Suppose we have a negatively charged balloon with total charge $Q=-5.0\cdot 10^{-9} \text{ C}$. What is the electric potential at a point $P$, which is a distance $R=20 \text{ m}$ from the center of mass of the balloon? A diagram below shows a rough sketch.

• The balloon has total charge $Q=-5.0\cdot 10^{-9} \text{ C}$.
• The point $P$ is a distance $R=20 \text{ m}$ away from the center of mass of the balloon.

• The electric potential at $P$.

• The electric potential at $P$ is due entirely to the balloon.
• $P$ lies outside of the balloon.
• The balloon's electric field outside the balloon acts like a point charge centered at the center of mass of the balloon.
• The electric potential infinitely far away from the balloon is $0 \text{ V}$. Read here for why this is important.

• The electric potential due to a point charge (to which we are approximating the balloon) can be written as $$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r},$$ where $q$ represents our charge and $r$ is our distance.

The electric potential at $P$ is given by

$$\begin{align*} V &= \frac{1}{4\pi\epsilon_0}\frac{q}{r} \\ &= \frac{1}{4\pi\cdot 8.85\cdot 10^{-12} \frac{\text{C}}{\text{Vm}}}\frac{-5.0\cdot 10^{-9} \text{ C}}{20 \text{ m}} \\ &= -2.2 \text{ V} \end{align*}$$ Notice how the magnitude of charge on the balloon is the same as in the “positively charged balloon” example. The reason the magnitude of the voltage is so much smaller, is because the distance is so much greater. The closer you get to a point charge, the higher the magnitude of electric potential.