Electrostatic potential energy has a similar form to gravitational potential energy. You can think of the electrostatic energy as the work to move two charges to a distance, r, from each other.

## $$ U_{e} = \frac{k_{e}q_{1}q_{2}}{r} $$

\(U_e\) = electrostatic potential energy [J, joules, kg m²/s²]

\(k_e\) = 8.987 × 10^{9} = Coulomb's constant [N m²/C²]

\(q\) = charge [C, Coulomb]

\(r\) = distance between the center of each charge [m, meters]

Only valid for stationary point charges.

Like all energy, electrostatic potential energy is a scalar, but it can go negative.

**Question:**How far apart do you need to bring two charges for there to be zero electric potential energy between them?

## answer

When the distance increases the energy decreases. As the distance approaches infinity the energy approaches zero.

**Example:**You rub a balloon on a dry erase board and pull -200 nC off the board onto the balloon. How much energy does it take to pull the balloon horizontally off a dry erase board if the centers of the charges are 5 mm apart?

## solution

$$\text{n = nano} = 10^{-9} \quad \quad \text{m = milli} = 10^{-3}$$ $$ U_{e} = \frac{k_{e}q_{1}q_{2}}{r} $$ $$ U_{e} = \frac{(8.987 \times 10^{9}) (200 \times 10^{-9})(-200 \times 10^{-9})}{5 \times 10^{-3}} $$ $$ U_{e} = -0.0719 \, \mathrm{J}$$The system has -0.0719J of energy. To separate the objects we will have to cancel out that energy, so it will take **0.0719J**.

**Example:**You use 200 J of energy to move a +1 mC charge towards another +1 mC charge. How close are they when you run out of energy?

## solution

$$ U_{e} = \frac{k_{e}q_{1}q_{2}}{r} $$ $$ r = \frac{k_{e}q_{1}q_{2}}{U_{e}} $$ $$ r = \frac{(8.987 \times 10^{9})(1 \times 10^{-3})(1 \times 10^{-3})}{200} $$ $$ r = \frac{8.987 \times 10^{3}}{200} $$ $$ r = 44.935 \, \mathrm{m} $$**Example:**Which will take more work/energy?

Moving two 1 C charges from 4 meters to 2 meters apart?

Moving a 1 C and a -1 C charge from 5 meters to 100 meters apart?

## solution

$$ U_{e} = \frac{k_{e}q_{1}q_{2}}{r} $$$$ \Delta U_{e} = \text{final - initial} $$ $$ \Delta U_{e} = \frac{k_{e}(1)(1)}{2} - \frac{k_{e}(1)(1)}{4} $$ $$ \Delta U_{e} = 0.5k_{e} - 0.25k_{e} $$ $$ \Delta U_{e} = 0.25k_{e} $$

$$ \Delta U_{e} = \text{final - initial} $$ $$ \Delta U_{e} = \frac{k_{e}(1)(-1)}{100} - \frac{k_{e}(1)(-1)}{5} $$ $$ \Delta U_{e} = -0.01k_{e} + 0.20k_{e} $$ $$ \Delta U_{e} = 0.19k_{e}$$

$$\text{4 m to 2 m takes slightly more energy}$$

Chemistry is most accurately described by quantum mechanics, but chemical bonds and chemical reactions can be loosely explained with electrostatic potential energy. Let's see how far classical physics will take us.

**Na: Sodium**

1 valence electron

mass = 3.8 × 10^{-26} kg

ionic radius = 227 × 10^{-12} m

**Cl: Chlorine**

7 valence electrons

mass = 5.9 × 10^{-26} kg

ionic radius = 175 × 10^{-12} m

**Example**: Calculate what Coulomb's law predicts for the energy holding together an atom of sodium and chlorine, NaCl.

## strategy

In nonionic atoms the numbers of electrons and protons are equal so the electrostatic forces are balanced to zero. When NaCl ionically bond one electron leaves Na and joins Cl. Na gains a +1 charge and Cl gains a -1 charge. The unbalanced charges produce an attractive force.

## solution

$$ U_{e} = \frac{k_{e}q_{1}q_{2}}{r} $$ $$ U_{e} = \frac{(8.987 \times 10^{9})(1.602 \times 10^{-19})(-1.602 \times 10^{-19})}{402 \times 10^{-12}} $$ $$ U_{e} = -5.737 \times 10^{-19} \, \mathrm{J}$$Our answer is negative because it would take added positive energy to get the ions to separate.

The measured NaCl dissociation energy is slightly higher at -6.82 × 10^{-19} J. Coulomb's law isn't perfect, but it is a close approximation.

