Concept of the Electric Field(picture included)

In summary, the electric field at the position indicated by the dot in the figure is strong and pointing in the direction of the positive x-axis.
  • #1
zha28
4
0
I did search for this problem found a few but they were all different and I couldn't follow the math


Homework Statement


http://session.masteringphysics.com/problemAsset/1169191/1/jfk.Figure.20.P24.jpg

What is the strength of the electric field at the position indicated by the dot in the figure?

What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line.


Homework Equations



E=K*Q/r^2
E1+E2= total

The Attempt at a Solution


8.99*10^9*1*10^-9/.05^2 = 3596 =E1
same equation for e2 get 3596 again add them together i got 7192nC

for the second problem i took a stab at it and got 0 degrees i used some logic and got that not exactly sure for the math to back that up but 0 is correct

Would appreciate any help or general tips, thanks

I'm not going to lie to you guys I have no idea what I'm doing
 
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  • #2
Hi zha28 and welcome to PF (you're the 3rd person I've welcomed tonight!) Anyway...,

Your equations are correct. The main problem with your calculation for the first question is your distance is incorrect. What is the distance between one of the charges and the dot?
 
  • #3
Thanks for the warm welcome and thanks for your quick response


The distance between the point and one of the charges is 5 cm isn't it?
 
  • #4
Not quite. If you look the point is 5cm up/down and 5cm to the right from the charges. Do you know how to use Pythagoras' theorem?
 
  • #5
A^2+B^2=c^2?


or am i going in the wrong direction.
 
  • #6
8.99*10^9*1*10^-9/.071^2 = 1783.38 times that by 2 since the formula is the same I get 3566.76 Does that sound about right? I only have 1 more guess on this problem so I'm trying to get some positiving assurance that I'm doing this properly
 
  • #7
You're correct up until multiplying the formula by 2. The issue is that, because both charges are positive, and either side of the point, the field doesn't perfectly add up. As a result, you need to calculate the x and y components of the electric field from each charge at the point. I'm guessing you're not familiar with using vectors, so you'll find this easiest if you use trigonometry.
 

FAQ: Concept of the Electric Field(picture included)

What is an electric field?

An electric field is a region in space surrounding an electrically charged object where another charged object will experience a force.

How is an electric field represented?

An electric field is represented by electric field lines, which show the direction and strength of the electric field at every point in space.

What factors affect the strength of an electric field?

The strength of an electric field is affected by the magnitude of the charged object creating the field, as well as the distance from the charged object. The strength also depends on the medium the field is passing through.

How is the direction of an electric field determined?

The direction of an electric field is determined by the direction of the force that a positive test charge would experience if placed in the field. The direction of the electric field lines always points away from positive charges and towards negative charges.

What is the equation for calculating the electric field?

The electric field can be calculated using the equation E = F/q, where E represents the electric field, F represents the force on a test charge, and q represents the magnitude of the test charge.

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