Electricity/Electric Field Questions

1. Apr 19, 2005

seiferseph

I've attached the questions from a worksheet.

http://i2.photobucket.com/albums/y15/seiferseph/1.jpg

For starters, how do i do the first one? i know Fe = q*E, but what is the charge? is it just 2 x 1.6x10^16 (because there are two protons?)

for the rest, here are the answers i got, could anyone confirm any? i'm not sure about these.

2) a) Fe = 0.45 N
b) E = 4.5 x 10^9 N/C
3) E = 250 N/C
4) Q = 3.0 x 10^-6 C
5) E = 5.4 x 10^10 N/C to the right

Thanks again!

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Last edited: Apr 19, 2005
2. Apr 19, 2005

seiferseph

3. Apr 19, 2005

dextercioby

1.$q_{p}=e\simeq 1.6\cdot 10^{-19}C$

Please show us how u did them.Numbers are least important when solving a problem.

Daniel.

4. Apr 20, 2005

Andrew Mason

1. check your value for e. The charge of an alpha particle is 2e
2. a) correct: F = kqQ/r^2
b) correct: E = F/q
3. correct: E = F/e
4. correct magnitude but check the sign.
5. incorrect. check distance value.

AM

5. Apr 20, 2005

ChickenChakuro

Andrew, why is the charge of an alpha particle 2e? I thought an alpha particle was electron-less.

6. Apr 20, 2005

dextercioby

It is.It's +2e,because of the 2 protons inside the nucleus...

Daniel.

7. Apr 20, 2005

ChickenChakuro

Ok. The plus sign makes it all better :)

8. Apr 20, 2005

seiferseph

ok, so

1) Fe = 2*1.6x10^-19*75
and i get 2.4 x 10^17 N

4) isn't the sign + ? if the field is directed away from it, doesn't it mean its a positive charge in Q?

5), i used the radius of 1, when it should be 1/2. using 1/2, i get
2.16 x 10^11 N/C right

thanks again for all the help!

9. Apr 20, 2005

Andrew Mason

Just a small slip on the exponent: 2.4 x 10^-17 N (but a huge difference). What is Fe? Shd be: F = qE where q = 2e.
A trick question. The first sentence is irrelevant and designed to confuse - it succeeded. You are right. The field direction is given by the direction of motion of a + charge, which is away from Q so Q is +.

Right.

AM

10. Apr 20, 2005

seiferseph

thanks for all this help, i was able to do all the problems from today (and get them right).