Resolving Vectors for Electric Charges

AI Thread Summary
The discussion focuses on calculating the electric field strength at point P due to two point charges of 4x10^-6C, positioned 1m apart. The user is unsure about using trigonometric functions to resolve vectors and seeks clarification on whether to apply sine or cosine. After some back-and-forth, it is confirmed that the x components of the electric field cancel out, while the y components can be doubled for the final calculation. The user recalculates the electric field strength and realizes a previous error in their digits, arriving at a corrected value of 6.2284x10^4 N/C. The conversation emphasizes the importance of precise calculations and understanding vector components in electric field problems.
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Homework Statement


Hi there, I am having a lot of trouble with this problem. Any help much appriciated.

Thanks inadvance.

Two point charges, each of 4x10-6C are placed 1m away at points A and B respectively. Calculate the electric field strength at a point P, which is a distance of 1m from both A and B in the cases when:

i) both charges are positive
ii) one is positive and one is negative


Homework Equations


E=kQ/r2


The Attempt at a Solution


Ok well I know that I need to work out a resultant, however I am not sure whether to use sin or cos, and even then how to use them to get the correct answer!

Thanks again for any input

Mike
 
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When you separate a single vector into two perpendicular components, the three lines form a triangle, yes? Then you can easily apply your trigonometric relations easily with reference to the diagram. E sin theta would give the component opposite to the angle, while E cos theta gives the component adjacent to the angle used.
 
Oh I see, so since the x components would cancel, you would just double the y component which is cos30*the elctric field strength calculated

i.e. 6.28104NC-1

Thanks alot, do you know if the answer I have calculated is correct?
 
That's for part i) I presume?
My answer appears to differ from yours from the 3rd significant digit; what value are you using for k?
 
Hello, I used 8.99x109 for k.

Thanks again!

Mike
 
Hmm...press your calculator again, you appear to have read some digits wrongly in the answer you provided above.
 
You are quite right!

I got 6.2284x104 this time!
 
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