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An electric field gets stronger the closer you get to a charge exerting that field. Distance and field strength are inversely proportional. When distance is increased, field strength decreases. The opposite is true as well. Additionally, field strength varies as the inverse square of the distance between the charge and the observer. Double the distance and you will find that there is 1/22 or 1/4th the electric field strength as there was at the start of your experiment.
The strength of an electric field depends on the charge that causes it, and on the distance from the charge.
Electric field strength depends on direction and magnitude because it is a vector quantity.
The amount of charge that produces the field and on the distance from the charge. (Novanet)
distance between charged particles.
As the distance from a charged particle increases the strength of its electric field DECREASES.
An electric field gets stronger the closer you get to a charge exerting that field. Distance and field strength are inversely proportional. When distance is increased, field strength decreases. The opposite is true as well. Additionally, field strength varies as the inverse square of the distance between the charge and the observer. Double the distance and you will find that there is 1/22 or 1/4th the electric field strength as there was at the start of your experiment.
The strength of an electric field depends on the charge that causes it, and on the distance from the charge.
Electric field strength depends on direction and magnitude because it is a vector quantity.
The amount of charge that produces the field and on the distance from the charge. (Novanet)
distance between charged particles.
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Electric field intensity is related to electric potential by the equation E = -dV/dx, where E is the electric field intensity, V is the electric potential, and x is the distance in the direction of the field. Essentially, the electric field points in the direction of decreasing potential, and the magnitude of the field is related to the rate at which the potential changes.
The strength of the electric field approaches zero
No, the strength of the electric field decreases with distance from a charged object. The electric field follows an inverse-square law, meaning it decreases with the square of the distance from the source charge. So, the closer you are to the charged object, the stronger the electric field.
The strength of the electric field is a scalar quantity. But it's the magnitude of thecomplete electric field vector.At any point in space, the electric field vector is the strength of the force, and thedirection in which it points, that would be felt by a tiny positive charge located there.
I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.