Where Did I Go Wrong in Calculating the Electric Field?

AI Thread Summary
The discussion centers on calculating the electric field at a specific point between two charges. The initial assumption was that the electric field strengths from both charges would cancel each other out, leading to an incorrect conclusion of zero net strength. The correct approach involves using vector addition to account for both the magnitude and direction of the electric fields from each charge. The participants emphasize the importance of breaking down the electric field into its x and y components for accurate calculation. The conversation highlights the need for clear problem descriptions and visual aids to facilitate understanding in physics problems.
rayhan619
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Homework Statement



a) What is the strength of the electric field at the position indicated by the dot in the figure?
b) 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.
Figure attached

Homework Equations



E = Kq/r^2


The Attempt at a Solution



At first I thought strength of the electric field on both side would be same and the net strength would be O
then I calculated the distance from the point to the charge using Pythagoras theorem and then using the above equation which give me E as 1836 N/C. so for both side it would be 1836*2 = 3673 N/C

but the answer was wrong?
where did I mess up?
 
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rayhan619 said:
a) What is the strength of the electric field at the position indicated by the dot in the figure?
...
Figure attached

where did I mess up?

I'm wondering the same thing.
 
ya i tried both 0 and 3.6*10^3
but didnt work
 
Well let's see here.

There's no figure.
There's no calculations.
There's no real description of what you are trying to calculate.

Can you imagine the difficulties others might have in trying to help you yet?
 
sorry i thought i attached the figure.

using Pythagoras theorem i figured out the distance from the point to the charge which is 7.07 cm = 0.07 m

and then i used the equation,
E = kq/r^2 = (9*10^9 Nm^2/C^2)(1*10^9 C)/(0.07 m)^2 = 1836.7 N/C
so for both side it would be 2*1836.7 N/C

sorry for the inconvenience.
 

Attachments

  • jfk.Figure.20.P24.jpg
    jfk.Figure.20.P24.jpg
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I can't see the figure yet. So do you have multiple charges? Can you describe it a little more?
 
ya i have two charges with with +1 nC charge which is 10 cm away from each other. there is a poitn in the middle 5 cm away. its like a triangle. they ask me to measure the strength and direction of the electric field.

Kinda look like this. I have attached the pic in last reply

O +1 nC
.
.
.
5 cm
.
.
.
------5cm----- o
.
.
.
5 cm
.
.
.
O +1 nC
 
Well there's the problem.

Remember the E field is a vector field. When using superposition you can't just add the |E| of both vectors.

Add x,y components only.
 
how do we figure out x component?
 
  • #10
rayhan619 said:
how do we figure out x component?

You have the angle for both E's
 
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