Electrostatic Forces on Charges at Different Locations

[Pietjuh]

I've got the following problem:
1) We've got a charge -Q at location at the origin and two charges of the same magnitude but opposite sign at (a,0,0) and (0,a,0). Determine the total force on the charge at the origin.

2) We've got the same charge -Q at the origin but now a charge of 2Q at (a,a,0). Again, determine the force on the charge at the origin.
Then compare your answer to the previous question

The first thing I did was to calculate the separate forces and then add them together.

\vec{F}_1 = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2} \hat{x}
\vec{F}_2 = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2} \hat{y}

So the total force is given by
\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2} (\hat{x}+\hat{y})

Now for the second question the distance between the two charges is given by \sqrt{a^2+a^2}, so the force is given by
\vec{F} = \frac{2Q^2}{4\pi\epsilon_0}\frac{a\hat{x} + a\hat{y}}{(a^2+a^2)^{\frac{3}{2}}} = \frac{2Q^2}{4\pi\epsilon_0}\frac{a\hat{x} + a\hat{y}}{2\sqrt{2}a^3} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{a^2}\frac{1}{\sqrt{2}}(\hat{x} + \hat{y})

Now I would like to ask you if these calculations are really correct?

 

[Doc Al]

So the total force is given by
\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2} (\hat{x}+\hat{y})

Perfectly correct. You may want to write your answer as a magnitude times a unit vector:
\vec{F} = \sqrt{2} \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2} [\frac{1}{\sqrt{2}}(\hat{x}+\hat{y})]

Now for the second question the distance between the two charges is given by \sqrt{a^2+a^2}, so the force is given by
\vec{F} = \frac{2Q^2}{4\pi\epsilon_0}\frac{a\hat{x} + a\hat{y}}{(a^2+a^2)^{\frac{3}{2}}} = \frac{2Q^2}{4\pi\epsilon_0}\frac{a\hat{x} + a\hat{y}}{2\sqrt{2}a^3} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{a^2}\frac{1}{\sqrt{2}}(\hat{x} + \hat{y})

Again, perfectly correct.

 

[wais]

Yes, your calculations are correct! In the first scenario, the charges at (a,0,0) and (0,a,0) are both a distance of a away from the charge at the origin, resulting in a force of Q^2/(4πε0a^2) along the x and y directions. Adding these forces together gives a total force of Q^2/(4πε0a^2)(x+y).

In the second scenario, the charge at (a,a,0) is a distance of √(a^2+a^2) = √2a away from the charge at the origin, resulting in a force of 2Q^2/(4πε0(√2a)^2) = Q^2/(4πε0a^2√2) along the x and y directions. Adding these forces together gives a total force of Q^2/(4πε0a^2√2)(x+y).

Comparing the two scenarios, we can see that the total force in the second scenario is larger due to the closer distance between the charges. This shows that the force between two charges is inversely proportional to the distance between them.

Great job on your calculations! Keep up the good work!