I keep getting this wrong but i think im doing it right can some1help

  • Thread starter s.bala
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In summary: C, so when you multiply them together you get a negative value. In summary, the conversation discusses the calculation of the net force on a third charge placed between two other charges on the x-axis. The charges and their positions are given, along with the necessary equations and conversion factors. There is confusion about the direction and magnitude of the calculated forces, but it is determined that the force directions should both be to the left. The calculated magnitudes are -0.000020143 and 0.000013501, with the first one being negative possibly due to the negative value of one of the given charges.
  • #1
s.bala
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i keep getting this question wrong can some1 help

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two point charges located on the x axis: one charge, q1 = -16.0 , is located at x1= -1.650m ; the second charge, q2= 38.5nC , is at the origin x=0. A
3rd charge is at x=-1.080m and its charge is 45.5nC
wat is the netforce of the charges acting on charge3?
(E_o=8.854*10^-12 therefore K=8.99*10^9)

i know net force on 3 would be to the left and all you have to do is use faradays law twice( Force 1 on 3 and Force 2 on 3) and add those up to get the net force. I keep getting -6.64*10^-6 but its wrong.. can some1help please

Fnet= Force 1 on 3+ Force 2 on 3
=(Kq_1q_3)/D^2(fromx1 to x3) + (Kq_2q_3)D^2(from x2 to x3)

and yes i converted nC int C (factor of 1*10^-9 rii?)
 
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  • #2
s.bala said:
i know net force on 3 would be to the left and all you have to do is use faradays law twice

Faraday's Law relates time varying magnetic fields to the electric fields they induce. You are thinking of Coulomb's Law.

Fnet= Force 1 on 3+ Force 2 on 3
=(Kq_1q_3)/D^2(fromx1 to x3) + (Kq_2q_3)D^2(from x2 to x3)

Careful, force is a vector. It has both a magnitude and a direction. What are the directions of F1on3 and F2on3?
 
  • #3
well, since the charge1 is negative and is on the leftside of charge3, and charge2 is positive and on the rightside of charge 3, charge 1 will exert a pull force and charge 2 will exert a push force.. so the force directions should both be to the left if I am not mistaken..ps ty for the reply
 
  • #4
Correct, so if you say that the positive direction is to the right (positive x-direction), then you would expect both F1on3 and F2on3 to be negative, correct?

Is that what you are getting when you calculate them?
 
  • #5
in terms of magnitude i get -0.000020143 and postive 0.000013501 if u consider the vecetors u would get postive 6.64*10^-6...im doing this for an online test at uft and I've treid the negative value for that and its wrong..i have only one chance left to answer it and the question says your answer can be either negative or postive so I am guessing my maginitude is wrong. soo... do u think I am correct?
 
  • #6
s.bala said:
in terms of magnitude i get -0.000020143 and postive 0.000013501

How are you getting a negative magnitude for the first one? Magnitudes of vectors are always positive by definition.
 
  • #7
gabbagabbahey said:
How are you getting a negative magnitude for the first one? Magnitudes of vectors are always positive by definition.

check my numbers please:

[(8.99x10^-12)(-16x10^-9)(45.5x10^-9)]/0.57^2
=-0.000020143
i beleieve it comes out to be negative becuase charge1 is given as -16nC
 

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What does it mean when I keep getting something wrong but I think I am doing it right?

This could mean that there is a misunderstanding or confusion about the topic or task at hand. It could also mean that there are gaps in your knowledge or approach that may need to be addressed.

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One way to improve is to practice regularly and actively engage with the material. It may also be helpful to seek feedback from others and to ask questions when something is unclear. Additionally, breaking down the task into smaller, manageable steps can also improve understanding and accuracy.

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How can I avoid getting things wrong in the future?

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