Electric potential problem(s)

So, the kinetic energy in electron volts will be equal to the potential difference in volts multiplied by the charge in Coulombs.In summary, an alpha particle starting at a potential of +250 V and accelerating towards a potential of -150 V will have a kinetic energy equal to the potential difference of -400 V multiplied by the charge of 2e (2 x 1.6 x 10^-19 C), resulting in a kinetic energy of -8.0 x 10^-19 J or -5.0 x 10^12 eV.
  • #1
Quincy
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0

Homework Statement



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 2 protons and 2 neutrons. An alpha particle starts from rest at A and accelerates toward B. When the alpha particle arrives at B, what Kinetic energy (in electron Volts) does it have?


Homework Equations



1 eV = 1.6 x 10^-19 J, W = -(EPEb - EPEa), V = EPE/q

The Attempt at a Solution



Vb - Va = -150 - 250

Vb - Va = -400
 
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  • #2
You know the potential difference and you have the charge. If the charge is accelerated through that PD then all the potential energy will be converted to kinetic energy.
 
  • #3


W = -(-400 x 1.6 x 10^-19)

W = 6.4 x 10^-17 J

KE = 6.4 x 10^-17 J

KE = (6.4 x 10^-17 J) / (1.6 x 10^-19 J/eV)

KE = 4 x 10^2 eV


As the alpha particle travels from Point A to Point B, it experiences a decrease in electric potential energy (EPE) due to the difference in potential between the two points. Using the equation W = -(EPEb - EPEa), we can calculate the work done on the alpha particle, which is equivalent to the change in EPE. This work done is then converted to kinetic energy (KE) using the equation KE = W/q. Since the alpha particle has a charge of +2e, we divide the work done by this charge to get the KE in joules. Finally, we convert the KE from joules to electron volts (eV) by dividing by 1.6 x 10^-19 J/eV. Therefore, the alpha particle has a KE of approximately 400 eV when it arrives at Point B.
 

Related to Electric potential problem(s)

1. What is electric potential?

Electric potential, also known as voltage, is a measure of the electric potential energy per unit charge at a point in an electric field. It is measured in volts (V) and is a fundamental quantity in the study of electricity and magnetism.

2. How is electric potential different from electric field?

Electric potential is a scalar quantity, meaning it has only magnitude, while electric field is a vector quantity, meaning it has both magnitude and direction. Electric potential is a measure of the energy a charge would have if placed at a specific point in an electric field, while electric field is a measure of the force that would be exerted on a charge placed at that point.

3. What is the relationship between electric potential and electric field?

The electric field at a point is equal to the negative gradient of the electric potential at that point. In other words, the electric field points in the direction of decreasing electric potential and its magnitude is proportional to the rate of change of electric potential.

4. How is electric potential calculated for a point charge?

The electric potential at a point in space due to a point charge can be calculated using the equation V = kq/r, where V is the electric potential, k is the Coulomb's constant, q is the magnitude of the charge, and r is the distance from the charge to the point.

5. What are some real-world applications of electric potential problems?

Electric potential problems are used in a variety of real-world applications, including designing electrical circuits, calculating the potential difference between two points in a circuit, and understanding the behavior of charged particles in electric fields. They are also used in fields such as electrochemistry, where electric potential is used to determine the direction and extent of chemical reactions.

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