Electric Potential Proton Problem

[milkyway200]

Hello there! I'm new to the forums. It seems like a lot of very knowledgeable people are here. I have my first question.

How can the question below be solved?

Thank you for your help.How much work is needed to assemble an atomic nucleus containing three protons (such as Be) if we model it as an equilateral triangle of side 2.00 x 10^-15 m with a proton at each vertex? Assume the protons started from very far away.

 

[Defennder]

0 work is done in placing the first charge. For the second charge, it's just the electric potential due to the first one. Finally the amount of work done for the placement of the last charge is the summation of electric potentials of the first 2 charges. Just sum all of it up.

 

[baba89]

To solve this question, we can use the equation for electric potential energy:

U = k(Q1Q2)/r

Where U is the potential energy, k is the Coulomb's constant, Q1 and Q2 are the charges of the protons, and r is the distance between them.

First, we need to calculate the distance between each proton and the center of the triangle. Since the triangle is equilateral, each side is equal to 2.00 x 10^-15 m, so the distance from the center to each proton is half of that, or 1.00 x 10^-15 m.

Next, we need to calculate the potential energy between each pair of protons. Since the protons are all the same charge (positive), we can use the equation above and plug in the values:

U = (9 x 10^9 Nm^2/C^2)(1.6 x 10^-19 C)^2 / (1.00 x 10^-15 m)

U = 2.304 x 10^-13 J

Since there are three protons, we need to multiply this value by 3 to get the total potential energy:

U_total = 3(2.304 x 10^-13 J) = 6.912 x 10^-13 J

This is the amount of work needed to assemble the atomic nucleus containing three protons in an equilateral triangle arrangement.