Electronics - Potential Difference & Electrical Potential

In summary, for problem 1, the electron would need to be accelerated by a potential difference of 1.50 x 10^8 volts to achieve a speed of 50.0% of the speed of light. In problem 2, the net force and electric field at the origin due to the two 2.50 µC charges are both 0, while the electrical potential at the origin is 5.62 x 10^13 volts. The change in electrical potential energy is related to the difference in voltage between two points, and the kinetic energy of the electron at half the speed of light can be calculated using the mass and energy conversion formula. In summary, these problems involve the principles of conservation of energy and
  • #1
Kawrae
46
0
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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  • #2
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:
  • #3
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.
 
  • #4
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).
~
 

1. What is potential difference?

Potential difference, also known as voltage, is the difference in electrical potential between two points in a circuit. It is measured in volts (V) and represents the amount of energy needed to move a unit charge from one point to another.

2. How is potential difference related to current?

Potential difference is directly related to current, as stated by Ohm's Law: V=IR. This means that the potential difference (V) is equal to the current (I) multiplied by the resistance (R) in a circuit. In other words, a higher potential difference will result in a higher current flow.

3. What is the difference between potential difference and electrical potential?

Potential difference and electrical potential are often used interchangeably, but they have slightly different meanings. Potential difference refers to the difference in electrical potential between two points, while electrical potential refers to the amount of potential energy per unit charge at a specific point in a circuit.

4. How is potential difference measured?

Potential difference is measured using a voltmeter, which is connected in parallel to the circuit. The voltmeter measures the difference in potential between two points in a circuit and displays the result in volts.

5. What factors affect potential difference?

The potential difference in a circuit is affected by several factors, including the voltage source, the resistance of the circuit, and the type of material used in the circuit. Additionally, the length and cross-sectional area of the circuit can also affect the potential difference.

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