Electric Potential & Kinetic Energy

In summary: Point A is at a potential of +250 V, and Point B is at a potential of -150 V. An alpha particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An alpha particle starts from rest at A and accelerates toward B. When the alpha particle arrives at point B, what kinetic energy( in electron volts) does it have?In summary, an alpha particle has a kinetic energy of 800eV when it arrives at point B.
  • #1
pinky2468
86
0
Here is the problem:
Point A is at a potential of +250 V, and Point B is at a potential of -150 V. An alpha particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An alpha particle starts from rest at A and accelerates toward B. When the alpha particle arrives at point B, what kinetic energy( in electron volts) does it have?

So here is what I know: alpha particle +2 charge and atomic mass of 4. KE=1/2mv^2. I know that E=KE + PE and E final = E initial.
I am going around in circles with this one any advice on where to begin?
 
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  • #2
the potential energy is [tex] q\Delta V [/tex], use this definition and conservation of energy...
the mass and velocity of the particel is irrelevent, since you are asking to find KE only... ... beware the unit you use... we are not doing SI unit here...
 
  • #3
So is KEf= KEi+EPEi-EPEf right? The answer is 800eV and if I do this...
KE=0 +(250)(+2)-(-150)(+2) ...I get 800J. So do I need to multiply the +2 charge by 1.6X10-19? If so then that changes the answer, I am missing a step somewhere right?
 
  • #4
pinky2468 said:
So is KEf= KEi+EPEi-EPEf right? The answer is 800eV and if I do this...
KE=0 +(250)(+2)-(-150)(+2) ...I get 800J. So do I need to multiply the +2 charge by 1.6X10-19? If so then that changes the answer, I am missing a step somewhere right?
Everything in the formula KEf= KEi+EPEi-EPEf must be expressed in the same units. You should express EPEi and EPEf in J. Then, when you determine KE, which will be in J, just divide that answer by 1.6X10-19 to get eV.
 
  • #5
If I do that it changes answer. I got 800 J and the answer is 800eV. If I divide that by 1.6E-19 I get 5E21?
 
  • #6
pinky2468 said:
If I do that it changes answer. I got 800 J and the answer is 800eV. If I divide that by 1.6E-19 I get 5E21?

you got the right idea... however though, the charge is not 2+. its +2e...

where e = 1.6 * 10 -19 Coulombs...
 
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  • #7
Point A is at a potential of +250 V, and Point B is at a potential of -150 V. An alpha particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An alpha particle starts from rest at A and accelerates toward B. When the alpha particle arrives at point B, what kinetic energy( in electron volts) does it have?

Ok, lt me try it.
(delta) V=EPE/q
400V = EPE/q
400V*q = Ekfinal (in J)
400V*3.2x10^(-19) = Ekfinal (in J) = 1.28x10^(-16) J

Convert this to eV (divide by the charge on 1 electron), and you get 800J.
Isn't that the answer?
 
  • #8
christinono said:
Point A is at a potential of +250 V, and Point B is at a potential of -150 V. An alpha particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An alpha particle starts from rest at A and accelerates toward B. When the alpha particle arrives at point B, what kinetic energy( in electron volts) does it have?

Ok, lt me try it.
(delta) V=EPE/q
400V = EPE/q
400V*q = Ekfinal (in J)
400V*3.2x10^(-19) = Ekfinal (in J) = 1.28x10^(-16) J

Convert this to eV (divide by the charge on 1 electron), and you get 800J.
Isn't that the answer?

well the units are wrong 800 eV is not the same as 800 J...
 
  • #9
No, the answer(according to the book) is 800eV
 
  • #10
christinono said:
Point A is at a potential of +250 V, and Point B is at a potential of -150 V. An alpha particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An alpha particle starts from rest at A and accelerates toward B. When the alpha particle arrives at point B, what kinetic energy( in electron volts) does it have?

Ok, lt me try it.
(delta) V=EPE/q
400V = EPE/q
400V*q = Ekfinal (in J)
400V*3.2x10^(-19) = Ekfinal (in J) = 1.28x10^(-16) J

Convert this to eV (divide by the charge on 1 electron), and you get 800J.
Isn't that the answer?
Sorry...small mistake. Change my last 2 lines (previous post) to:

Convert this to eV (divide by the charge on 1 electron), and you get 800eV.
Isn't that the answer?
 

FAQ: Electric Potential & Kinetic Energy

What is electric potential?

Electric potential is the amount of work needed to move a unit of positive charge from one point to another in an electric field. It is measured in volts (V).

How is electric potential different from electric potential energy?

Electric potential is a measure of the electric field's strength at a point, while electric potential energy is the energy a charged particle possesses due to its position in an electric field.

What is the relationship between electric potential and kinetic energy?

Electric potential is a form of potential energy, and when an electrically charged particle moves from one point to another, its potential energy is converted into kinetic energy. In other words, an increase in electric potential corresponds to an increase in kinetic energy.

How is kinetic energy calculated in relation to electric potential?

The kinetic energy of a charged particle can be calculated using the equation KE = qV, where q is the charge of the particle and V is the electric potential at its position.

Can electric potential and kinetic energy be converted into each other?

Yes, electric potential energy can be converted into kinetic energy when a charged particle moves in an electric field. Conversely, kinetic energy can be converted into electric potential energy when a charged particle comes to rest in an electric field.

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