Finding magnitude of electric field strength

In summary, when calculating the electric field strength at a point, you first need to determine the distance between the points. Next, you need to calculate the electric field strength based on the charges present. Finally, you need to add all three values together to get the final answer.
  • #1
Chi Futbol
3
0

Homework Statement


Consider three charges arranged as shown.


What is the magnitude of the electric field
strength at a point 2.7 cm to the left of the
middle charge? The value of the Coulomb
constant is 8.98755 × 109 N · m2/C2 . Answer
in units of N/C.

Homework Equations


I'm truly lost. I've been looking over some other places, trying to put things together, but nothing is becoming clear to me... How do the positive and negative charges affect each other? For example if I have 2 equal positive charges, the electric field should be 0 right? How do I determine it if it is 1 positive and 1 negative charge? (Both the same value -1 and +1)


The Attempt at a Solution


I was looking at some similar problems.

E1 = 8.98755e9 x 5.6e-6 / .064^2 = 1.228766602e7. I got that distance by finding the total distance between the point 2.7 cm to the left. So 3.7 + 2.7 = 6.4. That could be very wrong, I'm not sure.
E2 = 8.98755e9 x 3.2e-6 / .047^2 = 1.301953825e7
E3 = 8.98755e9 x -1.8e-6 / .017^2 = -5.597782007

Then i just decided to add all three of those together... As I said I'm really lost, and I'm quite sure none of what I did is correct. Any help?


Edit : Sorry... I inserted a picture to show the problem.
 

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  • #2
draw a picture of the field lines

define a poistive direction along the axis the charges are arranged
out from positive, into negative

work out the distance form each cahrge to the point you want to find

then based on whether the arrow aligns with you positive dircetion (+ or -), add them all up...
 
  • #3
I think I may have done my distances correctly? Can anyone confirm that? I guess I was confusing the difference between electric force and electric field. I didn't know that I should be subtracting the negatively charged field, I thought it would be attracting so you add it.

So I think the only thing possibly wrong are my distances. Assuming I did the distances correctly, and that I add the two positive and subtract the negative fields, I should find the correct answer right?
 
  • #4
Chi Futbol said:

Homework Statement


Consider three charges arranged as shown.


What is the magnitude of the electric field
strength at a point 2.7 cm to the left of the
middle charge? The value of the Coulomb
constant is 8.98755 × 109 N · m2/C2 . Answer
in units of N/C.

Homework Equations


I'm truly lost. I've been looking over some other places, trying to put things together, but nothing is becoming clear to me... How do the positive and negative charges affect each other? For example if I have 2 equal positive charges, the electric field should be 0 right? How do I determine it if it is 1 positive and 1 negative charge? (Both the same value -1 and +1)


The Attempt at a Solution


I was looking at some similar problems.

E1 = 8.98755e9 x 5.6e-6 / .064^2 = 1.228766602e7. I got that distance by finding the total distance between the point 2.7 cm to the left. So 3.7 + 2.7 = 6.4. That could be very wrong, I'm not sure.
E2 = 8.98755e9 x 3.2e-6 / .047^2 = 1.301953825e7
E3 = 8.98755e9 x -1.8e-6 / .017^2 = -5.597782007
first one looks right 3.7+2.7
shouldn't the 2nd be 2.7 as defined
then the last will be 4.4-2.7

best way to do these is draw the point on your diagram, then draw each line from each charg to the point, then you should be able to see what the distances are

Chi Futbol said:
Then i just decided to add all three of those together... As I said I'm really lost, and I'm quite sure none of what I did is correct. Any help?


Edit : Sorry... I inserted a picture to show the problem.
 
  • #5
Yes this worked out. Thanks for helping me clear this up.
 

1. What is the formula for finding the magnitude of electric field strength?

The formula for finding the magnitude of electric field strength is E = F/q, where E represents the electric field strength, F represents the force exerted on a charged particle, and q represents the magnitude of the charge of the particle.

2. How is the magnitude of electric field strength measured?

The magnitude of electric field strength is typically measured in units of Newtons per Coulomb (N/C) using an electric field meter or by calculating it using the aforementioned formula.

3. What factors affect the magnitude of electric field strength?

The magnitude of electric field strength is affected by the distance between the charged particles, the magnitude of the charge of the particles, and the medium through which the particles are interacting.

4. Can the magnitude of electric field strength be negative?

Yes, the magnitude of electric field strength can be negative. This indicates that the direction of the electric field is opposite to the direction of the force exerted on a charged particle.

5. How does the magnitude of electric field strength relate to the direction of the electric field?

The magnitude of electric field strength is a scalar quantity, meaning it only has magnitude and no direction. However, it is related to the direction of the electric field, which is a vector quantity. The direction of the electric field is given by the direction of the force exerted on a positively charged particle placed in the field.

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