Physics question (conceptual)
for an electric field. can someone explain this? \[E_{Q} = k(\frac{ |Q| }{ r^{2} })\]
@Michele_Laino if you aren't busy
your formula, expresses the magnitude of the vector electric field generated by an electric charge \(Q\) which is at rest
So when they say source charge that's only considering an electric field produced by just one charge?
here is the corresponding picture: |dw:1464119630539:dw|
An electric field can be generated by one charge or more than one charge
since we have to get a \(magnitude\), it is necessary to consider the absolute value of the charge, since, as you know, an electric charge can be positive or negative as well
I see. you know in the other formula k0(q1q2)/r^2 r^2 is the distance between the two charges right so r here is the distance between the source charge and point p?
yes! it is the distance between the center of mass of the charge \(Q\) and the point P of observation
how was q removed from the original formula?
i know we are only looking at the source charge
since the definition of the vector field, is: \[\left| {\mathbf{E}} \right| = \frac{{\left| {\mathbf{F}} \right|}}{q} = k\frac{{\left| Q \right|q}}{{{r^2}}}\frac{1}{q} = k\frac{{\left| Q \right|}}{{{r^2}}}\]
So we divided both sides by q \[\frac{ F }{ q } = \frac{ |Q|k_{0} }{ r^{2} }\]
wherein \(q\) is a little positive test charge
that's right!
okay great, I think I got this a little better now
:)
thank you
:)
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