## Section31.3Electric Potential of Point Charge

Electric potential of a point charge is the building block for electric potential by more complex charge distributions as we will see in the next section.

Electric potential of a point charge $q$ located at the origin has the same value at a distance $r$ from it, regardless of the direction.

\begin{equation} \phi = \dfrac{1}{4\pi\epsilon_0}\,\dfrac{q}{r}.\label{eq-electric-potential-point-charge-restate}\tag{31.3.1} \end{equation} ### Section31.3.1(Calculus) Derivation of the Formula for Electric Potential for Point Charge

To derive formula in (31.3.1) from the definition of electric potential given in Section 31.1, we need to perform an integral from the reference point, which is at $r=\infty$ to the field point P, which is at $r=r\text{.}$ The integral is required to obtain work done because the electric force varies along the path. Setting the test charge $Q$ to unity makes the integral equal to the electric potential at P.

\begin{equation} \phi_P = - \int_\text{ref}^{P} \vec E \cdot d\vec r.\tag{31.3.1} \end{equation}

Since electric field is radial, the dot product with $d\vec r$ gives

\begin{equation*} \vec E \cdot d\vec r = E\: dr = \frac{1}{4\pi\epsilon_0}\,\frac{q}{r^2}dr, \end{equation*}

where I have replaced the magitude of electric field $E$ by its formula obtained in previous chapter. With reference at $r=\infty\text{,}$ upon doing the integration we get the desired result.

\begin{align*} \phi_P \amp = - \int_\text{\infty}^{r}\frac{1}{4\pi\epsilon_0}\,\frac{q}{r^2}dr \\ \amp = \frac{1}{4\pi\epsilon_0}\,\frac{q}{r}. \end{align*}

(a) What is the electric potential of a proton at a distance of Bohr radius, $5.29\times 10^{−11}\text{ m}\text{,}$ from the proton?

(b) What is the potential energy of an electron when placed $5.29\times 10^{−11}\text{ m}$ from a proton? State both in $\text{eV}$ and $\text{J}\text{.}$

(c) What is the electric potential of an electron at a distance of Bohr radius, $5.29\times 10^{−11}\text{ m}\text{,}$ from the electron?

Data: $e = 1.60\times10^{-19}\text{ C}\text{.}$

Hint

(a) Use $\phi$ for a point charge, (b) $Q\times \phi$ gives potential energu of $Q\text{,}$ (c) No additional work required, just the sign needs to be right.

(a) $27.2\text{ V}\text{,}$ (b) $-27.2\text{ eV}\text{,}$ $-4.35 \times 10^{-18}\text{ J}\text{,}$ (c) $- 27.2\text{ V}\text{.}$

Solution 1 (a)

(a) Using the electric potential formula for a point charge we get

\begin{align*} \phi \amp = \dfrac{1}{4\pi\epsilon_0}\,\dfrac{e}{r},\\ \amp = 9\times 10^9\,\dfrac{1.60\times10^{-19}}{ 5.29\times 10^{−11}} \\ \amp = 27.2\text{ V}. \end{align*}
Solution 2 (b)

(b) By multiplying potential at the location of the electron by the charge of the electron we will get the potential energy of the electeron.

\begin{align*} \text{EPE} \amp = -e\,\phi = -27.2\text{ eV},\\ \amp = -1.60\times10^{-19}\text{C}\times 27.2\text{ V} = -4.35 \times 10^{-18}\text{ J}. \end{align*}
Solution 3 (c)

(c) This requires no additional work since absolute value of the charge of an electron is same as that of proton. The only difference from (a) is the sign.

\begin{equation*} \phi = - 27.2\text{ V}. \end{equation*}