# How is force applied on a charge moving through an electric field constant?

When a charged particle like an electron moves through an electric field (gravity-free space) with a uniform horizontal velocity, it is subject to a force given by the equation:
$$\vec F = q \vec E$$ where q is the charge carried by the electron or particle and E is the Electric Field Intensity.
Here, Electric Field Intensity is inversely proportional to the square of the distance from electron to the side applying the force.

$$\vec E = \frac{kQ}{d^2}$$

As a result, the electron tends to follow a parabolic path in the direction of the force applied as seen below.
(In this case, the electron has no initial vertical velocity and no horizontal acceleration)

Such a situation can also be seen in horizontal projectile motion under the influence of gravity where a ball is thrown horizontally from a height, where the force acting on the body is it's own weight or acceleration is due to gravity.

However, in this case, the force applied to the body is uniform throughout the motion of the projectile as acceleration due to gravity remains constant close to the surface of the Earth.

My question is as the electron moves in the direction of the force applied, doesn't the distance from the electron to the side applying the force continuously change throughout the motion of the particle and as a result isn't the electric field intensity continuously changing as well with the relation given above?
So, doesn't it affect the force as well such that it changes at every point of motion of electron unlike in the case of horizontal projectile motion where force due to gravity is always constant?

However, I have seen that the acceleration of electron in the vertical direction is considered to be constant with the relation: $$\vec a_{y} = \frac {\vec Eq}{m}$$
rather than it being variable. I would like an explanation as to why it is constant and not variable like I believe it to be.
(Source for formulas: MOTION OF A CHARGED PARTICLE IN ELECTRIC FIELD - A Detailed Analysis)

• The field between the plates is not the same as the field from a point charge. Commented Mar 24, 2021 at 10:45
• How do you answer the same question about gravitational attraction? Any bit of mass $M$ exerts a force $F=\frac{GMm}{r^2}$ on a particle with mass $m$. Commented Mar 24, 2021 at 11:36
• @BobD Yes, that is true since the plate is a uniform distribution of charges, however even if we consider the charge of each particle in the plate abstractly, aren't those charges responsible for applying the force and with changing distance from the particle, the charges are applying a force on, doesn't the magnitude of force change as a whole? Commented Mar 24, 2021 at 13:40
• @nwolijin Honestly speaking, I am not very sure about that as well now that I think about it, as it is causing a bit of confusion regarding the effect of force on the changing distances between 2 bodies, be it electrically charged or simply dude to the force of gravitation Commented Mar 24, 2021 at 13:50
• Except at the edges, the magnitude of the force between the plates does not change, if the separation of the plates is small relative to the area of the plates. This is the assumption for a parallel plate capacitor. This is the result of integrating the effect of the field contributed by each charge over the plate surface. Commented Mar 24, 2021 at 13:57

My question is as the electron moves in the direction of the force applied, doesn't the distance from the electron to the side applying the force continuously change throughout the motion of the particle and as a result isn't the electric field intensity continuously changing as well with the relation given above?

Once the electron enters the space between the plates, the force it experiences is constant.

First, the field between the plates can be considered constant except at the edges (depending on the plate area and separation distance). This is the result of integrating the field contributions of the charges on the plates. So the force on the electron after it enters the space between the plates can be considered constant upward given by

$$F=QE$$

$$E=\frac{V}{d}$$

$$F=\frac{QV}{d}$$

where $$d$$ is the plate separation and $$V$$ is the potential difference (voltage) between the plates.

Also, the force of gravity is 35 orders of magnitude less than the electrostatic force, so the gravitational force on the electron is miniscule compared to the electric force and can be ignored.

Hope this helps.

This is just an effect of computing the electric field due to an infinitely large plane.

It turns out to be generally true that the field due to a point charge decreases $$\propto \frac{1}{r^2}$$, the field due to an infinite line of charges decreases $$\propto \frac{1}{r}$$ and the field due to an infinite plane of charges is constant.

I'm not sure how familiar you are you electromagnetism, but the derivation of this result can be found in most introductory electromagnetism books.

This holds independently of whether you are considering gravity or electromagnetism. Note that when considering free fall close to the surface of the earth, we are essentially neglecting the curvature of earth and thinking of earth as just an infinitely large flat sheet of mass - which consequently leads to a constant gravitational field.