For conductors, the Continuity equation will beSelect one:a. J = σ/Eb. J = ωσEc. J = E/σd. J = σE

To solve this question, we need to determine the correct form of the Continuity equation for conductors.

The Continuity equation relates the current density (J) to the conductivity (σ) and the electric field (E) in a conductor.

Looking at the options provided, we can see that option a. J = σ/Eb, option b. J = ωσEc, option c. J = E/σ, and option d. J = σE are the possible forms of the Continuity equation.

To find the correct form, we need to consider the units and the physical meaning of each option.

Option a. J = σ/Eb does not have a clear physical meaning and does not match the units of current density.

Option b. J = ωσEc includes the term ω, which typically represents angular frequency and is not relevant to the Continuity equation.

Option c. J = E/σ does not match the units of current density, as it would result in current density being measured in volts per meter instead of amperes per square meter.

Option d. J = σE is the correct form of the Continuity equation for conductors. It matches the units of current density (amperes per square meter) and represents the relationship between current density, conductivity, and electric field in a conductor.

Therefore, the correct answer is d. J = σE.