A proton is placed in an electric field of intensity 700 N/C

1. Jun 17, 2012

Curious314

A proton is placed in an electric field of intensity 700 N/C. What is the magnitude and direction of the acceleration of this proton due to this field?

in a google I found that
1 proton= 1.6022 x 10 ^ -19
and 1.6726 x 10 ^ -27 kg

so I first get the force :
F= q*E
f= (1.6022 x 10 ^ -19 )(700 N/C)
f=1.1215 * 10^ -16N

f=ma

f/m=a
1.1215 * 10^-16 / 1.6726 * 10^-27 = a
6.71* 10^10 m/s^2 = a

am I right?

Thanks!

2. Jun 17, 2012

SammyS

Staff Emeritus
Looks like the correct magnitude.

What's the direction of the acceleration ?

3. Jun 17, 2012

Curious314

since it a positive charge and a positive field, i guess opposite. am I right??

4. Jun 17, 2012

Curious314

In second thought, it goes with the elctric field... because is positive and the proton is positive, so they go in the same direction... right?

5. Jun 17, 2012

Xisune

Yes, the proton is going in the direction of the electric field.

6. Jun 17, 2012

Curious314

thank you a lot SammyS and Xisune!!!