A proton is a distance d/2 directly above 2


A proton is a distance d/2 directly above the center of a square of side d. What is the magnitude of the electric flux through the square?

A proton is a distance d/2 directly above

Image from: J. Walker, D. Halliday, and R. Resnick, Fundamentals of physics: [extended], 10th ed. United States: Wiley, John & Sons, 2013.

Solution:

Show me the final answer↓

Let us assume that this is one side of a cube. Gauss’s law states the following:

\Phi=\dfrac{q}{\epsilon_0}

(Where \Phi is the flux, q is the charge in the enclosed Gaussian surface, and \epsilon_0 is the permittivity of free space.)

 
If we take the total flux through a closed Gaussian surface and divide it by 6, we get the flux through one side.

Thus, we can write the following:

\Phi=\dfrac{q}{6\epsilon_0}

 

Remember that the charge of a proton is 1.6\times10^{-19} C.

\Phi=\dfrac{1.6\times10^{-19}}{(6)(8.85\times 10^{-12})}

\Phi=3.01\times 10^{-9}\dfrac{N\cdot m^2}{C}

Final Answer:

\Phi=3.01\times 10^{-9}\dfrac{N\cdot m^2}{C}

 

This question can be found in Fundamentals of Physics, 10th edition, chapter 23, question 5.

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