Electric potential and potential energy

In summary, a proton placed at rest in a uniform electric field of magnitude 1.30x10^3 N/C experiences a force of 2.08x10^-16 N. Using Newton's second law, the acceleration of the proton is calculated to be 1.24x10^11 m/s^2. To find the time it takes for the proton to move 1 cm in this field, the equation x = 1/2at^2 can be used, with the acceleration found in part (b). The mistake in the calculation was using the electric field strength instead of the acceleration in the equation. The correct calculation yields a time of approximately 1.58x10^-6 seconds.
  • #1
dpogre
4
0
A proton is initially at rest (in a vacuum) in a uniform electric field of magnitude 1.30 10^3 N/C.
(a) Calculate the force applied to the proton by the field. ANSWER: 2.08E-16 N

(b) Apply Newton's second law to calculate the acceleration of the proton. (Don't be surprised if the acceleration is large. It isn't applied for a very long time.) ANSWER: 1.24E11 m/s^2

(c) Calculate the time it takes for the proton to move 1 cm in this field, starting from rest. (Hint: The acceleration is uniform.)
in part c i cannot figure out how to find the uniform acceleration.

the equation given to find the time it takes for the proton to move 1cm in this field is
x = 1/2at^2 and I'm solving for t.
0.01 m = 1/2 (Vf - Vi) t^2
0.01 m = 1/2 (1.30x10^3 - 0) t^2
but the answer i got was wrong. what am i doing wrong to find the time it takes for the proton to move 1 cm?
 
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  • #2
(c) Calculate the time it takes for the proton to move 1 cm in this field, starting from rest. (Hint: The acceleration is uniform.)
in part c i cannot figure out how to find the uniform acceleration.
I would think you just use the acceleration you found in (b).

the equation given to find the time it takes for the proton to move 1cm in this field is
x = 1/2at^2 and I'm solving for t.
Yes, this equation will work. Now you know what the acceleration is and don't need to be substituting for it in your calculations. Just solve.

0.01 m = 1/2 (Vf - Vi) t^2
0.01 m = 1/2 (1.30x10^3 - 0) t^2
Why did you but 1.30x10^3 in for Vf? That number is the strength of the electric field, it has nothing to do with velocity. Also Remember that a = Δv/Δt
 
  • #3
thank you
 

1. What is electric potential?

Electric potential is the amount of electric potential energy that a unit charge has at a certain point in an electric field. It is measured in volts (V).

2. How is electric potential different from electric potential energy?

Electric potential is a measure of the potential energy per unit charge, while electric potential energy is the total amount of energy that a charge possesses due to its position in an electric field.

3. What is the relationship between electric potential and electric field?

Electric potential is directly proportional to the electric field strength. This means that as the electric field increases, the electric potential also increases.

4. How is electric potential calculated?

Electric potential is calculated by dividing the electric potential energy by the amount of charge present at a certain point in an electric field. The formula for electric potential is V = U/Q, where V is the electric potential, U is the electric potential energy, and Q is the charge.

5. What are some real-life applications of electric potential and potential energy?

Electric potential and potential energy are used in many everyday devices, such as batteries, generators, and electric motors. They are also important in understanding and analyzing the behavior of electric circuits and the flow of electricity in power grids.

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