Question

(a) Find the direction and magnitude of an electric field that exerts a $4.80\times 10^{-17}\textrm{ N}$ westward force on an electron.
(b) What magnitude and direction force does this field exert on a proton?

- $3.00\times 10^{2}\textrm{ N/C}$
- $4.80\times 10^{-17}\textrm{ N, East}$

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Video Transcript

This is College Physics Answers with Shaun Dychko. A force of 4.80 times 10 to the minus 17 newtons is exerted to the west on an electron; the charge on an electron is the elementary charge— 1.60 times 10 to the minus 19 coulombs— and our question is what is the electric field? So in order for the force to be to the left that means the field must be to the right because the field direction is in the same direction as the force would be on a positive test charge. So this being negative, it will be a force in the opposite direction. So the force magnitude is the charge multiplied by the electric field and so we divide both sides by

*q*to solve for*E*. So electric field then is 4.80 times 10 to the minus 17 newtons divided by 1.60 times 10 to the minus 19 coulombs and that's 300 newtons per coulomb and so our final answer is 3.00 times 10 to the 2 newtons per coulomb to the east is the electric field and so that's to the right in this picture. Force on the proton would have the same magnitude because the charge on a proton has the same magnitude as the charge on an electron and so the force is gonna be just the opposite direction. So 4.80 times 10 to the minus 17 newtons is the same as that on an electron but in the opposite direction to the east instead of to the west.