kuruman said:You need to sharpen your thinking. The electric field is a vector, so to specify it you need to specify its three components. These are
[itex]E_{x}=-\frac{\partial V}{\partial x}[/itex]; [itex]E_{y}=-\frac{\partial V}{\partial y}; [/itex][itex]E_{z}=-\frac{\partial V}{\partial z}[/itex]
The electric field is equal to the negative derivative of the electric potential. This means that the electric field is a measure of how the electric potential changes over distance. In other words, the electric field is the rate of change of the electric potential.
The negative derivative of electric potential is calculated by taking the negative slope of the electric potential curve. This can be done using calculus by finding the derivative of the electric potential function.
This is due to the fundamental relationship between electric potential and electric field. The electric field is a measure of the force experienced by a charged particle in an electric field, while the electric potential is a measure of the potential energy per unit charge. The negative derivative of electric potential is necessary to calculate the electric field because it takes into account the direction of the force.
A negative electric potential means that the electric potential energy at that point is lower than the reference point. This could mean that the electric field is pointing in the opposite direction of the potential gradient, or that the charged particle would experience a decrease in potential energy if it moved towards that point.
Yes, the electric field can be zero at certain points, known as equipotential points. At these points, the electric potential does not change over distance, meaning there is no force acting on a charged particle. This could occur in symmetrical systems, such as a point charge, where the electric field is zero at the center.