Magnitude of electric force


by bastige
Tags: electric, force, magnitude
bastige
bastige is offline
#1
Mar23-08, 07:19 PM
P: 15
1. The problem statement, all variables and given/known data
Four point charges are placed at the four corners of a square. Each side of the square has a length L.
Find the magnitude of the electric force on q2 due to all three charges q1 , q2 , q3
and q4. Given L=1 and q= 1.38 μC. Answer in N


http://rs209gc2.rapidshare.com/files...f5e75220df.jpg
Photo of problem ^

2. Relevant equations



3. The attempt at a solution
http://rs220l32.rapidshare.com/files...3_19-17-38.jpg
Photo of work ^
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dynamicsolo
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#2
Mar24-08, 05:48 PM
HW Helper
P: 1,664
Since you give units of N and microcoulombs in your statement, you are presumably working in SI units. Right off, I'd say check your value for k: I believe it is off by three orders of magnitude... (In your statement, I think you want the net force on q2 from the other three charges, so you don't want to type 'q2' again...)

Also, watch your forces and components. For starters, what is the distance along the diagonal of the square? What is the force between the diagonally opposed charges?
bastige
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#3
Mar25-08, 06:07 PM
P: 15
Well the diagonal is also 1. K should = 8.99E9. I still get the wrong answer.

dynamicsolo
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#4
Mar25-08, 07:13 PM
HW Helper
P: 1,664

Magnitude of electric force


Quote Quote by bastige View Post
Well the diagonal is also 1. K should = 8.99E9. I still get the wrong answer.
If the side of a square is 1, how long is the diagonal?
bastige
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#5
Mar25-08, 08:50 PM
P: 15
A^2 + B^2 =c^2
1^2+1^2=c^2
2=c^2
1=c
Snazzy
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#6
Mar25-08, 09:06 PM
P: 459
Square root of 2 is 1?
bastige
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#7
Mar25-08, 09:49 PM
P: 15
Quote Quote by Snazzy View Post
Square root of 2 is 1?
hmm...good point, guess i shouldn't do this in my head. well, its 1.41421. Where Do i need to use it? I didn't have r in my final formula? where did i go wrong?
dynamicsolo
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#8
Mar25-08, 09:54 PM
HW Helper
P: 1,664
Quote Quote by bastige View Post
hmm...good point, guess i shouldn't do this in my head. well, its 1.41421. Where Do i need to use it? I didn't have r in my final formula? where did i go wrong?
The value for c you just calculate is what you will use in Coulomb's Law for the force between the diagonally opposite charges. Since the formula calls for r^2, you could use r^2 = 2 there.
bastige
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#9
Mar25-08, 10:11 PM
P: 15
Quote Quote by dynamicsolo View Post
The value for c you just calculate is what you will use in Coulomb's Law for the force between the diagonally opposite charges. Since the formula calls for r^2, you could use r^2 = 2 there.

??????I'm sorry, I'm physics illiterate
dynamicsolo
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#10
Mar26-08, 01:55 PM
HW Helper
P: 1,664
You have the equation for the force

kqq/(r^2) , which is Coulomb's Law.

In the image you show for your work, the force for the "southeast" diagonal should be using 2 in the denominator, instead of 1, since the distance squared between the two charges q2 and q4 is 2 .


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