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.