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 \(\vec{J}\). (This is the precise form of Ohm's Law.) Suppose in some material an electric field, \(\vec{E}\), produces current density, \(\vec{J},\) 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 \(\vec{E}\) and \(\vec{J}\) related via Ohm's Law, that is, in an ohmic material with conductivity \(\sigma\) or resistivity \(\rho .\)

(Recall the formula for power in terms of electric field and current density: \(P = \vec{J} \cdot \vec{E}\). The rate of energy dissipation per unit volume (\(W\)) is power per unit volume, so we simply divide the power by the volume: \(W = \frac{P}{V}\). Now, substituting the formula for power into the expression, we get \(W = \frac{\vec{J} \cdot \vec{E}}{V}\). Knowing that power has the unit of W (watt) and volume has the unit of m^3, we can deduce the unit of rate of energy dissipation: W/m^3.)

(For an ohmic material, we have \(\vec{J} = \sigma \vec{E}\). Plugging this into the expression for rate of energy dissipation, we get \(W = \frac{(\sigma \vec{E}) \cdot \vec{E}}{V} = \frac{\sigma \vec{E} \cdot \vec{E}}{V}\). Now, simplifying the expression gives \(W = \frac{\sigma |\vec{E}|^2}{V}\). With the rate of energy dissipation expressed in terms of the electric field alone, the unit remains the same: W/m^3.)

(To express the result in terms of the current density alone, we need to find how \(\vec{E}\) is related to \(\vec{J}\) in an ohmic material. We have \(\vec{E} = \rho \vec{J}\) for an ohmic material, where \(\rho\) is the resistivity of the material. Now, plugging this into the expression for rate of energy dissipation and replacing \(\sigma\) by \(\frac{1}{\rho}\), we have \(W = \frac{(\frac{1}{\rho} \rho \vec{J}) \cdot (\rho \vec{J})}{V} = \frac{|\vec{J}|^2 \rho}{V}\). With the rate of energy dissipation expressed in terms of the current density alone, the unit remains the same: W/m^3.) So, the rate of energy dissipation per unit volume \(W\) can be written in terms of: - \(\vec{E}\) and \(\vec{J}\): \(W = \frac{\vec{J} \cdot \vec{E}}{V}\) - \(\vec{E}\) alone (ohmic material): \(W = \frac{\sigma |\vec{E}|^2}{V}\) - \(\vec{J}\) alone (ohmic material): \(W = \frac{|\vec{J}|^2 \rho}{V}\)