Aproton is placed in an electric field of intensity 700 n/c. what are the magnitude and direction of the acceleration of this proton due to this field? (mproton = 1.67 × 10-27 kg, e = 1.60 × 10-19 c)

Acceleration of proton will be

Explanation:

We have given a proton is placed in an electric field of intensity of 700 N/C

So electric field E = 700 N/C

Mass of proton

Charge on proton

So electric force on the proton

This force will be equal to force due to acceleration of the proton

According to newton's law force is given by F = ma

So

So acceleration of proton will be

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