Question

What is the potential energy of an electron that is 22.0 cm from a charge of 43.0 nC?
How much work is required to move the electron very far from the charge?

Answer 1

I can give you a hint on the first part.

Potential energy is (I think opposite) the integral of the force with respect to displacement... Which doesn't mean anything if you're not in calculus.
\(U=k\dfrac{q_1\ q_2}r\)

Answer 2

One charge is given, the other charge is that of an electron (which you can probably find in your book).

Answer 3

Good luck! I have to get going!

Answer 4

yea thats exactly what i did but it wasnt the right answer

Answer 5

Did @theEric's help make sense to you? I think it was complete.

Answer 6

the two parts of the question ask the same thing...

Answer 7

i didnt get the right answer
i got -2.81x10^-16 J

Answer 8

i also tried +2.81x10^-16 J

Answer 9

It should be. because you need extra energy to move the electron more away from the other. it is the external work.

Answer 10

so how can i calculate it ? Because this one wouldn't work

Answer 11

I am doing some quantum problems and dont have time to write equations. I would like @theEric to help you further!

Answer 12

The charge of the electron is about \(-1.602\times10^{-19}\ \rm C\).
The other charge is \(43.0\ \rm{nC}=43.0\times10^{-6}\ \rm C\)

So \(q_1\ q_2=(-1.602)(10^{-19})(43.0)(10^{-6})=-68.9\times10^{-25}\).

Is that what you got to that point?

Answer 13

\(r=22.0\ \rm{cm}=22.0\times10^{-2}\ \rm m\)

And there's that.


The reason I'm going through this now is because I see all the different units and wonder if that is a point of possible error.

Answer 14

1n=10^-9m

Answer 15

Coulomb's constant is about \(\large8.988\times10^9\ \rm\frac{N\bullet m^2}{C^2}\) (the dot is just multiplication).

Answer 16

Thank you @Saeeddiscover !!! :) Bad mistake...

Answer 17

The charge of the electron is about \(−1.602×10^{−19}\ \rm C.\)
The other charge is \(43.0\ \rm{nC}=43.0\times10^{-\color{red}9}\ \rm C\)

So \(q_1\ q_2=(-1.602)(10^{-19})(43.0)(10^{-\color{red}9})(\rm{C^2})=-68.9\times10^{-\color{red}{28}}\ \rm{C^2}\)

Answer 18

Again,
\(r=22.0\ \rm{cm}=22.0\times10^{-2}\ \rm m\)
and
\(k=\large8.988\times10^9\ \rm\frac{N\bullet m^2}{C^2}\)

Answer 19

\[U=k\frac{q_1\ q_2}r\]\(=-28.1\times10^{(9)+(-28)-(-2)}\ \rm J=-28.1\times10^{-17}\ \rm J\\=-2.81\times10^{-16}\ \rm J\)
Interesting... And we would say that its potential is positive.... So \(-2.81\times10^{-16}\ \rm J\).

Answer 20

I meant, so \(2.81\times10^{-16}\ \rm J\).

Answer 21

Moving the electron to the positive charge.. The electron will do that naturally, so the electrostatic force does the work. The work done by it is positive. The work done on it is negative.

Other than that, I forget the formula. I could derive it, but I don't know if I should trust myself. Logically, it's the opposite the work to move it from where it is (22.0 cm away) to infinity (where the electrostatic force is theoretically \(0\)).

Answer 22

I have to go! Take care!

Answer 23

ohh i did opposite the Work is negative
alright thanks

Answer 24

Di you get them right?

Answer 25

did*

Answer 26

yea i did!! thanks

Answer 27

Congrats!