- #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.
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.
What's dV/dr for that potential? Compare that to E.shirel said:V=kq/r
An electric field is a region in space where an electrically charged particle experiences a force. It is created by a source charge and is represented by electric field lines that point away from positive charges and towards negative charges.
The strength of an electric field at a point is determined by the magnitude and direction of the source charge and the distance between the source charge and the point. The formula for calculating electric field is E = kQ/r^2, where k is the Coulomb's constant, Q is the source charge, and r is the distance between the source charge and the point.
Electric potential is the amount of work required to move a unit positive charge from infinity to a point in an electric field. It is measured in volts and is a scalar quantity, meaning it has magnitude but no direction.
While electric field represents the force experienced by a charged particle, electric potential represents the energy of the charged particle. Electric potential is related to electric field through the equation V = E*d, where V is electric potential, E is electric field, and d is the distance between the point and the source charge.
The relationship between electric field and electric potential is given by the equation E = -dV/dr, where E is electric field, V is electric potential, and r is the distance between the point and the source charge. This equation shows that electric field is the negative gradient of electric potential, meaning that it points in the direction of decreasing potential.