Electronics - Potential Difference & Electrical Potential

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Homework Help Overview

The discussion revolves around two problems in the context of electronics and electric potential. The first problem involves calculating the potential difference required for an electron to reach a specific speed, while the second problem deals with the forces, electric fields, and electrical potential created by two point charges positioned along the x-axis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between potential difference, kinetic energy, and velocity in the first problem, with some questioning the calculations and assumptions made. In the second problem, there is a focus on the principle of superposition for electric potential and field, with participants suggesting a review of calculations for accuracy.

Discussion Status

Some participants have provided hints and guidance regarding the relationships between potential energy, kinetic energy, and the calculations involved. There are indications of confusion and uncertainty, particularly around the calculations for electrical potential in the second problem, but no consensus has been reached.

Contextual Notes

Participants note potential confusion between voltage and velocity, and there are references to specific constants and equations relevant to the problems. There is also mention of the conservation of energy principle in the context of the first problem.

Kawrae
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1. Through what potential difference would an electron need to be accelerated for it to achieve a speed of 50.0% of the speed of light, starting from rest? The speed of light is c = 3.00 108 m/s.

>> I'm really not too sure what to do with this problem. I took .50*3.00e8 and got 1.50e8. Then I said deltaV = Vb-Va = 1.50e8 - 0 and got 1.50e8 as my answer. But that doesn't seem right... I'm confused :(

2. Two charges Q1 = Q2 = 2.50 µC are positioned along the x-axis, where Q1 is at point (-.800m, 0) and Q2 is at point (+.800, 0), with a positive test charge q = 1.22e18 C at the origin.
(a) What is the net force exerted on q by the two 2.50 µC charges?
(b) What is the electric field at the origin due to the two 2.50 µC charges?
(c) What is the electrical potential at the origin due to the two 2.50 µC charges?

>> I know for part (a) the answer must be 0 because they cancle each other out. I also know that for part (b) the answer must be 0 because the fields will be repelled. Part (c) I'm stuck on... I tried Vp=Ke(q1/r1 + q2/r2) and got Vp=(8.99e9)(2)(2.50e-6/.8) and a final answer of 5.62e13. But that's not right... how should I solve part c??
 
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1. Don't confuse the V for voltage with the v for velocity. Remeber voltage is potential energy / charge, so a change in voltage results in a change in PE, which leads to an opposite change in KE (by conservation of energy), from which you can get the velocity change. And by the way, c is slightly less that 3.00 e8 m/s, not more.

2. The principle of superposition applies to potential as well as field, so get the potential due to each charge individually from the 1/r formula and add them. It looks like this is what you did, so just go back and make sure you did the calculations right.
 
Last edited:
for problemm (1) the speed you calculated is correct.

Here are some questions for you to help you on your way
1. how is Va - Vb related to the change in electrical potential energy ( there is an equation for this )
2. How much kinetic energy does the electron have at half the speed of light
3. ( not a question ) energy is conserved

For problem 2,
a and b are right
the equation for c is right... must be calculational error... can't possibly be E
13.
 
Problem #1:
Another quick hint:
1.0 electron-Volt = 1.60x10^(-19) Joules
{electron mass} = 9.11x10^(-31) kg

Problem #2:
Technique correct; exponent should be (+4).
~
 

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