What is the magnitude & direction of the eletric field?

In summary, the direction of the electric field at the dot in the figure is 9.32 degrees above the horizontal with a magnitude of 1.2*10^4 N/C. This was determined by calculating the field contributions from each charge and adding them as vectors, taking into account the direction and magnitude of each field component. Using similar triangles, the angle was found to be 9.32 degrees instead of the previously assumed 60 degrees.
  • #1
sunnnystrong
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6

Homework Statement


Two +3.0nC charges are shown in the diagram below which are spaced 10cm apart. What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?

What is the electric field in N/C?
What is the direction of the electric field? (above the horizontal)

Homework Equations


E = kq/r^2

The Attempt at a Solution



For the top charge:
E1 = (8.99*10^9)(3*10^-9C)/(.05m)^2 = 10788 N/C
E2 = (8.99*10^9)(3*10-9C)/(.1118m)^2 = 2157.7 N/C

Add them together and I got 1.3 * 10 ^4 N/C

My answer was wrong... was it because the two charge have the same signed charge so really the y-components cancel so the magnitude of the electric field is just 2*E1?
 

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  • #2
The electric field is a vector quantity. You need to be adding vectors, not magnitudes.
 
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  • #3
sunnnystrong said:
Add them together and I got 1.3 * 10 ^4 N/C
The field contribution from each charge is a vector. You must add them as vectors, not just numbers. Direction counts!
 
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  • #4
Oh, okay so for the E1 (Top charge):

E1 = ((8.99*10^9)(3*10-9)/(.05^2) i, 0j)

E2 (Bottom charge):
E2 = ((8.99*10^9)(3*10-9)/(.05^2) i, (8.99*10^9)(3*10-9)/(.1^2) j)

Etotal = (21576 i, 2697j)

|Etotal| = 21743.909 N/C ? I already tried this and they said it was wrong
 
  • #5
gneill said:
The electric field is a vector quantity. You need to be adding vectors, not magnitudes.

Doc Al said:
The field contribution from each charge is a vector. You must add them as vectors, not just numbers. Direction counts!

^^ see reply. I am a little confused about the directions.
 
  • #6
sunnnystrong said:
E2 (Bottom charge):
E2 = ((8.99*10^9)(3*10-9)/(.05^2) i, (8.99*10^9)(3*10-9)/(.1^2) j)
This calculation is not correct. You can't take the components of the distance vector and use them individually as distances for the calculation of the field components. You need to use the scalar distance between the charge and the location of interest in the Coulomb equation when finding the magnitude of the field, then break that magnitude into x and y components according to the vector direction. The radial distances are ##r_1## and ##r_2## in the figure below:

upload_2017-4-16_17-49-35.png
 
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  • #7
gneill said:
This calculation is not correct. You can't take the components of the distance vector and use them individually as distances for the calculation of the field components. You need to use the scalar distance between the charge and the location of interest in the Coulomb equation when finding the magnitude of the field, then break that magnitude into x and y components according to the vector direction. The radial distances are ##r_1## and ##r_2## in the figure below:

View attachment 195077

Okay so I am a little confused but for Q1 the electric field would be:
10788 N/C...

for Q2 it would be 2157.73 N/C...

The components of Q1 would be (10788i, 0j) & for Q2 it would be (2157.7cos(60) i, 2157.7sin(60) j) ?
 
  • #8
gneill said:
This calculation is not correct. You can't take the components of the distance vector and use them individually as distances for the calculation of the field components. You need to use the scalar distance between the charge and the location of interest in the Coulomb equation when finding the magnitude of the field, then break that magnitude into x and y components according to the vector direction. The radial distances are ##r_1## and ##r_2## in the figure below:

View attachment 195077

& The magnitude of the field would be 1.2*10^4 N/C & the direction would be 8.95 degrees above the horizontal?
 
  • #9
sunnnystrong said:
Okay so I am a little confused but for Q1 the electric field would be:
10788 N/C...

for Q2 it would be 2157.73 N/C...

The components of Q1 would be (10788i, 0j) & for Q2 it would be (2157.7cos(60) i, 2157.7sin(60) j) ?
Close, but is the angle really exactly 60 degrees? How did you determine the angle?

Note that when you have similar triangles (in the geometric sense of "similar"), it's often easier to just work with the ratios of known side lengths to form the trig functions. So, for example, the cosine of your angle might be formed as:

##cos(θ) = \frac{5}{\sqrt{5^2 + 10^2}}##
 
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  • #10
gneill said:
Close, but is the angle really exactly 60 degrees? How did you determine the angle?

Note that when you have similar triangles (in the geometric sense of "similar"), it's often easier to just work with the ratios of known side lengths to form the trig functions. So, for example, the cosine of your angle might be formed as:

##cos(θ) = \frac{5}{\sqrt{5^2 + 10^2}}##
ohhh! thank you so much!

the direction of the field will be 9.32 degrees above the horizontal
 
  • #11
sunnnystrong said:
the direction of the field will be 9.32 degrees above the horizontal
Yes! :approve:
 
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FAQ: What is the magnitude & direction of the eletric field?

1. What is the definition of magnitude of an electric field?

The magnitude of an electric field is a measure of the strength of the electric field at a particular point. It is represented by the symbol E and is measured in units of Newtons per Coulomb (N/C).

2. How is the magnitude of an electric field calculated?

The magnitude of an electric field can be calculated by dividing the force exerted on a test charge by the magnitude of the test charge itself. This can be written mathematically as E = F/q, where E is the electric field, F is the force, and q is the test charge.

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

The magnitude of an electric field is affected by the amount of charge present, the distance between the charges, and the medium through which the charges are interacting. The magnitude also depends on the type of charge (positive or negative) and the direction of the field.

4. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction that a positive test charge would move if placed in the field. The field lines always point in the direction that a positive charge would be pushed or pulled by the field, and the direction is represented by arrows on a diagram.

5. Can the direction of an electric field change?

Yes, the direction of an electric field can change depending on the location and the configuration of the charges. Electric fields always point away from positive charges and towards negative charges, so the direction can vary based on the arrangement of charges in a given system.

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