**Example:**Two protons are fired at each other. Both protons have a velocity of 1 m/s, but in opposite directions. What is the minimum distance the protons could reach before they stop? (mass of a proton = 1.6726219 × 10

^{-27}kg)

## strategy

We can use conservation of energy. The sum of both kinetic energies is equal to the electric potential energy. We can then solve for the radius in the potential energy.

## solution

$$E_i = E_f $$ $$K_1 + K_2 = U_e$$ $$\tfrac{1}{2}mv^2 + \tfrac{1}{2}mv^2 = \frac{k_{e}q_{1}q_{2}}{r}$$ $$mv^2 = \frac{k_{e} q^2}{r}$$ $$r = \frac{k_{e} q^2}{mv^2}$$ $$r = \frac{(8.987 \times 10^{9})(1.602 \times 10^{-19})^2}{(1.67 \times 10^{-27})(1)^2}$$ $$r = 0.138 \, \mathrm{m}$$**Example:**How close could the protons from the previous example get if they were both moving at 800 000 m/s towards each other?

## solution

$$r = \frac{k_{e} q^2}{mv^2}$$ $$r = \frac{(8.987 \times 10^{9})(1.602 \times 10^{-19})^2}{(1.67 \times 10^{-27})(800\,000)^2}$$ $$r = 2.158 \times 10 ^{-11} \, \mathrm{m}$$## Electric Potential

Electric potential is mathematically similar to electric fields. Electric fields are the force per unit charge, and electric potential is the energy per unit charge.

$$ V = \frac{U_e}{q}$$You can think of electric potential as the energy needed to bring a +1C test charge from very far away to a distance, r, from the other charge.

## $$V = \frac{k_{e}q}{r} $$

\(V\) = electric potential [V, volts, J/C]

\(k_e\) = 8.987 × 10^{9} = Coulomb's constant [N m²/C²]

\(q\) = charge [C, Coulomb]

\(r\) = distance between the center of each charge [m, meters]

Electric potential has units of volts, a unit that shows up again when dealing with electric circuits.

charges ≈ Here is a Coulomb's law simulation of electric potential as a scalar field. The mountains indicate positive potential, and the valleys indicate negative potential.

Click the simulations a few times to push the charges around. These simulations help develop a vague intuition for electric potential, but they don't capture the nuances of quantum mechanics.

**Example:**A scanning electron microscope can achieve resolution better than 1 nanometer. It produces images by scanning with a focused beam of electrons. The electrons are propelled at a sample target with a voltage between 5 000 V and 25 000 V. Generally the higher voltage gives better resolution.

Let's estimate the voltage needed for a 1 nanometer resolution. We can start by finding the energy to potentially bring an electron very close to another electron. What is the potential at 1 nanometer from an electron?

## solution

$$ V = \frac{k_{e}q}{r} $$ $$ V = \frac{(8.987 \times 10^{9})(-1.6 \times 10^{-19})}{10^{-9}} $$ $$ V = -1.440\, \mathrm{volts}$$What is the potential at 0.001 nm from a single electron?

## solution

$$ V = \frac{(8.987 \times 10^{9})(-1.6 \times 10^{-19})}{10^{-12}} $$ $$ V = -1440 \, \mathrm{volts}$$**Example:**The electric potential at 0.1 m from a charge is 10 J/C. If I were to bring another 4 μC charge to 0.1 m away from the original charge how much energy would that take?

## solution

$$ V = \frac{\color{red}{k_{e}q_{1}}}{\color{red}{r}} $$ $$ U_{e} = \frac{ {\color{red}{k_{e}q_{1} } }{q_2}}{\color{red}{r}} $$ $$ U_{e} = V q $$ $$ U_{e} = (10)(4 \times 10^{-6}) $$ $$ U_{e} = 40 \times 10^{-6} \, \mathrm{J} $$**Example:**A Van de Graaff generator produces an electric potential difference of 40 000 V. If an electron started from rest, what is the maximum speed it could gain from the potential difference?

## strategy

Use conservation of energy. The electron will start with only potential energy (U = Vq). It will end with only kinetic energy.

## solution

$$E_i = E_f$$ $$U_e = K$$ $$qV = \tfrac{1}{2}mv^2$$ $$v^2 = \frac{2qV}{m}$$ $$v = \sqrt{\frac{2qV}{m}}$$ $$v = \sqrt{\frac{2(1.6 \times 10^{-19})(40\,000)}{9.1 \times 10^{-31}}}$$ $$v = 1.19 \times 10^{8} \, \mathrm{\tfrac{m}{s}}$$This result is near the speed of light. We need to use relativistic kinetic energy to improve the accuracy.

These are Coulomb's law based simulation of the flow of charges with electric potential shown as height.