# Simple electrostatics

1. Feb 12, 2005

### 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.

The first thing I did was to calculate the seperate 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?

2. Feb 12, 2005

### Staff: Mentor

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

Again, perfectly correct.