?

The direction of the electric field due to a positive point charge is radially outward, away from the point charge. This means that a positive test charge placed in the field would experience a repulsive force away from the point charge.

4. How does the electric field change as the distance from a point charge increases?

The electric field strength decreases as the distance from a point charge increases. This is because the force exerted by the point charge on a test charge decreases with distance, according to the inverse square law. Therefore, as the distance increases, the electric field becomes weaker.

5. Can multiple point charges create a combined electric field?

Yes, the electric field due to multiple point charges can be calculated by vector addition. Each point charge creates its own electric field, and the total electric field at any point is the sum of the individual electric fields. The direction and strength of the combined electric field will depend on the direction and magnitude of each individual point charge.

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