Question

A proton 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)

Answer 1

Answer:

Acceleration of proton will be $a=0.67\times 10^{11}m/sec^2$

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 $m=1.67\times 10^{-27}kg$

Charge on proton $e=1.6\times 10^{-19}C$

So electric force on the proton $F=qE=1.6\times 10^{-19}\times 700=1.120\times 10^{-16}N$

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 $1.67\times 10^{-27}\times a=1.120\times 10^{-16}$

$a=0.67\times 10^{11}m/sec^2$

So acceleration of proton will be $a=0.67\times 10^{11}m/sec^2$