Finding the position of an electric field.

In summary: Now calculate the magnitude along this line as (160,000)2 + (80,500)2 with your -10nC charge and you will have your distance.The reason you got the first one right was because it was on a line || to the x-axis passing through 2,1 and you didn't need to account for the distance as the sum of the squares.
  • #1
jonyamikun
4
0

Homework Statement


A -10.0 nC charge is located at position (x,y) = (2cm, 1cm).

At what (x,y) position(s) is the electric field:

Here is where they give me the different electric fields, one of which is: 161,000[tex]\hat{i}[/tex] - 80,500[tex]\hat{j}[/tex] N/C

The answers must be in cm.

Homework Equations


[tex]\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}[/tex]

The Attempt at a Solution


What I first did was plug in the the numbers I'm given:

E = 161000
epsilon_0 = 8.85x10^-12
q = 10x10^-9

Then I try to get r^2 by itself by multiplying it with both sides and then dividing both sides by 161000.

I now have: [tex]r^2 = \frac{10\times10^{-9}}{4\pi(8.85\times10^{-12})(161000)}[/tex]

Which gives me r^2 = 5.85*10^-4

Squarerooting the answer gives me 0.0236 meters which converted to cm is 2.36 rounded to 2.4.

Taking 2.4, I have added to the initial 2cm in the beginning of the problem to get 4.4 and have tried subtracting from 2 to get -0.4 but both are wrong. I did the first problem (E = -225,000 i-hat) the exact same way and I got the correct answer. My friend tried it out and also got the same answers as me.

http://i385.photobucket.com/albums/oo297/jonyamikun/2664.jpg
 
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  • #2
I don't understand your problem statement.

You have a charge.

What is the other electric field? What is the field you cite? What is the 225,000 from you mention at the end?

What exactly are you trying to find?
 
  • #3
LowlyPion said:
I don't understand your problem statement.

You have a charge.

What is the other electric field? What is the field you cite? What is the 225,000 from you mention at the end?

What exactly are you trying to find?

I'm sorry if my description was too vague. Here is a picture if it helps.

http://i385.photobucket.com/albums/oo297/jonyamikun/2664.jpg
 
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  • #4
jonyamikun said:
I'm sorry if my description was too vague. Here is a picture if it helps.

http://i385.photobucket.com/albums/oo297/jonyamikun/2664.jpg

Ohhh. I see. You have a point charge at 2,1. If the field strength is xxx, where are you measuring the field from? Got it.

What you have then is an r-hat. The direction is given by the ratio of the i,j components passing through 2,1 and the magnitude of r or |r| is given by the sum of the squares away from 2,1.
 
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  • #5
LowlyPion said:
Ohhh. I see. You have a point charge at 2,1. If the field strength is xxx, where are you measuring the field from? Got it.

What you have then is an r-hat. The direction is given by the ratio of the i,j components passing through 2,1 and the magnitude of r or |r| is given by the sum of the squares away from 2,1.

Sorry but I can't seem to figure out how this will help me solve the problem.

I clicked on "show answer" for parts C and D and the answers are x=0, and y=2. This just doesn't make sense to me. Maybe I'm interpreting something wrong.
 
  • #6
jonyamikun said:
Sorry but I can't seem to figure out how this will help me solve the problem.

I clicked on "show answer" for parts C and D and the answers are x=0, and y=2. This just doesn't make sense to me. Maybe I'm interpreting something wrong.

Your direction is determined by 1) the sign of the charge. It will be pointing toward 2,1 at all points. And 2) in the case of the 160i and -80.5j it lays on a line sloped 2x/1y pointing in from the top left. (Draw a diagram in the x,y plane.) This means that it intersects the y-axis at (0,2).

Now calculate the magnitude along this line as (160,000)2 + (80,500)2 with your -10nC charge and you will have your distance.
 
  • #7
The reason you got the first one right was because it was on a line || to the x-axis passing through 2,1 and you didn't need to account for the distance as the sum of the squares.
 

Related to Finding the position of an electric field.

1. What is an electric field?

An electric field is a physical phenomenon created by electrically charged particles. It is a vector field, meaning it has both magnitude and direction, and is responsible for the force exerted by one charged particle on another.

2. How do you find the position of an electric field?

The position of an electric field can be found by using a process called field mapping. This involves measuring the electric field at different points in space using specialized instruments, and then creating a visual representation of the field lines. The position of the field can also be calculated using mathematical equations, taking into account the charges and distances involved.

3. What is the unit of measurement for electric fields?

The unit of measurement for electric fields is volts per meter (V/m). This measures the strength of the field at a specific point in space.

4. How does the position of an electric field affect the strength of the field?

The position of an electric field can greatly affect its strength. As the distance from the source of the field increases, the strength of the field decreases. This is known as the inverse square law, where the intensity of the field is inversely proportional to the square of the distance from the source.

5. What are some practical applications of finding the position of an electric field?

Knowing the position of an electric field has many practical applications. It is essential in designing and operating electrical equipment, such as motors and generators. It is also used in medical imaging, such as MRI machines, and in telecommunications for signal transmission. Understanding the position of an electric field is crucial in ensuring the safety and efficiency of various technological devices.

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