Question
Suppose that the force exerted on an electron is $5.6 \times 10^{−17} \textrm{ N}$, directed to the east.
1. Find the magnitude of the electric field that exerts the force.
2. What will be the direction of the electric field?
3. If the electron is replaced by a proton, what will be the magnitude of force exerted?
4. What will be the direction of force on the proton?
1. $350 \textrm{ N/C}$
2. West
3. The same. ($5.6 \times 10^{-17} \textrm{N})$
4. West
Solution Video

# OpenStax College Physics for AP® Courses Solution, Chapter 18, Problem 37 (Test Prep for AP® Courses) (1:12)

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Video Transcript

This is College Physics Answers with Shaun Dychko. Electrostatic force is the electric field strength multiplied by the charge. So we can solve for electric field by dividing both sides by <i>q</i> and we get <i>E</i> is the force divided by the charge. So that's 5.6 times ten to the minus seventeen Newtons divided by the charge on an electron which is the elementary charge of 1.6 times ten to the minus nineteen coulombs, and that gives and electric field of 350 Newtons per coulomb. Now the direction of electric field is defined as the direction of force on a positive charge. So because we're talking about an electron which is a negative charge, that force on the electron is to the east and so the force that would be on a positive charge is in the opposite direction, to the west. So this electric field is pointing to the west. In part C it says if the electron is replaced by a proton, will there be -- what will be the magnitude of force exerted? The answer is it'll be the same because it'll have the same magnitude charge, the elementary charge. So, still 5.6 times ten to the minus seventeen Newtons. But the direction of the force will be different. The direction of the force will be in the same direction as the electric field which will be to the west.