Three-charge Electric Field Triangle.

[physphun531]

Homework Statement

http://img183.imageshack.us/img183/7473/56823155cm1.th.jpg
Lets all the top positive 6 microC charge, 'A', and the bottom right negative -4.2 charge 'B'.

Homework Equations

E = F / q

F= Ke q1q2
--------
r²

The Attempt at a Solution

I calculated the Force exerted on charge 'q' by 'A' and came up with 0.153 N and then used the cosine of 60 to obtain -0.77 as the x-component force and the sin of 60 to obtain -0.133 as the y-component force. Then I calculated the force exterted by 'B', the negative charge and obtained 0.107 in the 'i' direction giving me a sum force of 0.03 in the x direction. This SHOULD have given me part B, except two reasons why it was wrong. One, the units there say mN, but that did not work either, Two, I have done something conceptually wrong and cannot see it.
As for part A, I was going to use the above in E = F / q, but I first need the above part correct and even then I am not sure which charge is q.

 

[physphun531]

Picture is up.

If anyone can see where I have gone astray can you help me out please?

 

[alphysicist]

Hi physphun531,

Homework Statement

http://img183.imageshack.us/img183/7473/56823155cm1.th.jpg
Lets all the top positive 6 microC charge, 'A', and the bottom right negative -4.2 charge 'B'.

Homework Equations

E = F / q

F= Ke q1q2
--------
r²

The Attempt at a Solution

I calculated the Force exerted on charge 'q' by 'A' and came up with 0.153 N and then used the cosine of 60 to obtain -0.77 as the x-component force and the sin of 60 to obtain -0.133 as the y-component force. Then I calculated the force exterted by 'B', the negative charge and obtained 0.107 in the 'i' direction giving me a sum force of 0.03 in the x direction.

How did you get this number for the force in the x direction? It seems rather small to me (and in the wrong direction).

 

[physphun531]

I used the 0.153 N force calculated and multiplied it by (0.45/0.9) for 'A' and then just combined with 'B'. They did it the same way in my textbook.

 

[alphysicist]

I used the 0.153 N force calculated and multiplied it by (0.45/0.9) for 'A' and then just combined with 'B'. They did it the same way in my textbook.

That's right; I was lead astray by a typo in your post (you said the x component of the first vector was -0.77, but it should be -0.077). I should have noticed that was a typo.

Unless I'm reading it incorrectly your procedure looks right to me. Are you keeping enough digits in your answer? The error in using 30mN (for example) instead of the more accurate answer seems to be almost 2 percent. Do you know how close you have to be?

 

[physphun531]

http://img240.imageshack.us/img240/7827/75926219lv4.th.jpg

Updated. You have to pretty darn close. Cengage is picky. one one problem I was slightly over 1% off and it didn't take my answer.

 

[alphysicist]

http://img240.imageshack.us/img240/7827/75926219lv4.th.jpg

Updated. You have to pretty darn close. Cengage is picky. one one problem I was slightly over 1% off and it didn't take my answer.

Does that mean it is saying the 30.7mN is incorrect? If so, try not rounding anything until the very end. On some other systems I have also heard instructions to input at least four or five digits in your answer (I think it was four) so that might make a difference.

 

[physphun531]

http://img155.imageshack.us/img155/3967/17263012kt7.th.jpg

Done... I hate cengage

 

[alphysicist]

http://img155.imageshack.us/img155/3967/17263012kt7.th.jpg

Done... I hate cengage

That does seems strict to me. I think your last incorrect response was only a bit over two-tenths of one percent off. I would definitely suggest keeping four digits in your answers.