Problem: Find the force on a charge of 39 Coulombs in the presence of a 27000 V/m potential gradient, and use it to determine the force per Coulomb acting on the charge. Find a symbolic expression for force/Coulomb on a charge q in the presence of a potential gradient dV/dx. Explain why we say that a volt/meter is identical to a Newton/Coulomb.
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Solution:
The force on the 39 Coulomb charge (see preceding problem) is F = q (dV/dt) = 39 C ( 27000 V/m) = 1053000 C V/m = 1053000 C (J/C) / m = 1053000 J/m = 1053000 N m / m = 1053000 N.
The force per Coulomb is 1053000 N / ( 39 C) = 27000 N / C. Actually we nearly came to the conclusion in the preceding problem that V / m = (J/C) / m = (N m / C) / m = N / C.
The force on a charge q in the presence of a potential gradient dV/dx is F = q (dV/dx), so the force per unit charge is
F/q = q (dV/dx) / q = dV/dx.
F / q clearly has units of Newtons / Coulomb; dV/dx has units of Volts / meter.
Generalized Response: As found in the preceding problem, the force on a charge of (q Coulombs) in a potential gradient dV/dx is (q Coulombs) dV/dx. The force per Coulomb is found by dividing by the number q of Coulombs, which gives force per Coulomb of just dV / dx. Thus dV / dx, which measures volts per meter, is also a measure of Newtons per Coulomb. Since a volt is a Joule per Coulomb, a volt per meter turns out to be identical to a Newton per Coulomb.
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Figure description: The figure below shows a charge q moving through a displacement `ds in the direction of a potential gradient dV / dx. The potential difference will be the product of the potential gradient and the displacement: `dV = dV / dx * `ds. The work done on the charge is therefore `dW = q `dV; this work is also the product `dW = Fave * `ds of the average force on the charge and the displacement. Setting these two expressions for work `dW equal we arrive at the conclusion that Fave = q dV/dx = q E.
E = dV / dx is called the electric field. Since E = Fave / q, we see that the electric field E is the force per unit charge experienced by a charge.
