Suppose we have a distribution of point charges in a plane near a point $P$. There are three point charges: Charge 1 with charge $-Q$, a distance $2R$ to the left of $P$; Charge 2 with charge $Q$, a distance $R$ above $P$; and Charge 3 with charge $Q$, a distance $2R$ to the right of $P$. 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 $P$
• 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 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): $V_{tot}=V_1+V_2+V_3$.

• Find the electric field and electric potential at P.

• The electric potential infinitely far away from the point charge is $0 \text{ V}$.

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 $V_1$. Before we show the calculation, though, we need to make an assumption about the electric potential.

Now, we are ready to find the contributions from Charge 1 on the electric field and electric potential. $$\vec{E}_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{4R^2}\hat{x} = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}$$ $$V_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{2R} = -\frac{1}{8\pi\epsilon_0}\frac{Q}{R}$$ 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.