A charge of 29 `microC is surrounded by an electric field of 73 volts / meter. Determine the potential gradient dV/dx of the field, the force exerted on the charge, the work done by the field in moving this charge .13 meters in the direction of the field, and the force per unit charge exerted by the field.
An electric field of 73 volts/meter is equivalent to 73 N / C. Thus the 29 `microC charge will experience a force of 29 `microC ( 73 N/C) = .002117 N.
The work done over a .13 meter displacement in the direction of the field is
The potential gradient is just the 73 volts / meter; the potential gradient and the electric field strength are identical.
The force per unit charge is just the 73 N/C calculated above. The electric field strength is the force per unit charge, which is what this quantity represents.
A charge q in an electric field E will experience a force of q E, as is obvious from the units (E is measured in volts/meter, the same as Newtons/Coulomb, while q is measured in Coulombs).
The work done in moving the charge through a displacement `ds in the direction of the field will have magnitude |W| = |q E | `ds.
The potential difference has magnitude |W| / q = | E | `ds.
The potential gradient is potential difference/displacement, so has magnitude | E | `ds / `ds = | E |.
Since E measures Newtons/Coulomb, it measures force/unit charge.
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.
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