understand how the concept of electric potential leads to the electric potential energy of two point charges and use EP = Qq / (4πε0 r)

Electric Potential

Electric potential (\$V\$) is a scalar quantity that describes the amount of electric potential energy per unit charge at a point in an electric field. It provides a convenient way to discuss the energy changes that occur when charges move, without having to track the vector nature of the electric field directly.

Key Definitions

• Electric Potential (\$V\$): The work done by an external agent in bringing a unit positive test charge from infinity to a point in the field, without acceleration. Measured in volts (V), where \$1\ \text{V}=1\ \text{J C}^{-1}\$.
• Electric Potential Energy (\$U\$): The energy associated with a configuration of charges. For a system of two point charges \$Q\$ and \$q\$ separated by a distance \$r\$, the potential energy is given by

  \$U = \frac{Qq}{4\pi\varepsilon_0 r}.\$

• Permittivity of Free Space (\$\varepsilon0\$): \$\varepsilon0 = 8.854\,\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}\$.

From Electric Potential to Potential Energy

Consider a test charge \$q\$ placed in the electric potential \$V\$ created by another charge \$Q\$. The electric potential energy \$U\$ of the pair is the product of the test charge and the potential at its location:

\$U = qV.\$

Since the potential \$V\$ due to a single point charge \$Q\$ at a distance \$r\$ is

\$V = \frac{Q}{4\pi\varepsilon_0 r},\$

substituting this into \$U = qV\$ yields the familiar expression for the potential energy of two point charges:

\$U = \frac{Qq}{4\pi\varepsilon_0 r}.\$

Sign Conventions

• If both charges have the same sign (both positive or both negative), \$U\$ is positive – the system has stored energy and work must be done to bring the charges together.
• If the charges have opposite signs, \$U\$ is negative – the system releases energy as the charges attract each other.

Worked Example

Two point charges, \$Q = +5.0\ \mu\text{C}\$ and \$q = -2.0\ \mu\text{C}\$, are separated by \$r = 0.10\ \text{m}\$. Calculate the electric potential energy of the system.

1. Convert microcoulombs to coulombs: \$Q = 5.0\times10^{-6}\ \text{C}\$, \$q = -2.0\times10^{-6}\ \text{C}\$.
2. Insert values into the formula:

   \$U = \frac{(5.0\times10^{-6})(-2.0\times10^{-6})}{4\pi(8.854\times10^{-12})(0.10)}.\$

3. Calculate the denominator: \$4\pi\varepsilon_0 = 1.112\times10^{-10}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}\$.
4. Compute \$U\$:

   \$U = \frac{-1.0\times10^{-11}}{1.112\times10^{-11}} \approx -0.90\ \text{J}.\$

The negative sign indicates that the system has lower potential energy than the separated charges at infinity; energy would be released if the charges were allowed to move together.

Summary Table

Common Pitfalls

• Confusing electric potential (\$V\$) with electric potential energy (\$U\$). Remember \$U = qV\$.
• Neglecting the sign of charges when determining whether the potential energy is positive or negative.
• Using the distance \$r\$ measured from the centre of one charge to the centre of the other; for point charges this is straightforward, but for extended objects the appropriate separation must be used.

Practice Questions

1. Two identical positive charges of \$3.0\ \mu\text{C}\$ are placed \$0.05\ \text{m}\$ apart. Calculate the electric potential energy of the system.
2. A charge \$q = +1.5\ \mu\text{C}\$ is moved from a point where the electric potential due to a fixed charge \$Q\$ is \$200\ \text{V}\$ to a point where the potential is \$50\ \text{V}\$. Determine the change in potential energy of \$q\$.
3. Explain why the electric potential energy of a system of like charges increases as the separation \$r\$ decreases, using the formula for \$U\$.