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Mathematics 82 Online
OpenStudy (greatlife44):

Physics question (conceptual)

OpenStudy (greatlife44):

for an electric field. can someone explain this? \[E_{Q} = k(\frac{ |Q| }{ r^{2} })\]

OpenStudy (greatlife44):

@Michele_Laino if you aren't busy

OpenStudy (michele_laino):

your formula, expresses the magnitude of the vector electric field generated by an electric charge \(Q\) which is at rest

OpenStudy (greatlife44):

So when they say source charge that's only considering an electric field produced by just one charge?

OpenStudy (michele_laino):

here is the corresponding picture: |dw:1464119630539:dw|

OpenStudy (michele_laino):

An electric field can be generated by one charge or more than one charge

OpenStudy (michele_laino):

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

OpenStudy (greatlife44):

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?

OpenStudy (michele_laino):

yes! it is the distance between the center of mass of the charge \(Q\) and the point P of observation

OpenStudy (greatlife44):

how was q removed from the original formula?

OpenStudy (greatlife44):

i know we are only looking at the source charge

OpenStudy (michele_laino):

since the definition of the vector field, is: \[\left| {\mathbf{E}} \right| = \frac{{\left| {\mathbf{F}} \right|}}{q} = k\frac{{\left| Q \right|q}}{{{r^2}}}\frac{1}{q} = k\frac{{\left| Q \right|}}{{{r^2}}}\]

OpenStudy (greatlife44):

So we divided both sides by q \[\frac{ F }{ q } = \frac{ |Q|k_{0} }{ r^{2} }\]

OpenStudy (michele_laino):

wherein \(q\) is a little positive test charge

OpenStudy (michele_laino):

that's right!

OpenStudy (greatlife44):

okay great, I think I got this a little better now

OpenStudy (michele_laino):

:)

OpenStudy (greatlife44):

thank you

OpenStudy (michele_laino):

:)

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