# Magnitude of an electric field

• Larissa McNeil
In summary: To add to that, another good habit to get into is solving problems algebraically as far as you can. It has many benefits.
Larissa McNeil

## Homework Statement

An electron is projected with an initial speed v0 = 4.65×106 m/s into the uniform field between the parallel plates in (Figure 1). The direction of the field is vertically downward, and the field is zero except in the space between the two plates. The electron enters the field at a point midway between the plates. If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
There is a diagram showing the horizontal distance as 2.00cm and the vertical distance as 1.00cm but the electric is situated at half of the vertical distance so 0.5cm.

## Homework Equations

kinematics equation: y = vt + 0.5at^2
E=F/q=ma/q[/B]

## The Attempt at a Solution

I solved for the horizontal and vertical parts using kinematics.
For horizontal: 0.02=(4.65*10^6)t which equals 4*10^-9
For vertical: (2*0.005)/(4*10^-9)^2 which equals 6.25*10^14
Then I used E=F/q=ma/q:
(9.11*10^-31)(6.25*10^14)/(1.602*10^-19)=3.55*10^3
I've had friends look at my work and they agree with it, but mastering physics keep saying that my answer is wrong. It is due this Sunday and my professor does not like to be emailed. [/B]

Please get into the habit of quoting units everywhere.
Larissa McNeil said:
For horizontal: 0.02=(4.65*10^6)t which equals 4*10^-9
A little inaccurate.

My professor showed us how to solve the problem using different numbers. I’m solving it exactly like he did but apparently it’s wrong. My friends who also have to do the homework solved it the way I am and they got it correct. I thought it might be math error but I’ve typed it into the calculator at least ten times.

Larissa McNeil said:
For horizontal: 0.02=(4.65*10^6)t which equals 4*10^-9
As @haruspex noted, you've rounded off too much here. So, your value for t is inaccurate. Later, you square t which just compounds the error. It helps to have your calculator in "scientific notation mode".

TSny said:
As @haruspex noted, you've rounded off too much here. So, your value for t is inaccurate. Later, you square t which just compounds the error. It helps to have your calculator in "scientific notation mode".
To add to that, another good habit to get into is solving problems algebraically as far as you can. It has many benefits.
Normally this means you can get the answer as a single algebraic expression before plugging in any numbers. You can then do all the calculation in one sequence on your calculator, avoiding the rounding errors that result when you type back in numbers from previous steps.

## 1. What is the definition of the magnitude of an electric field?

The magnitude of an electric field is a measure of the strength of an electric field at a specific point in space. It is typically represented by the symbol E and is measured in units of volts per meter (V/m).

## 2. How is the magnitude of an electric field calculated?

The magnitude of an electric field is calculated by dividing the electric force acting on a test charge by the magnitude of the test charge. It can also be calculated using the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the charge creating the field, and r is the distance from the charge to the point in the field.

## 3. How does the magnitude of an electric field affect charged particles?

The magnitude of an electric field determines the strength of the force experienced by charged particles in the field. The larger the magnitude of the electric field, the stronger the force on the particles will be. This force can cause charged particles to accelerate, change direction, or come to a stop depending on the direction and strength of the field.

## 4. What factors can affect the magnitude of an electric field?

The magnitude of an electric field can be affected by the distance from the source charge, the amount of charge creating the field, and the presence of other charges in the surrounding area. It can also be affected by the dielectric properties of the surrounding medium, such as its permittivity and conductivity.

## 5. What is the significance of the magnitude of an electric field in everyday life?

The magnitude of an electric field is important in various aspects of everyday life, such as in the functioning of electronic devices, the transmission of electricity through power lines, and the behavior of lightning. It is also crucial in understanding and predicting the behavior of charged particles in various natural phenomena, such as the aurora borealis and solar flares.

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