Electronics - Potential Difference & Electrical Potential

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To accelerate an electron to 50% of the speed of light, a potential difference must be calculated using the relationship between kinetic energy and potential energy. The initial calculations for the speed were correct, but confusion arose regarding the potential energy and voltage relationship. For the two charges positioned along the x-axis, the net force on a test charge at the origin is zero due to equal and opposite forces from the charges. The electric field at the origin is also zero for the same reason, but there was uncertainty regarding the calculation of electrical potential at that point. The principle of superposition applies to potential, and recalculating using the correct formula should yield the correct result.
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.
 
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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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