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

  • Thread starter Thread starter s.bala
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating the net force acting on a third charge due to two other point charges using Coulomb's Law. The user is confused about their calculations, repeatedly arriving at a value of -6.64*10^-6, which they believe is incorrect. Participants clarify that force is a vector quantity, emphasizing that magnitudes should always be positive, and they discuss the directions of forces exerted by the charges. The user is advised to check their calculations and ensure they are considering the correct signs for the forces based on charge types and positions. Ultimately, the conversation highlights the importance of correctly applying Coulomb's Law and understanding vector directions in electrostatics.
s.bala
Messages
7
Reaction score
0
i keep getting this question wrong can some1 help

--------------------------------------------------------------------------------

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?)
 
Physics news on Phys.org
s.bala said:
i know net force on 3 would be to the left and all you have to do is use faradays law[/color] 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?
 
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
 
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?
 
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?
 
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.
 
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
 
Back
Top