A small object, which has a charge q = 7.3 µC and mass m = 9.0 10-5 kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 2.0 103 m/s in a time of 0.91 s. Determine the magnitude of the electric field.
2006-08-28 08:27:42 · 4 answers · asked by onewify 1 in Science & Mathematics ➔ Physics
Hmmm?
E=F/q
F=ma
F=m(v/t)
E=m v/(t q)=
=(9.0 E-5 )( 2.0 E3)/((.91 )(7.3 E-6)= 27,096 Newton / Coulomb
However as given the force on the charge particle is constant (constant electric field), in reality it is not and diminishes as the charge moves with a square of the distance between E producing source.
2006-08-28 08:56:15 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
Hi,
Since i am at work, i can only help you to this extent, i tried to find conversion factors during my lunch break, but my luck ran out.....this is how i think this problem can be solved (but please do not completely rely on this answer)
F=ma = 9X10e-5X2.0103 = 18.0927e-5 Kg-m/S
E=F/q = 18.0927e-5/7.3 = 2.478e-5 kg-m/s/micro coloumb.
I am sorry i do not have enough time to deduce the units....got to get back to work.
Krishna
2006-08-28 09:13:18 · answer #2 · answered by Novice 1 · 1⤊ 0⤋
Find the equation in your textbook, plug in numbers, you'll have answer. Faster than asking here.
2006-08-28 09:15:52 · answer #3 · answered by Ken H 4 · 0⤊ 0⤋
A little larger than a bread box.
2006-08-28 08:34:16 · answer #4 · answered by nothing 6 · 0⤊ 0⤋