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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.

Respuesta :

Answer:

Electric field, [tex]E=5.4\times 10^6\ N/C[/tex]

Explanation:

It is given that,

Electromagnetic force acting on the proton when it is at rest, [tex]F=8.7\times 10^{-13}\ N[/tex] (in +x direction)

Speed of proton, [tex]v=1.5\times 10^6\ m/s[/tex]

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

[tex]E=\dfrac{F}{q}[/tex]

q is charge on proton

[tex]E=\dfrac{8.7\times 10^{-13}\ N}{1.6\times 10^{-19}\ C}[/tex]

E = 5437500 N/C

or

[tex]E=5.4\times 10^6\ N/C[/tex]

So, the magnitude of electric field is [tex]E=5.4\times 10^6\ N/C[/tex]. hence, this is the required solution.

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