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anmolnanda
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in which case potential is distance times the electric field..please explain so that i am cleared the the basics of this.
For simple cases, where the field is constant, you can find the potential difference by multiplying the field times the distance:anmolnanda said:in which case potential is distance times the electric field..please explain so that i am cleared the the basics of this.
silmaril89 said:[tex]
\vec{E} = \vec{\nabla}V
[/tex]
Potential energy is the energy that an object possesses due to its position or configuration in a force field. In the case of electric potential energy, it is the energy that a charged object possesses due to its position in an electric field.
Electric potential energy is directly related to electric potential. Electric potential is the potential energy per unit charge at a given point in an electric field. It is a measure of the intensity of the electric field at that point.
The unit of electric potential is voltage, which is measured in volts (V). One volt is equal to one joule per coulomb (J/C).
An electric field is a force field created by electric charges. It is a region in which a charged particle experiences a force due to the presence of other charged particles.
Electric field strength is directly related to electric potential. The electric field strength at a point is equal to the negative of the rate of change of electric potential with distance at that point.