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 × 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.
answer
When the distance increases the energy decreases. As the distance approaches infinity the energy approaches zero.
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.
solution
$$ U_{e} = \frac{k_{e}q_{1}q_{2}}{r} $$ $$ r = \frac{k_{e}q_{1}q_{2}}{U_{e}} $$ $$ r = \frac{(8.987 \times 10^{9})(1 \times 10^{-3})(1 \times 10^{-3})}{200} $$ $$ r = \frac{8.987 \times 10^{3}}{200} $$ $$ r = 44.935 \, \mathrm{m} $$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
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.
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
$$E_i = E_f $$ $$K_1 + K_2 = U_e$$ $$\tfrac{1}{2}mv^2 + \tfrac{1}{2}mv^2 = \frac{k_{e}q_{1}q_{2}}{r}$$ $$mv^2 = \frac{k_{e} q^2}{r}$$ $$r = \frac{k_{e} q^2}{mv^2}$$ $$r = \frac{(8.987 \times 10^{9})(1.602 \times 10^{-19})^2}{(1.67 \times 10^{-27})(1)^2}$$ $$r = 0.138 \, \mathrm{m}$$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 × 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.
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.
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
$$ V = \frac{(8.987 \times 10^{9})(-1.6 \times 10^{-19})}{10^{-12}} $$ $$ V = -1440 \, \mathrm{volts}$$solution
$$ V = \frac{\color{red}{k_{e}q_{1}}}{\color{red}{r}} $$ $$ U_{e} = \frac{ {\color{red}{k_{e}q_{1} } }{q_2}}{\color{red}{r}} $$ $$ U_{e} = V q $$ $$ U_{e} = (10)(4 \times 10^{-6}) $$ $$ U_{e} = 40 \times 10^{-6} \, \mathrm{J} $$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 simulations of the flow of charges with electric potential shown as height.