# Electric Fields and Some Kinematics

• KAT444
In summary, the electron has an initial speed of 2.7 x 10^7 m/s as it enters region I. The magnitude of the electric field E is 12500 V/m directed from the positive plate towards the negative plate. The magnitude of the electric force F is -2.0025 x 10^-15 N directed opposite to the electric field.
KAT444

## Homework Statement

An electron initially moves in a horizontal direction and has a kinetic energy of 2 x 10^3 electron volts when it is in the position shown above. It passes through a uniform electric field between two oppositely charged horizontal plates (region I) and a field free region (region II) before eventually striking a screen at a distance of .08 meters from the edge of the plates. The plates are .04 meters long and are separated from each other by a distance of .02 meters. The potential difference across the plates is 250 volts.

a. Calculate the initial speed of the electron as it enters region I.
b. Calculate the magnitude of the electric field E between the plates, and indicate its
direction
c. Calculate the magnitude of the electric force F acting on the electron while it is in
region I

## Homework Equations

KE=(1/2)mv^2
1 eV=1.6 x 10^-19 J
V=Ed
F=qE
mass of electron is 9.11 x 10^-31
charge of electron is 1.602 x 10^-19

## The Attempt at a Solution

I attempted all three parts of the solution, and am looking for clarification on my work.

a. First I changed 2 x 10^3 into 3.2 x 10^-16 J
KE=.5mv^2
3.2 x 10^-16 = .5(9.11 x 10^-31)(v^2)
2.7 x 10^7 m/s=v
b. V=Ed
250=E(.02)
12500 V/m=E
c. F=qE
F=(-1.602 x 10^-15)(12500)
F=2 x 10^-15

Any help anyone could give me would be great, especially if I did parts b and/or c wrong.

Hello, thank you for your post. Let me review your work and provide some feedback.

a. Your calculation for the initial speed of the electron is correct. Good job!

b. Your calculation for the magnitude of the electric field is also correct. However, you did not indicate the direction of the electric field. Remember, the direction of the electric field is from positive to negative charges. In this case, since the electron is negatively charged, the electric field would be directed from the positive plate towards the negative plate. So, the complete answer would be: The magnitude of the electric field E= 12500 V/m directed from the positive plate towards the negative plate.

c. Your calculation for the magnitude of the electric force is incorrect. First, you used the wrong value for the charge of the electron. It should be -1.602 x 10^-19 C. Also, the units for force should be in Newtons (N), not Coulombs (C). So, the correct calculation would be:

F = (-1.602 x 10^-19 C)(12500 V/m)
F = -2.0025 x 10^-15 N

Note that the negative sign indicates that the electric force is directed in the opposite direction of the electric field, which makes sense since the electron is negatively charged and the electric field is directed towards the positive plate.

Overall, your work is good and you have the right approach to solving these types of problems. Just be careful with units and make sure to indicate the direction of the electric field. Keep up the good work!

Your calculations for parts a and b are correct. However, for part c, the magnitude of the electric force should be F = qE = (-1.602 x 10^-19)(12500) = -2 x 10^-15 N. The negative sign indicates that the force is in the opposite direction of the electric field, which makes sense since the electron has a negative charge. Overall, your solution is correct and shows a clear understanding of the concepts of electric fields and kinematics. Good job!

## 1. What is an electric field?

An electric field is a physical quantity that describes the influence that an electric charge has on other charges within its vicinity. It is represented by a vector and is measured in units of newtons per coulomb (N/C).

## 2. How is an electric field created?

An electric field is created by a charged object. When an electric charge is present, it creates an electric field in the space surrounding it. This electric field exerts a force on other charged objects within its vicinity.

## 3. What is the relationship between electric fields and kinematics?

Electric fields can affect the motion of charged particles through the force they exert. This can be described using the principles of kinematics, such as acceleration, velocity, and displacement.

## 4. How do electric fields and kinematics relate to each other in practical applications?

In practical applications, electric fields are often used to manipulate and control the motion of charged particles, such as in particle accelerators or electronic devices. Understanding the relationship between electric fields and kinematics is crucial in designing and optimizing these applications.

## 5. How can we calculate the strength of an electric field?

The strength of an electric field can be calculated by dividing the force exerted on a test charge by the magnitude of the test charge. This is represented by the equation E = F/q, where E is the electric field strength, F is the force exerted on the test charge, and q is the magnitude of the test charge.

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