Problem: A charge of 8 C moves from point A to point B, a distance of 27 m. No energy is dissipated. 14 Joules of work are done on the charge. What is the average electric field strength between A and B (give the magnitude of the average field)?
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Solution: The average strength of an electric field nis measured as the average force per unit charge. The average force associated with 14 Joules of work over 27 meters is ( 14 J)/( 27 m) = ( 14 N m)/( 27 m) = .5185 N. The field strength associated with a force of .5185 N on a charge of 8 C is |( .5185 N)/( 8 C)| = .06481 N/C.
Generalized Response: If a charge Q moves between two points and `dW Joules of work are done as the charge moves through a distance `ds, then we easily determine the average force F = `dW / `ds on the charge. The electric field is the force per unit charge (measured in Newtons/Coulomb), or in this case
E = F / Q = (`dW / `ds) / Q = `dW / (Q `ds).
Note that since `dW / Q = `dV, the potential difference, we have
E = `dW / (Q `ds) = `dV / `ds.
This shows that the electric field can also be expressed in units of volts per unit distance (i.e., volts / meter).
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Figure description: The figure below shows a charge Q moving from a low potential V1 to a high potential V2. The potential difference is `dV = V2 - V1, and the work required to move the charge is W = Q `dV (`dV is measured in Joules of work required per Coulomb of charge, so work must be the product of Q (Coulombs) and `dV (Joules / Coulomb) ).
