anonymous
  • anonymous
vector addition in coulomb's law is confusing for me..to find the point where E is zero for a configuration of +3c and -1c ,should I put two minus signs for E due to -1c(one for negative charge and one for negative direction of the field there)?
MIT 8.02 Electricity and Magnetism, Spring 2002
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1378656634548:dw| is this correct..considering the value of R is different from each other..
anonymous
  • anonymous
@.Sam.
anonymous
  • anonymous
When you put a vector in an equation (thus adding those minus signs that are confusing you) you have to be careful: The sign of a vector is not uniquely defined by his sole existence. It depends updon the frame of reference. Think of this analogue problem for example, if you wish to find the push that you have to apply to a weight of 2kg in order for it to stay still on the Earth (at surface height, so let's assume g=costant) you could setup a frame of reference, with the positive vertical axis pointing UP at the sky. The equation for equilibrium in this case would be \[|\vec F_{\text{hand}} | - |\vec P| = 0\] But if you oriented the frame of reference, so that the vertical axis would point towards the center of the Earth, the equation would become: \[-|\vec F_{\text{hand}} | + |\vec P | = 0\] So, is I understood your problem, you must fix a frame of reference first. (a clever one for your purpose) Then, assign the plus or minus signs of the values of the vectors in the equation, according to their verses. I sketched this in 1 minute: (see file) For me, drawing down some field lines helps visualizing the problem a lot. I saw for example that the point you're looking for could never be in between the charges! Hope i understood the problem and explained clearly.
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anonymous
  • anonymous
I even forgot k in the equation because I was 'forethinking' it would have cancelled out, hehe
anonymous
  • anonymous
ok ..vector sign understood..but what abt the negative sign of -1 coulomb?
anonymous
  • anonymous
What you can say about an E field without setting up a frame of reference is that it always points toward the charge if the charge generating it is negative, and away from the charge if it's negative. With the frame of reference that I've chosen in the picture, I've put a minus sign before it in the equation, because it pointed in the opposite direction of the x-axis. Maybe I understood what you're asking, it's quite radical isn't it? The fact that we consider a negative charge the one that attracts or is attracted by a positive charge is a mathematical thing. In fact you could switch the two and nothing would change, it would be like rotating your sheet of paper; nature behaves in the same way, we try to make a mathematical model for it. You are being correct when you put the minus sign in the expression of the E-field generated by the negative charge \[E_{\ominus} = k_0 \frac{-1C}{r^2}\] but the important thing is that it REVERSES direction when you switch the sign. The fact that a field with a minus sign points in a direction or in the opposite, depends merely on your frame of reference. Are you convinced? The best thing to do is waste some sheets of paper drawing down charges and fields. Maybe you could start from Coulomb's force, and then find the field by doing F/q. You'll notice that your questions is very deep indeed. \[E=\lim_{q_2\to 0} \frac{Fc}{q_2}\]
anonymous
  • anonymous
ok I think Iam getting a hang of this..in my book..they are assuming the distance vector as always pointing towards the charge which we are considering and substituting the negative sign for unlike charges and positive for like charges in numerator...so all is in choosing the frame of reference and consistently maintaining a particular convention...
anonymous
  • anonymous
what do you mean by 'the distance vector'?
anonymous
  • anonymous
r(the distance btwn the charges)
anonymous
  • anonymous
remember that distance is also +ive

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