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, UB - UA 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 VB - VA 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 = keq/r2. Replacing in equation A), changing
ds by dr, we obtain
VB- VA = keq(1/rB - 1/rA)
Defining VA = 0 at rA = infinite, the potential energy due to a point charge q1 at any distance r from it, is
V = keq1/r
The electric potential energy associated to a pair of point charges separated by a distance r12 is then
U = Vq2 = keq1q2/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
U = keq1q2/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: Ufinal - Uinitial + Kfinal - Kinitial = 0 (K: kinetic energy).
Conservation of momentum: 0 = (0.2 kg)v1 + (0.8 kg)v2
or v1 = - 4v2
Uinitial = 9x109 [Nm2/C2](-3x10-6 C)(4x10-6)/0.9 m = - 0.12 J
Ufinal = 9x109 [Nm2/C2](-3x10-6 C)(4x10-6)/0.01 m = - 10.8 J
Kfinal- Kinitial = ½ (0.2 kg)v12 + ½ (0.8 kg)v22 (Kinitial = 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.
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