How Is Kinetic Energy Calculated After an Electron Crosses Parallel Plates?

[x86]

Homework Statement

An electron is accelerated from rest across a set of parallel plates that have a potential difference of 150V and are separated by 0.80 cm.

a) Determine the kinetic energy of the electron after it crosses the plates

Homework Equations

N/C = J/C/d

The Attempt at a Solution

+++++++++++++++++ delta => {(150 V), (0.8 cm)}

e (going up)
----------------------

150 J/C / (0.8 cm * 1 m / 100 cm) = 18750 N/C

So at the start it has about 18750 N/C * e of potential energy, all which transfers to kinetic energy at the end, meaning it has 18750 N/C * e of kinetic energy at the end point

18750 N/C * 1.6 * 10^-19 C

and I get an answer of 3e-15 J.

However, the book lists 2.4 * 10^-17 J. What did I do wrong?

 

[SammyS]

Homework Statement

An electron is accelerated from rest across a set of parallel plates that have a potential difference of 150V and are separated by 0.80 cm.

a) Determine the kinetic energy of the electron after it crosses the plates

Homework Equations

N/C = J/C/d

That should be N/C = J/C/m

The Attempt at a Solution

+++++++++++++++++ delta => {(150 V), (0.8 cm)}

e (going up)
----------------------

150 J/C / (0.8 cm * 1 m / 100 cm) = 18750 N/C

So at the start it has about 18750 N/C * e of potential energy, all which transfers to kinetic energy at the end, meaning it has 18750 N/C * e of kinetic energy at the end point

A quantity with units of N/C time a quantity with units of charge, gives a quantity with units of N, i.e. a Force, not energy.

18750 N/C * 1.6 * 10^-19 C

and I get an answer of 3e-15 J.

However, the book lists 2.4 * 10^-17 J. What did I do wrong?

 

[x86]

That should be N/C = J/C/m


A quantity with units of N/C time a quantity with units of charge, gives a quantity with units of N, i.e. a Force, not energy.

Thank you for your help. I can't believe I didn't notice that. I guess it's time for a break! :P

 

[SammyS]

Thank you for your help. I can't believe I didn't notice that. I guess it's time for a break! :P

Yes.

An object of charge, Q, when accelerated through a potential difference of X Volts, will gain (Q times X) Joules of Kinetic Energy.

 

[Nickie]

Your solution is incorrect because you have used the wrong units for distance. In the equation N/C = J/C/d, the distance should be in meters, not centimeters. So your calculation should be 150 J/C / (0.008 m) = 18750 N/C.

Furthermore, the equation you used is for electric field strength, not potential energy. To calculate the kinetic energy of the electron, you need to use the equation KE = 1/2 * m * v^2, where m is the mass of the electron and v is its velocity.

Using this equation, we can calculate the velocity of the electron after crossing the plates. We know that the potential difference, V, is equal to the change in potential energy, so we can write the equation as V = KE/e. Solving for KE, we get KE = e * V.

Plugging in the values, we get KE = (1.6 * 10^-19 C) * (150 V) = 2.4 * 10^-17 J, which is the correct answer given in the book.

In conclusion, to solve this problem correctly, you need to use the correct units and the correct equation for calculating kinetic energy.