Magnitude of electric field - Problem check

[duki]

Hey everyone. I've been getting tons of help on here and it's paying off. I really appreciate your efforts. Could someone check this problem for me and tell me if I have the right answer?


Book: Cutnell & Johnson Physics
pp. 571 #66


A small object, which has a charge q = 7.5uC and mass m = 9.0 x 10^-5kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 2.0 x 10^3m/s in a time of 0.96 s. Determine the magnitude of the electric field.



My Answer:

I found the equation for the magnitude of an electric field: E= k|q| / r^2

I wasn't sure how to find r but I noticed I had a and t so I did the following: r = (2.0 x 10^5) x (9.6)^2 / 2 = 1.8 x 10^8

I did that because, in order to find the time you do sqrt{2(distance)/(accel)} correct?

Then I did E = (8.99 x 10^9) x (7.5 x 10^-6) / (1.8 x 10^8)^2 = 2.08 x 10^-12


I figure if I messed up it's probably doing the reverse acceleration equation. Can someone check for me? Thanks!

 

[nrqed]

Hey everyone. I've been getting tons of help on here and it's paying off. I really appreciate your efforts. Could someone check this problem for me and tell me if I have the right answer?


Book: Cutnell & Johnson Physics
pp. 571 #66


A small object, which has a charge q = 7.5uC and mass m = 9.0 x 10^-5kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 2.0 x 10^3m/s in a time of 0.96 s. Determine the magnitude of the electric field.



My Answer:

I found the equation for the magnitude of an electric field: E= k|q| / r^2

No, you cannot use this equation here. The E field is uniform which tells you that it is *not* produced by a point charge. It is produced by a large uniformly charged surface or some other charge distribution. The only thing you can use then is F=q E.
So you must use the information about the motion to find the acceleration, then use F=m a to find the force and then use F=qE to find the magnitude of the E field.

I wasn't sure how to find r but I noticed I had a and t so I did the following: r = (2.0 x 10^5) x (9.6)^2 / 2 = 1.8 x 10^8
[/quot]e
I am not sure where the 2.0 x10^5 comes from and shouldn't you have 0.96 instead of 9.6?

I did that because, in order to find the time you do sqrt{2(distance)/(accel)} correct?
[/quote]
Yes (if an onbject starts from rest and moves along a straight line) BUT you do not know the acceleration here! Just the final speed!