# The electric field at a point in space is E=(400i 800j)nC. [Solved] Similarly , acceleration of the electron ,
ae = F / 9.1 x 10^-31 = ( 70.33 i 140.66 j ) x 10^12 m/s^2

The unit you have given for electric intensity is not correct . It must have been N/C or N/nC.
Anyway, I assume that it is N/C. ( If it is N/nC. the answers should be multiplied by 10^9 )

X component of E is 400 N/C
Y component of E is 800 N/C
Force on a proton of charge q = 1.6 x 10^-19 C is
Fx = Ex q = 640 x 10^-19 N towards positive X axis
Fy = Ey q = 1280 x 10^-19 N towards positive Y axis

Similarly , acceleration of the electron ,
ae = F / 9.1 x 10^-31 = ( 70.33 i 140.66 j ) x 10^12 m/s^2

Acceleration of proton,
ap = F / Mp = ( 640 i 1280 j ) x 10^-19 / 1.6725 x 10^-27= ( 382.66 i 765.32 j ) x 10^8 m/s^2

Similarly , acceleration of the electron ,
ae = F / 9.1 x 10^-31 = ( 70.33 i 140.66 j ) x 10^12 m/s^2

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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