Electric field and potential

  • Thread starter shirel
  • Start date
  • #1
shirel
38
0
Hi,

I don't understand why dV/dr=-E

As far as I know:
V=kq/r
E=Kq/(r^2)
so V/r becomes to be E and not -E.
 

Answers and Replies

  • #2
Doc Al
Mentor
45,425
1,876
V=kq/r
What's dV/dr for that potential? Compare that to E.
 
  • #3
shirel
38
0
ohh I should derivate the function!! thank you :)
 

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