Electric potential, potential difference, and potential energy

In summary, the conversation involves three problems related to potential difference and acceleration of particles. The first problem involves finding the potential difference needed for an electron to reach a certain speed, which was solved to be 135.159V. The second problem involves finding the potential difference between two points for a decelerating electron, which was attempted using the same approach but resulted in an incorrect answer. And the third problem involves finding the speed of a proton accelerated through a potential difference, which was also attempted using the same formula but yielded an imaginary number.
  • #1
btaylor
2
0
1. Here is a problem that I know how to solve
Through what potential difference would an electron need to be accelerated for it to achieve a speed of 2.3% of the speed of light (2.99792x10^8 m/s), starting from rest? Answer in units of V.

For this problem I used:
Code:
deltaK + deltaU = 0
(1/2)mv^2 - 0 = -qdeltaV

It works out to be 135.159V.

2. Now here is a similar problem that I can't seem to solve
An electron moving parallel to the x-axis has an initial speed of 2x10^6 m/s at the origin. Its speed is reduced to 500000 m/s at the point p, 1cm away from the origin. The mass of the electron is 9.10939x10^-31 kg and the charge of the electron is -1.60218x10^-19 C. Calculate the magnitude of the potential difference between this point and the origin. Answer in units of V.

I tried to use the same approach for this problem:
Code:
(1/2)m2v2^2 - (1/2)m1v1^2 = -qdeltaV
m1 and m2 are the same, so the equation becomes:
Code:
(1/2)m(v2^2 - v1^2) / -q = deltaV
(1/2)(9.10939x10^-31)(500000^2 - (2x10^6)^2) / 1.60218x10^-19 = deltaV
-10.66054142V = deltaV

This is not the right answer, and I don't know what I could be doing wrong.

3. Here is something else that I can't seem to solve
Calculate the speed of a proton that is accelerated from rest through a potential difference of 69V. Answer in units of m/s.

I attempt to use the same formula:
Code:
(1/2)mv^2 - 0 = -qdeltaV
(1/2)(1.67262158x10^-27)v^2 = (-1.60218x10^-19)(69)

However, this yields an imaginary number.


Any hint as to what concepts I'm missing here would be greatly appreciated. :)
 
Physics news on Phys.org
  • #2
I figured it out.
 
  • #3
how do u do it?
 

1. What is electric potential?

Electric potential, also known as voltage, is the amount of electric potential energy per unit charge at a given point in an electric field.

2. What is potential difference?

Potential difference is the difference in electric potential between two points in an electric field, which determines the direction and magnitude of the electric current.

3. How are electric potential and potential energy related?

Electric potential energy is the energy that a charged particle possesses due to its position in an electric field, while electric potential is the potential energy per unit charge at a given point in the field. They are directly proportional to each other.

4. How is electric potential measured?

Electric potential is measured in units of volts (V) using a voltmeter, which measures the potential difference between two points in an electric field.

5. What factors affect electric potential?

The electric potential at a point is affected by the amount of charge present, the distance from the source of the electric field, and the surrounding medium's dielectric constant.

Similar threads

  • Introductory Physics Homework Help
Replies
21
Views
559
  • Introductory Physics Homework Help
Replies
5
Views
738
  • Introductory Physics Homework Help
Replies
1
Views
955
  • Introductory Physics Homework Help
Replies
3
Views
983
  • Introductory Physics Homework Help
Replies
23
Views
286
  • Introductory Physics Homework Help
Replies
34
Views
2K
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
278
  • Introductory Physics Homework Help
Replies
2
Views
133
  • Introductory Physics Homework Help
Replies
7
Views
306
Back
Top