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A proton is at the origin and an electron is at the point x = 0.35 nm, y = 0.33 nm. Find the...

Question:

A proton is at the origin and an electron is at the point x = 0.35 nm, y = 0.33 nm. Find the electric force on the proton.

Coulomb's law:

The force between two charges {eq}q_{1} {/eq} and {eq}q_{2} {/eq} seperated by a distance r is given by

{eq}F =\frac{kq_{1}q_{2}}{r^{2}} {/eq}

where k is constant

Answer and Explanation:

Charge of proton = {eq}1.6 \times 10^{-19} {/eq}

Charge of electron = {eq}-1.6 \times 10^{-19} {/eq}

Given points x=0.35nm and y=0.33nm

so Distance r={eq}\sqrt{(0.35^{2}+0.33^{2}) \times 10^{-18}} {/eq}

=0.489nm

Now net force F=

{eq}F =\frac{kq_{1}q_{2}}{r^{2}} {/eq}

{eq}F =\frac{9\times 10^{9} \times (1.6 \times 10^{-19})^2}{(0.489 \times 10^{-9})^{2}} {/eq}

={eq}9.68A ^{0} {/eq}N


Learn more about this topic:

Strength of an Electric Field & Coulomb's Law

from UExcel Physics: Study Guide & Test Prep

Chapter 12 / Lesson 4
11K

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