What is the relationship between the maginitude of the radial electric field an

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  • #1

Homework Statement


What is the relationship between the magnitude of a radial electric field and and its associated potential? Choose the correct general relationship?


Homework Equations



The following alternatives are given:
a) E(r)= dV(r)/dr
b) E(r)= V(r)/r
c) E(r)= -dV(r)/dr
d) E(r)= -V(r)/r

The Attempt at a Solution



I know the electric field is proportional to the inverse of (r^2) and the potential is proportional to the inverse of (r). My problem lies in understanding the calculus of the above formulas. Can anyone guide me through them please?
 
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Answers and Replies

  • #2
Delphi51
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Welcome to PF, physicsMad!
Think of any electric field, not necessarily the one from a point charge.
You have a tiny test charge q that you want to push from point A to point B. A is considered to have an electric potential of zero volts. What is the potential at B? By definition it is the work done (energy given) per unit charge to move q from A to B. So work is W = F*r where r is the distance or, since the F = qE may well be changing as you go,
dW = F*dr = qE*dr for each little increment in distance dr
The potential is the work done per charge or
dV = dW/q = E*dr
You can turn this around to get E = dV/dr.
 
  • #3
Thank you for the kind welcome Delphi51.

I can now understand much better than before due to your explanation. I really appreciate it. However, i forgot to mention that the point charge has positive charge. So, the work would be negative the change occurs from higher potential to lower potential. Am I on the right track? If my thinking is correct, I would choose equation E(r)= -dV(r)/r.

Thanks again :-)
 
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