0
$\begingroup$

Work done by electric field in moving a unit charge is the potential drop. In wires voltage (potential drop) is given by ohms law. Can somebody explain to me why work done by electric field can not be used to calculate the potential drop.

$\endgroup$
0
$\begingroup$

The work done by the electric field can indeed be used to determine the voltage drop. However, in circuits the primary quantities of interest are voltage and current. Finding the electric field is not typically useful in circuit theory.

$\endgroup$
0
$\begingroup$

The voltage in a electric circuit with resistances can be computed starting from its defintition as path integral over the electrical field:

$$ V = \int_0^l E\cdot dr$$

where the path integral can be considered along a straight line. We can benefit from the close relationship of the electric field in a wire to the current flux J:

$$ J= \sigma E$$ where $\sigma$ is the conductivity of the wire material. Moreover we use the relationship between the current flux and the current $I$ through the wire: $I = J A$ where $A$ is the cross section surface of the wire. Putting all together we get (assuming that the properties of the wire don't change along the length of the wire):

$$V= \int_0^l E\cdot dr = E\,l =\frac{1}{\sigma}J l =\frac{1}{\sigma} I \frac{l}{A} $$

Recalling the definition of an Ohm's resistance $R = \rho \frac{l}{A}$ where $\rho=\frac{1}{\sigma}$ is the resistivity of the wire material, we get finally:

$$V= \int_0^l E\cdot dr = I\rho \frac{l}{A} = RI$$

This demonstrates that the definition of the voltage drop within a circuit by the means of the electrical field is equivalent with Ohm's law. However, in circuit theory it is more practical to use $R$ and $I$, in particular $R$, because it gathers in its definition quantities which all are properties of the wire. On the other hand the wire properties making up $R$ are distributed over both the electrical field $E$ and the path length $l$ which makes the use of the latter in circuit theory unhandy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.