Direction and magnitude question again

AI Thread Summary
A charge of -6mC is placed at each corner of a square with a side length of 0.1m, and the task is to determine the force on each charge. The initial calculations for the forces between the charges were incorrect due to neglecting the influence of all charge interactions, particularly F13. After recalculating the forces and their components, the correct resultant force was found to be 2.96 x 10^7 N, with an angle of 315 degrees towards the center of the square. The discussion highlights the importance of considering all charge interactions and accurately calculating vector components. The problem was ultimately resolved through persistence and detailed recalculation.
excelsion
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Homework Statement


A charge of -6mC is placed at each corner of a square .1m on a side. Determine the magnitude and direction of the force on each charge.
upper left is 1, bottom left is 2 upper right is 4 bottom right is 3
http://img208.imageshack.us/img208/7318/payxhbx6.jpg

Homework Equations


F=k (Q1Q2)/r^2

The Attempt at a Solution


sigh I am back again..
I start with F12 so i did:
9 X 10^9(6 X 10^-3 X 6 X 10^-3/.1^2) and i got
3.24 X 10^7

since all the charges are the same the answer is going to be the same for all the charges so F14 is also 3.24 X 10^7.

Now i break F12 into x and y components:
since 1 and 2 is repulsive there is no x component so the y component is is +3.24 X 10^7N

as for F14 there is no y component so x is -3.24 X 10^7.

now i do resultant formula:
total x is -3.24 X 10^7 and total y is +3.24 X 10^7
so the resultant should be... 4.58 X 10^7 right the book get the answer as 2.96 X 10^7

where did i go wrong?
 
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You didn't do F13.
 
F13 affects it too??! wow how did i miss that ok so i do F13 is the same thing but the x and y comps are different:
3.24 X 10^7cos45=2.3 X 10^7
3.24 X 10^7sin45=2.3 X 10^7

the x component is negative and the y is postive since its moving north west..
now total x is -5.72 X 10^7 and total y is 5.53 X 10^7 then i do resultant formula and i get... 7.95 X 10^7 argghh what did i do wrong books answer is 2.96
 
excelsion said:
F13 affects it too??! wow how did i miss that ok so i do F13 is the same thing but the x and y comps are different:
3.24 X 10^7cos45=2.3 X 10^7
3.24 X 10^7sin45=2.3 X 10^7

the x component is negative and the y is postive since its moving north west..
now total x is -5.72 X 10^7 and total y is 5.53 X 10^7 then i do resultant formula and i get... 7.95 X 10^7 argghh what did i do wrong books answer is 2.96

For F13 the total force is not 3.24x10^7... because the r isn't 0.1.

Still the force is going to be higher... does it say 2.96x10^7, or 2.96x10^8?
 
its 10^7
and i did the F13 over:
9 X 10^9(6 X 10^-3 X 6 X 10^-3/.14^2) and i get 1.63 X 10^7
find the component:
1.63 X 10^7cos45=1.15 X 10^7
1.63 X 10^7sin45=1.15 X 10^7

now the total X is
-1.15X10^7 + 0 + -3.24X10^7 = -4.39 X 10^7

total y is
1.15X10^7 + 0 + 3.24X10^7= 4.39 X 10^7
making the resultant:
6.2 X 10^7 ... why can't i get this thing!
 
You sure the question is exactly as you've posted it? I'm getting the same answer as you.
 
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physics is making go mad I am sorry... the NE and SW charges are positve...so let me try this again..
F12 is 3.24 X 10^7
x comp:0
y comp:-3.24 X 10^7

F14 is 3.24 X10^7
x comp:+3.24 X 10^7
y comp:0

F13 is 1.63 X 10^7
x comp:-1.15 X 10^7
y comp:1.15 X 10^7

Total X is 2.09 X 10^7
Total Y is -2.06 X 10^7

Resultant is 2.96 X 10^7(Finally got it)
and angle is found by doing tan inverse total y/total x and that is -45 deg
subtract that from 360 give me 315 degs pointing towards middle of square!

thanks for your patience lol see you tomorrow night most likely!
 
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