Potential Difference. Electric Potential. Potential Energy. Conceptual Example.

When a test charge q₀ is placed in an electric field E, the electric force acting on the test charge is q₀E.

For an infinitesimal displacement ds the work done by the electric field on the charge is Fds = q₀Eds. As this work is done by the field, the potential energy of the field-charge system is reduced in a quantity dU = - q₀Eds.

For a charge displacement between points A and B, the change in potential energy of the system, U_B - U_A is

The integration is performed along the path described by q₀. This integral is called line integral.

The potential energy per unit charge U/q₀ does not depend on the value of q₀, and has a unique value in each point in an electric field. U/q₀ receives the name electric potential, or simply potential, and is denoted by V, V = U/q₀.

The potential difference V_B - V_A between any two points A and B in an electric field is defined as the change in potential energy divided by the test charge q₀:

The SI unit for the electric potential (and for the potential difference) is joule/coulomb, a unit called a volt, abbreviated V,
1 V = 1 J/C.

Electric Potential and Potential Energy Due to Point Charges.

We know the electric field E due to a point charge q is E = k_eq/r². Replacing in equation A), changing ds by dr, we obtain

V_B- V_A = k_eq(1/r_B - 1/r_A)

Defining V_A = 0 at r_A = infinite, the potential energy due to a point charge q₁ at any distance r from it, is

V = k_eq₁/r

The electric potential energy associated to a pair of point charges separated by a distance r₁₂ is then

U = Vq₂ = k_eq₁q₂/r₁₂

Electric Potential Energy Example.

Two non conductor spheres with 0.4 cm and 0.6 cm radius, 0.2 kg and 0.8 kg mass and -3 micro Coulombs and 4 micro Coulombs charge, respectively, are released from rest when their centers are separated 0.9 m. Calculate each sphere speed when they collide. The charge is uniformly distributed on each sphere.

The electric potential energy for point charges is

U = k_eq₁q₂/r₁₂.

We can use this relation because the charge is uniformly distributed on each sphere so all the charge can be considered as concentrated in their centers.

Also, two conservation principles can be used:

Conservation of energy: U_final - U_initial + K_final - K_initial = 0 (K: kinetic energy).

Conservation of momentum: 0 = (0.2 kg)v₁ + (0.8 kg)v₂
or v₁ = - 4v₂

U_initial = 9x10⁹ [Nm²/C²](-3x10^-6 C)(4x10^-6)/0.9 m = - 0.12 J
U_final = 9x10⁹ [Nm²/C²](-3x10^-6 C)(4x10^-6)/0.01 m = - 10.8 J
K_final- K_initial = ½ (0.2 kg)v₁² + ½ (0.8 kg)v₂² (K_initial = 0)

Replacing, ½ (0.2 kg)v₁²+ ½ (0.8 kg)v₂² = 10.68 J and using
v₁ = - 4v₂ to solve this system of equations, we finally obtain:
v₁ = 9.24 m/s, v₂= 2.31 m/s.

Exercises, Potential and
Electric Potential Energy