Electric Field at a distance

In summary: The direction wasn't changed, so what could make the magnitude bigger?In summary, when the proton is moved closer to the disk, the electric field due to electron Ec increases. However, the magnitude of the net electric field Es,net due to electrons Es decreases.
  • #1
exitwound
292
1

Homework Statement



disk.jpg


Homework Equations



E=kq/r^2

The Attempt at a Solution



The electric field due to 1 of the electrons is:

E=kq/r^2
E=(9e9)(1.6E-19) / (.0325^2)
E=1.363E-6

The electric field due to the other electron is the same, correct? They should be.

So, the net electric field due to the 2 electrons is twice this, or 2.72E-6.

When I submit this, it comes back incorrect. What am I doing wrong? It's only asking for magnitude, not direction, so I'm submitting the absolute value, disregarding the direction.
 
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  • #2
exitwound said:
The electric field due to 1 of the electrons is:

E=kq/r^2
E=(9e9)(1.6E-19) / (.0325^2)
E=1.363E-6
What direction does it have?

The electric field due to the other electron is the same, correct? They should be.
The magnitude is the same.

So, the net electric field due to the 2 electrons is twice this, or 2.72E-6.
An electric field is a vector, so you must add them like vectors.

When I submit this, it comes back incorrect. What am I doing wrong? It's only asking for magnitude, not direction, so I'm submitting the absolute value, disregarding the direction.
You can certainly disregard the direction of the net field when presenting your answer, but you must consider the direction of each contribution as you add them up.
 
  • #3
So since the x-components of the Efield vectors cancel, the resulting Efields are only in the y direction. So the net magnitude from the Es vectors is 1.36E-6(cos45)*(2)? And direction would be -y?
 
  • #4
exitwound said:
So since the x-components of the Efield vectors cancel, the resulting Efields are only in the y direction. So the net magnitude from the Es vectors is 1.36E-6(cos45)*(2)? And direction would be -y?
Right!
 
  • #5
Okay I got the answer to (b) as 1.93E-6 and it's correct.

However, moving on to (d), I don't think I'm doing something right, although I can't find any error.

E=kQ/d^2 //Coulombs Law
E=(9e9)(1.6e-19) / d^2 = 1.44E-9 / d^2

d^2=z^2+r^2
d^2=.023^2 + .0023^2 = 5.34e-4

1.44E-9/5.34e-4 = 2.70E-6

2.70e-6(cos45)=1.91e-6 //This is the force along the y-axis from one electron.

1.91e-6*2=3.81e-6 //Multiply by two for two electrons. X-components cancel out.

The final answer I get, 3.81e-6 does NOT correlate with the next question, however, in which it states that the field from (d) should be less than the field from (b). 3.81e-6 > 1.93e-6.

What am I missing?

(I absolutely HATE these online homework submissions.)
 
  • #6
exitwound said:
Okay I got the answer to (b) as 1.93E-6 and it's correct.
Good.

However, moving on to (d), I don't think I'm doing something right, although I can't find any error.
Better tell me what (d) is asking for. (May as well post the entire problem.)
 
  • #7
OH! HA. I forgot I cropped the image :)

(a) At the proton's location, what is the magnitude of the electric field Ec due to electron Ec?

Answer: 2.72E-6 which is correct.

(b) At the proton's location, what is the magnitude of the net electric field Es,net due to electrons Es?

Answer: 1.93E-6 which is the one we worked on and is correct.

(c) The proton is then moved to a distance z = R/10. What then is the magnitude of Ec at the proton's location?

Answer: 2.72E-4 which is correct.

(d) What is the magnitude of Es,net at the proton's location after it has moved to a distance z = R/10?



(e) From (a) and (c) we see that as the proton moves nearer to the disk, the magnitude of Ec increases. Why does the magnitude of Es,net decrease, as we see from (b) and (d)?
 
  • #8
This looks like a problem from Serway.

Anyway you would have to calculate individual vector forces on the proton, resolve them into components and then add .
 
  • #9
Here's the problem:
exitwound said:
2.70e-6(cos45)=1.91e-6 //This is the force along the y-axis from one electron.
When the proton moves, the angle is no longer 45 degrees.
 
  • #10
Understood. But it still doesn't work out right:

tan theta = .0023/.023
theta = 5.71 degrees

2.696E-6 (cos 5.71) * 2 = 5.36E-6

According to that answer, it's INCREASING from the original 1.93E-6 which contradicts what (e) is asking.
 
  • #11
exitwound said:
Understood. But it still doesn't work out right:

tan theta = .0023/.023
theta = 5.71 degrees

2.696E-6 (cos 5.71) * 2 = 5.36E-6

According to that answer, it's INCREASING from the original 1.93E-6 which contradicts what (e) is asking.
You have the wrong component: cos(theta) gives you the x-component. (When the angle was 45 it didn't matter, since sin(45) = cos(45) )
 
  • #12
Ooooookay. Again.

tan theta = .0023/.023
theta = 5.71 degrees
90-5.71=84.29 //The other angle in the triangle. I HATE upside down triangles.

2.696E-6 (cos 84.29) * 2 = 5.36E-6 \\I could have used (sin 5.71) too, right?

The correct answer is: 5.37E-7, which indeed is smaller than the original.

To answer (e) though, I'm not so sure why. The reason why (c) is bigger than (a) is because the distance is decreasing. The distance is limiting to zero as P-->Ec. As the denominator gets closer to zero, the value of the efield would be increase.

However, I can't grasp why the efield due to Es got smaller.
 
  • #13
exitwound said:
Ooooookay. Again.

tan theta = .0023/.023
theta = 5.71 degrees
90-5.71=84.29 //The other angle in the triangle. I HATE upside down triangles.

2.696E-6 (cos 84.29) * 2 = 5.36E-6 \\I could have used (sin 5.71) too, right?
Yes, since sinθ = cos(90-θ).

The correct answer is: 5.37E-7, which indeed is smaller than the original.

To answer (e) though, I'm not so sure why. The reason why (c) is bigger than (a) is because the distance is decreasing. The distance is limiting to zero as P-->Ec. As the denominator gets closer to zero, the value of the efield would be increase.

However, I can't grasp why the efield due to Es got smaller.
For Es, what matters is the y-component. As the angle with the horizontal gets smaller, so does the y-component.
 
  • #14
That makes sense now. I was thinking about it mathematically, and not visually.

Thanks for all the help. I'm sure I'll need more.
 

What is an electric field at a distance?

An electric field at a distance refers to the strength and direction of the electric force that a charged object exerts on another charged object at a distance. It is represented by vectors and is dependent on the magnitude and distance of the charges.

How is the electric field at a distance calculated?

The electric field at a distance is calculated by dividing the electric force by the charge of the object. This is represented by the equation E = F/q, where E is the electric field, F is the electric force, and q is the charge of the object.

What factors can affect the electric field at a distance?

The electric field at a distance can be affected by the magnitude of the charges, the distance between the charges, and the medium between the charges. Other factors such as the presence of other charges, conductors, and insulators can also affect the electric field.

How does the direction of the electric field at a distance relate to the direction of the electric force?

The direction of the electric field at a distance is always in the direction of the electric force. This means that if the electric field is positive, the electric force is also positive, and vice versa. The direction of the electric field can be determined by the direction that a positive test charge would move when placed in the field.

What are some real-life applications of understanding the electric field at a distance?

Understanding the electric field at a distance is crucial in many areas of science and technology. It is used in the design of electronic devices, power grids, and medical equipment. It is also important in understanding natural phenomena such as lightning and the behavior of charged particles in space.

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