Question

Review When at rest, a proton experiences a net electromagnetic force of magnitude 8.7×10−13 N pointing in the positive x direction. When the proton moves with a speed of 1.5×106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.5×10−13 N , still pointing in the positive x direction. You may want to review (Pages 773 - 777) . Part A Find the magnitude of the electric field. Express your answer using two significant figures.

Answer 1

Electric field,
`E=5.4* 10^6\ N/C`

Step-by-step explanation:

It is given that,

Electromagnetic force acting on the proton when it is at rest,
`F=8.7* 10^(-13)\ N` (in +x direction)

Speed of proton,
`v=1.5* 10^6\ m/s`

We need to find the magnitude of the electric field. We know that when the charged particle is at rest it experiences electric force which is given by :

F = q E

`E=(F)/(q)`

q is charge on proton

`E=(8.7* 10^(-13)\ N)/(1.6* 10^(-19)\ C)`

E = 5437500 N/C

or

`E=5.4* 10^6\ N/C`

So, the magnitude of electric field is
`E=5.4* 10^6\ N/C`. hence, this is the required solution.