Electric Potential Energy

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.

q 1 q 2 q 2 r ∞

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

\(U_e\) = electrostatic potential energy [J, joules, kg m²/s²]
\(k_e\) = 8.987 × 109 = 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

Na: Sodium
1 valence electron
mass = 3.8 × 10-26 kg
ionic radius = 227 × 10-12 m

Cl

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 Na +1 Cl -1

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.

q +1C +1C r ∞

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

\(V\) = electric potential [V, volts, J/C]
\(k_e\) = 8.987 × 109 = 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.

Use the mouse to move the camera. Double click for full screen.
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.
This 2D simulation uses color to show electric potential. Right click adds positive charge, middle mouse adds negative charge.

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.

Use the mouse to move the camera. Double click for full screen.

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

Practice printout.pdf

In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

Question: A positive source charge creates electric potential at a nearby point. Is that potential positive, negative, or zero?
answer

The potential is positive. Electric potential has the same sign as the source charge in the equation V = ke q / r.

Example: A +2.0 μC charge and a -3.0 μC charge are 0.20 m apart. What is the electric potential energy of the pair?
solution $$U_e=\frac{k_e q_1 q_2}{r}$$ $$U_e=\frac{(8.99\times10^9)(2.0\times10^{-6})(-3.0\times10^{-6})}{0.20}$$ $$U_e=-0.270\,\mathrm{J}$$

The energy is negative because opposite charges are bound together. It would take positive work to separate them very far apart.

Example: A balloon gains 150 nC of negative charge from a wall, leaving the nearby wall patch with +150 nC. If their charge centers are 4.0 mm apart, how much energy would it take to separate them very far apart?
solution

Find the electric potential energy of the close pair. The work to separate them is the positive amount needed to bring that energy up to zero.

$$q_1=150\times10^{-9}\,\mathrm{C}$$ $$q_2=-150\times10^{-9}\,\mathrm{C}$$ $$r=0.0040\,\mathrm{m}$$ $$U_e=\frac{k_e q_1 q_2}{r}$$ $$U_e=\frac{(8.99\times10^9)(150\times10^{-9})(-150\times10^{-9})}{0.0040}$$ $$U_e=-0.0506\,\mathrm{J}$$

It takes 0.0506 J of work to separate them very far apart.

Example: Two identical +4.0 μC charges have 0.72 J of electric potential energy. How far apart are their centers?
solution $$U_e=\frac{k_e q_1 q_2}{r}$$ $$r=\frac{k_e q_1 q_2}{U_e}$$ $$r=\frac{(8.99\times10^9)(4.0\times10^{-6})(4.0\times10^{-6})}{0.72}$$ $$r=0.20\,\mathrm{m}$$
Example: What electric potential does a +8.0 nC source charge create at a point 12 cm away?
solution $$q=8.0\times10^{-9}\,\mathrm{C}$$ $$r=0.12\,\mathrm{m}$$ $$V=\frac{k_e q}{r}$$ $$V=\frac{(8.99\times10^9)(8.0\times10^{-9})}{0.12}$$ $$V=599\,\mathrm{V}$$
Example: What electric potential does a -6.0 nC source charge create at a point 3.0 cm away?
solution $$q=-6.0\times10^{-9}\,\mathrm{C}$$ $$r=0.030\,\mathrm{m}$$ $$V=\frac{k_e q}{r}$$ $$V=\frac{(8.99\times10^9)(-6.0\times10^{-9})}{0.030}$$ $$V=-1800\,\mathrm{V}$$

The potential is negative because the source charge is negative.

Example: A small charged sphere on an insulating stand creates a potential of +4.5 × 103 V at a point 10 cm from its center. What is the sphere's charge?
solution $$r=0.10\,\mathrm{m}$$ $$V=\frac{k_e q}{r}$$ $$q=\frac{Vr}{k_e}$$ $$q=\frac{(4.5\times10^3)(0.10)}{8.99\times10^9}$$ $$q=5.0\times10^{-8}\,\mathrm{C}$$

The sphere's charge is +50 nC.

