Question

A material is said to be ohmic if an electric field, \(\vec{E}\), in the material gives rise to current density \(\vec{J}=\sigma \vec{E},\) where the conductivity, \(\sigma\) is a constant independent of \(\vec{E}\) or \(\bar{J}\). (This is the precise form of Ohm's Law, ) Suppose in some material an electric field, \(\vec{E}\), produces current density, \(\bar{J},\) not necessarily related by Ohm's Law; that is, the material may or may not be ohmic. a) Calculate the rate of energy dissipation (sometimes called ohmic heating or joule heating) per unit volume in this material, in terms of \(\vec{E}\) and \(\vec{J}\). b) Express the result of part (a) in terms of \(\vec{E}\) alone and \(\vec{J}\) alone, for \(\overrightarrow{\boldsymbol{E}}\) and \(\vec{J}\) related via Ohm's Law, that is, in an ohmic material with conductivity \(\sigma\) or resistivity \(\rho\).

Short answer

Answer: In ohmic materials, the rate of energy dissipation per unit volume can be expressed as \(P = \sigma |\vec{E}|^2\) in terms of the electric field, and as \(P = \rho |\vec{J}|^2\) in terms of current density, where \(\sigma\) is the conductivity and \(\rho\) is the resistivity of the material.

Step-by-step solution

Find the power

We'll find the power (P) given by the product of current (I) and voltage (V) per unit volume. Since current density is given by \(\vec{J}\), the power per unit volume can be obtained as: \(P = \vec{J} \cdot \vec{E}\)

Express energy dissipation using E and J

Now, we already have the power per unit volume in terms of the electric field and current density, that is: $$ P = \vec{J} \cdot \vec{E} $$ b) Express energy dissipation in terms of \(\vec{E}\) alone and \(\vec{J}\) alone for ohmic materials.

Express energy dissipation using E alone

We have the Ohm's Law relationship: \(\vec{J} = \sigma \vec{E}\) Substitute this into the equation for power per unit volume obtained earlier: $$ P = (\sigma \vec{E}) \cdot \vec{E} = \sigma (\vec{E} \cdot \vec{E}) $$ Since \(\vec{E} \cdot \vec{E}\) is the square of the magnitude of \(\vec{E}\), we can write: $$ P = \sigma |\vec{E}|^2 $$

Express energy dissipation using J alone

The relationship between resistivity and conductivity is given by: \(\rho = \frac{1}{\sigma}\) So, the Ohm's Law relationship can also be expressed as: \(\vec{E} = \rho \vec{J}\) Substitute this into the equation for power per unit volume obtained earlier: $$ P = \vec{J} \cdot (\rho \vec{J}) = \rho (\vec{J} \cdot \vec{J}) $$ Since \(\vec{J} \cdot \vec{J}\) is the square of the magnitude of \(\vec{J}\), we can write: $$ P = \rho |\vec{J}|^2 $$ Now we have the expressions for the rate of energy dissipation per unit volume in terms of \(\vec{E}\) and \(\vec{J}\) for ohmic materials: $$ P = \sigma |\vec{E}|^2 \quad \text{and} \quad P = \rho |\vec{J}|^2 $$

Key concepts

Electrical Conductivity

Electrical conductivity is a measure of how well a material can allow the flow of electric current. It can be thought of as the traffic rules for electrons moving through a substance. High conductivity means that electrical current can pass through the material with ease, much like a car on a clear highway. Metals like copper and aluminum are known for their high conductivity, which is why they are commonly used in wires and cables.

In technical terms, conductivity, often symbolized as \(\sigma\), is the reciprocal of resistivity \(\rho\). While resistivity quantifies how strongly a material opposes the current, conductivity illustrates its ability to facilitate electrical flow. For instance, imagine turning on a light; the electricity travels from the switch to the light bulb through wires with high conductivity, allowing the bulb to illuminate without significant resistance.

Understanding conductivity is crucial when studying materials and designing electrical components, as choosing the right substance is key to ensuring efficiency and safety in electrical systems.

Current Density

Current density is an important concept in the realm of electricity, and can be compared to the density of traffic on a road. It's a way of describing the electric current - the flow of charge - per unit area through a material. The symbol for current density is \(\vec{J}\), and it's expressed in amperes per square meter (A/m^2).

Just as more cars on a street means higher traffic density, a higher current density means more electrical current is flowing through a given cross-section of a conductor. If you have a wire and increase the voltage - which could be likened to stepping on the gas pedal - you typically increase the current density, assuming the cross-sectional area of the wire and other conditions remain the same.

Current density isn't just about quantity; it also has a direction, making it a vector quantity. This aspect is particularly important when trying to understand and design complex electrical circuits and components, as the direction of current can affect magnetic fields and other electrical properties.

Energy Dissipation

Energy dissipation in electrical systems, often referred to as Joule heating, is like the heat generated when rubbing your hands together. In electrical circuits, as electrons move through a conductor, they encounter resistance, which converts some of their electrical energy into thermal energy, heating up the material. The rate of this energy conversion is what we call power.

The formula for power - the rate of energy dissipation per unit volume - in terms of current density \(\vec{J}\) and electric field \(\vec{E}\) is \(P = \vec{J} \cdot \vec{E}\). This can be further refined in an ohmic material, where Ohm's Law applies, as \(P = \sigma |\vec{E}|^2\), using conductivity, and \(P = \rho |\vec{J}|^2\), using resistivity. Both these expressions show that power is proportional to either the square of the electric field for a given conductivity, or the square of the current density for a given resistivity.

Understanding how and why energy is dissipated in electrical systems is key, not just for calculating efficiency, but also for designing safe systems. Without proper management, excessive energy dissipation can lead to overheating, which can damage components and pose fire hazards. Therefore, it's vital to understand and manage energy dissipation in any electrical system.