Suppose we have a distribution of point charges as shown in the figure. Find the electric potential and the electric field at the point P.

• All charges in the distribution are point charges.
• There are three point charges:
  • -Q, a distance 2R to the left of P
  • Q, a distance R above P
  • Q, a distance 2R to the right of P
• We can use superposition to add electric field contributions from the point charges (vector superposition): $\vec{E}_{tot}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_3$.
• We can use superposition to add electric potential contributions from the point charges (scalar superposition): $V_{tot}=V_1+V_2+V_3$.

• The electric field at P.
• The electric potential at P.

• The electric field and electric potential at $P$ is due entirely to the three point charges.
• The electric potential infinitely far away from the point charge is $0 \text{ V}$.
• Charges are fixed in place and won't move.
• Charge is constant (not discharging)

• The electric field from a point charge can be written as $$\vec E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}.$$
• The electric potential from a point charge can be written as $$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}.$$
• We can set up coordinate axes to define direction and unit vectors.

In this example, it makes sense to set up coordinate axes so that the $x$-axis stretches from left to right, and the $y$-axis stretches from down to up. To make it easier to discuss the example, we'll also label the point charges as Charge 1, Charge 2, and Charge 3, as shown in the diagram.

First, let's find the contribution from Charge 1. The vector $\vec{r}_1$ points from the source to P, so $\vec{r}_1 = 2R\hat{x}$, and $$\hat{r_1}=\frac{\vec{r}_1}{|r_1|}=\frac{2R\hat{x}}{2R}=\hat{x}$$ Visually, this is what we know about $\hat{r_1}$, and what we expect for $\vec{E}_1$, since Charge 1 is negative: Now, we can find $\vec{E}_1$ and $V_1$: $$\vec{E}_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{4R^2}\hat{x} = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}$$ $$V_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{2R} = -\frac{1}{8\pi\epsilon_0}\frac{Q}{R}$$ These answers should make sense. We have a negative electric potential, which we would expect from a negative charge. We also found that our electric field points toward Q1, which we would again expect because Q1 is negative.

For Charge 2, we expect the following visual to be accurate, again since $\hat{r_2}$ points from source to P, and Charge 2 is positive: We can determine that $\hat{r_2}=-\hat{y}$ (Pointing from Q2 to P), and $$\vec{E}_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}(-\hat{y}) = -\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\hat{y}$$ $$V_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$ These answers should also make sense for a few reasons: 1) we now have a positive electric potential (coming from the positive Q2 charge), 2) the electric field points away from Q2 (which we could expect since Q2 is positive), and 3) both the magnitude of the electric field and the electric potential are larger than those from Q1, which makes sense because Q2 is closer to Point P.

For Charge 3, we expect the following visual to be accurate, since Charge 3 is positive: We can determine that $\hat{r_3}=-\hat{x}$ (Pointing from Q3 to P), and $$\vec{E}_3 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{4R^2}(-\hat{x}) = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}$$ $$V_3 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{2R} = \frac{1}{8\pi\epsilon_0}\frac{Q}{R}$$ Again, we get a positive electric potential from the positive charge (so this is good), and we get an electric field that points away from the positive charge, so this also makes sense.

Now that we have the individual contributions from the point charges, we can add together the potentials and use the principle of superposition to add together the electric field vectors (see above!): $$\vec{E}_{tot}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_3 = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}-\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\hat{y}-\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x} = \frac{1}{8\pi\epsilon_0}\frac{Q}{R^2}(-\hat{x}-2\hat{y})$$ $$V_{tot}=V_1+V_2+V_3 = -\frac{1}{8\pi\epsilon_0}\frac{Q}{R}+\frac{1}{4\pi\epsilon_0}\frac{Q}{R}+\frac{1}{8\pi\epsilon_0}\frac{Q}{R} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$ So our total electric potential is positive - this should make sense as we have two positive charges and only one negative charge, and one of the positive charges is much closer to P than the other charges so it should have a stronger effect. We found that our electric points down and to the left, which makes sense as it points away from both of the positive charges and towards the negative charge.