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A proton is released in a uniform electric field, and it experiences an electric force of $1.86 \times 10^{-14} \textrm{ N}$ toward the south. Find the magnitude and direction of the electric field.

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer: 

$1.16 \times 10^5 \textrm{ N/C, South}$

Giancoli 7th Edition, Chapter 16, Problem 20


Chapter 16, Problem 20 is solved.

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Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. The electric field points in the same direction as the force it applies on a positive charge. Because this proton experiences a force towards the South, that means electric field must be to the South, because electric field points in the direction of force on a positive charge. The electric field strength is the force divided by the charge. That's gonna be 1.86 times 10 to the minus 14 Newtons divided by 1.602 times 10 to the minus 19 Coulombs, magnitude of a charge in a proton. This gives us 1.16 times 10 to the five Newtons per Coulomb electric field strength which is directed South.