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

[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!

 

[SammyS]

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!

Looks like the correct magnitude.

What's the direction of the acceleration ?

 

[Curious314]

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

 

[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?

 

[Xisune]

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?

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

 

[Curious314]

thank you a lot SammyS and Xisune!