## Electric Potential Energy

Electric potential energy is similar to gravitational potential energy. We need to calculate the energy to completely separate two charges instead of two masses.

The main mathematical difference from Coulomb's law is that the radius is no longer squared and we get a scalar instead of a vector as our result.

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

\(U_e\) = electric 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]

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

## solution

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

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

$$ \Delta U_{e} = \text{final - initial} $$ $$ \Delta U_{e} = \frac{k_{e}(1)(-1)}{100} - \frac{k_{e}(1)(-1)}{5} $$ $$ \Delta U_{e} = -0.01k_{e} + 0.25k_{e} $$ $$ \Delta U_{e} = 0.24k_{e}$$

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

## Chemistry and Coulomb's Law

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

**Example**: Calculate what Coulomb's law predicts for the force that holds together regular table salt, NaCl.

## lengthy investigation

**Na: Sodium**

1 valence electron

mass = 3.8 x 10
^{-26} kg

ionic radius = 227 x 10
^{-12} m

**Cl: Chlorine**

7 valence electrons

mass = 5.9 x 10
^{-26} kg

ionic radius = 175 x 10
^{-12} m

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 becomes Na+1 and Cl becomes Cl-1. This unbalanced positive and negative charge produces an attractive force.

$$r = \left(227 \times 10^{-12}\right) + \left(175 \times 10^{-12}\right) = 402 \times 10^{-12}m $$ $$ F = \frac{k_{e}q_{1}q_{2}}{r^{2}} $$ $$ F = \frac{(8.987 \times 10^{9})(1.602 \times 10^{-19})(1.602 \times 10^{-19})}{(402 \times 10^{-12})^{2}} $$ $$ F = 1.427 \times 10^{-9} \, \mathrm{N}$$That's a small force, but atoms don't have much mass. Let's calculate the acceleration that Na feels.

$$F = ma$$ $$a = \frac{F}{m}$$ $$a = \frac{1.427 \times 10^{-9}}{3.8 \times 10^{-26}}$$ $$a = 3.75 \times 10^{16} { \, \mathrm{\tfrac{m}{s^{2}}}}$$That acceleration is very large! Let's calculate the electric potential energy between the ions.

$$ U_{e} = \frac{k_{e}q_{1}q_{2}}{r} $$ $$ U_{e} = \frac{(8.987 \times 10^{9})(1.602 \times 10^{-19})(1.602 \times 10^{-19})}{402 \times 10^{-12}} $$ $$ U_{e} = 5.737 \times 10^{-19} \, \mathrm{J}$$The internet has the "NaCl Dissociation Energy" slightly higher at 6.82 x 10
^{-19} J. Not bad Coulomb!

## Electric Potential

Electric potential is similar to electric potential energy, with one difference. We assume one of the charges is a +1C test charge. The potential is the energy needed to bring the +1C test charge from very far away to some distance from the other charge.

Electric potential has units of volts.

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

\(V\) = 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]

Here is a simulation of electric potential as a scalar field. You can think of the charges like mountains and valleys. A similar scalar field for gravity would look like a topographical map.

Click the simulations a few times to push the electrons around.

**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.

Find the electric potential one nanometer away from just one electron.

What about 100 electrons at 0.01 nanometers?

## solution

$$ V = \frac{k_{e}q}{r} $$ $$ V = \frac{(8.987 \times 10^{9})(-1.6 \times 10^{-19})}{10^{-9}} $$ $$ V = -1.440\, \mathrm{volts}$$$$ V = \frac{(8.987 \times 10^{9})(100 \times -1.6 \times 10^{-19})}{10^{-11}} $$ $$ V = -14,400 \, \mathrm{volts}$$

**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

$$ V = \frac{\color{red}{k_{e}q_{1}}}{\color{red}{r}} $$ $$ U_{e} = \frac{ {\color{red}{k_{e}q_{1} } }{q_2}}{\color{red}{r}} $$ $$ U_{e} = V q $$ $$ U_{e} = (10)(4 \times 10^{-6}) $$ $$ U_{e} = 40 \times 10^{-6} \, \mathrm{J} $$## Electric Potential Difference, Voltage

Electric potential difference is often just called voltage. It plays a significant role in electronics. Voltage is simply a change in electric potential. It can be thought of as the work it would take to move a +1C test charge between two points.

## $$ \Delta V = V_{f} - V_{i}$$ $$ \Delta V = \frac{k_{e}q}{r_{f}} - \frac{k_{e}q}{r_{i}}$$

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

\(V_f\) = final voltage [V, volts]

\(V_i\) = initial voltage [V, volts]

A voltage applied to a wire will cause the electrons to flow, possibly doing something useful.

Voltage increases can be produced by: batteries and electrical sockets.

Voltage drops can be caused by: resistors, processors, lights, motors, speakers.

**Example:**What is the electric potential difference between 1 m and 4 m away from a 2 C charge?

## solution

$$ V = \frac{k_{e}q}{r} $$$$\Delta V = V_{f} - V_{i} $$ $$\Delta V = \frac{k_{e}2}{1} - \frac{k_{e}2}{4} $$ $$\Delta V = k_{e} \left(\frac{2}{1} - \frac{2}{4} \right) $$ $$\Delta V = 1.5k_{e} $$ $$\Delta V = 1.348 \times 10^{10} \, \mathrm{volts} $$