Example: A +3.0 μC bead is placed at a location where the electric potential is -1200 V. The bead is taped to a plastic holder, but its potential energy depends on charge and potential. What is the bead's electric potential energy there?
solution $$V=\frac{U_e}{q}$$ $$U_e=Vq$$ $$U_e=(-1200)(3.0\times10^{-6})$$ $$U_e=-3.6\times10^{-3}\,\mathrm{J}$$
Example: A +2.0 μC bead is placed 0.30 m from a charged sphere. What is the bead's electric potential energy?
answer This cannot be solved from the information given. Electric potential energy needs the electric potential at that point, or enough information to calculate it. The sphere's charge is missing.
Example: An electron is at a location where the electric potential is +1500 V. What is the electron's electric potential energy?
solution

An electron has charge -1.60 × 10-19 C.

$$U_e=Vq$$ $$U_e=(1500)(-1.60\times10^{-19})$$ $$U_e=-2.40\times10^{-16}\,\mathrm{J}$$
Example: A +4.0 nC charge is 20 cm west of a point, and a -2.0 nC charge is 10 cm east of the same point. What is the electric potential at the point?
answer

Electric potential is a scalar, so add the positive and negative contributions.

$$V_1=\frac{(8.99\times10^9)(4.0\times10^{-9})}{0.20}$$ $$V_1=180\,\mathrm{V}$$ $$V_2=\frac{(8.99\times10^9)(-2.0\times10^{-9})}{0.10}$$ $$V_2=-180\,\mathrm{V}$$ $$V_{net}=V_1+V_2$$ $$V_{net}=0\,\mathrm{V}$$
Example: A +6.0 nC charge is 30 cm to the left of a point. A -3.0 nC charge is 10 cm to the right of the point. What is the electric potential at the point?
solution $$V_1=\frac{(8.99\times10^9)(6.0\times10^{-9})}{0.30}$$ $$V_1=180\,\mathrm{V}$$ $$V_2=\frac{(8.99\times10^9)(-3.0\times10^{-9})}{0.10}$$ $$V_2=-270\,\mathrm{V}$$ $$V_{net}=V_1+V_2$$ $$V_{net}=-90\,\mathrm{V}$$

The point has negative potential because the closer negative charge contributes more strongly.

Example: A +4.0 nC charge is 20 cm west of a point, and a -2.0 nC charge is 10 cm east of the same point. The electric potential at the point is zero. What is the electric field at that point?
solution

The potential cancels, but the field does not. At the point, the positive charge makes a field east, and the negative charge also makes a field east.

$$E_1=\frac{(8.99\times10^9)(4.0\times10^{-9})}{(0.20)^2}$$ $$E_1=899\,\mathrm{N/C}$$ $$E_2=\frac{(8.99\times10^9)(2.0\times10^{-9})}{(0.10)^2}$$ $$E_2=1800\,\mathrm{N/C}$$ $$E_{net}=E_1+E_2$$ $$E_{net}=2700\,\mathrm{N/C}$$

The electric field is 2700 N/C east.

Example: A +2.0 μC bead moves from a 50 V location to a 300 V location. What is the change in its electric potential energy?
solution $$\Delta V=V_f-V_i$$ $$\Delta V=300-50$$ $$\Delta V=250\,\mathrm{V}$$ $$\Delta U_e=q\Delta V$$ $$\Delta U_e=(2.0\times10^{-6})(250)$$ $$\Delta U_e=5.0\times10^{-4}\,\mathrm{J}$$

The bead's electric potential energy increases by 0.00050 J.

Example: An electron starts from rest and is accelerated through a 1200 V potential difference. Using the nonrelativistic kinetic energy equation, what speed could it reach?
solution

The electric potential energy becomes kinetic energy.

$$qV=\tfrac{1}{2}mv^2$$ $$v^2=\frac{2qV}{m}$$ $$v=\sqrt{\frac{2qV}{m}}$$ $$v=\sqrt{\frac{2(1.60\times10^{-19})(1200)}{9.11\times10^{-31}}}$$ $$v=2.1\times10^7\,\mathrm{m/s}$$

This is fast, but still far below the speed of light.

Example: Two protons move directly toward each other in a simplified head-on model. Each proton starts with speed 4.0 × 105 m/s. What is the closest distance between their centers if all their kinetic energy becomes electric potential energy?
solution

Use conservation of energy. There are two moving protons, so the total starting kinetic energy is mv².

$$K_i=U_f$$ $$mv^2=\frac{k_e q^2}{r}$$ $$r=\frac{k_e q^2}{mv^2}$$ $$r=\frac{(8.99\times10^9)(1.60\times10^{-19})^2}{(1.67\times10^{-27})(4.0\times10^5)^2}$$ $$r=8.6\times10^{-13}\,\mathrm{m}$$