In a uniform electric field of 41 volts / meter, determine the force exerted on a charge of 59 `microC in the vicinity of this field, the work done by the field in moving this charge .1 meters in the direction of the field, the potential gradient dV/dx of the field and the force per unit charge exerted by the field.
An electric field of 41 volts/meter is equivalent to 41 N / C. Thus the 59 `microC charge will experience a force of 59 `microC ( 41 N/C) = .002419 N.
The work done over a .1 meter displacement in the direction of the field is
The potential gradient is just the 41 volts / meter; the potential gradient and the electric field strength are identical.
The force per unit charge is just the 41 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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