# Acceleration in electric field

• charmedbeauty
In summary: Well, what's potential energy in an electric field?I feel that you approach is correct. It's a kinematics type problem only. Some where you might be making mistake in units.U = kQq/r
charmedbeauty

## Homework Statement

An electron with initial velocity vx0 =1.0 *10^4 m/s enters a region of width 1.0 cm where its electrically accelerated. it emerges with velocity vx = 4.0 *10^6
what was its acceleration, assumed constant?

F = ma
v^2 = u^2 + 2as

## The Attempt at a Solution

Well I thought I would try this

(v^2 - u^2)/2s =a

but this did not work.

Im guessing I am going to need to find E?

But don't know how to do this with given information.

can someone point me in the direction of the formula needed for the question?

thanks!

Last edited:
I don't think you need to actually find the field. You can think of this one as a conservation problem where it's final kinetic energy is using the final velocity and so on. You do know the mass of the electron afterall?

As far as the 1cm goes, do they specify that to be the width, as in, the x-component (the direction the electron is traveling?) or is it 1cm as in, the width of a loop or wire or something that it travels through? In which case, it's length would be mostly arbitrary and you could treat it as an instantaneous point where the velocity changes.

QuarkCharmer said:
I don't think you need to actually find the field. You can think of this one as a conservation problem where it's final kinetic energy is using the final velocity and so on. You do know the mass of the electron afterall?

As far as the 1cm goes, do they specify that to be the width, as in, the x-component (the direction the electron is traveling?) or is it 1cm as in, the width of a loop or wire or something that it travels through? In which case, it's length would be mostly arbitrary and you could treat it as an instantaneous point where the velocity changes.

not sure about the width I think it is in the x direction
as for the kinetic energy and conservation how do I link 1/2 m v^2 to acceleration?

charmedbeauty said:
not sure about the width I think it is in the x direction
as for the kinetic energy and conservation how do I link 1/2 m v^2 to acceleration?

Well, what's potential energy in an electric field?

I feel that you approach is correct. It's a kinematics type problem only. Some where you might be making mistake in units.

U = kQq/r
Still not understanding how that factors into 1/2 mv^2?

## 1. What is acceleration in an electric field?

Acceleration in an electric field refers to the rate of change of velocity of a charged particle when it is subjected to an electric field. It is a vector quantity that is affected by the strength and direction of the electric field.

## 2. How is acceleration in an electric field calculated?

The acceleration in an electric field can be calculated using the formula a = qE/m, where a is the acceleration, q is the charge of the particle, E is the strength of the electric field, and m is the mass of the particle.

## 3. What is the relationship between acceleration and electric field strength?

The acceleration of a charged particle is directly proportional to the strength of the electric field it is subjected to. This means that as the electric field strength increases, the acceleration of the particle also increases.

## 4. What factors affect the acceleration of a charged particle in an electric field?

The acceleration of a charged particle in an electric field is affected by the strength and direction of the electric field, as well as the charge and mass of the particle.

## 5. How does acceleration in an electric field differ from acceleration due to gravity?

The acceleration in an electric field is caused by the interaction between the electric field and the charged particle, whereas acceleration due to gravity is caused by the force of gravity on an object's mass. Additionally, acceleration in an electric field is affected by the charge of the particle, while acceleration due to gravity is not.

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