As we know $$c^2\text{curl}\;\bf B= \frac{\bf J}{\epsilon_0}$$ is incomplete & in many cases like capacitor give contradictory result against Law of conservation of charge.

As Feynman writes;

Until Maxwell’s work [...] the equation for the magnetic field of steady currents was known only as $$\boldsymbol ∇×\boldsymbol B=\frac{\boldsymbol j}{ϵ_0c^2}.\tag{18.1}$$ Maxwell began by considering these known laws and expressing them as differential equations, as we have done here. (Although the $\bf ∇$ notation was not yet invented, it is mainly due to Maxwell that the importance of the combinations of derivatives, which we today call the curl and the divergence, first became apparent.) He then noticed that there was something strange about Eq. (18.1). If one takes the divergence of this equation, the left-hand side will be zero, because the divergence of a curl is always zero. So this equation requires that the divergence of $\boldsymbol j$ also be zero. But if the divergence of $\boldsymbol j$ is zero, then the total flux of current out of any closed surface is also zero. The flux of current from a closed surface is the decrease of the charge inside the surface. This certainly cannot in general be zero because we know that the charges can be moved from one place to another. The equation $$\boldsymbol ∇\cdot \boldsymbol j=−\frac{∂ρ}{∂t}\tag{18.2}$$ has, in fact, been almost our definition of $\boldsymbol j.$ This equation expresses the very fundamental law that electric charge is conserved—any flow of charge must come from some supply. Maxwell appreciated this difficulty and proposed that it could be avoided by adding the term $∂\boldsymbol E/∂t$ to the right-hand side of Eq. (18.1); he then got the fourth equation in Table 18–1: $$\text{IV}.c^2\boldsymbol ∇×\boldsymbol B=\frac{\boldsymbol j}{ϵ_0}+\frac{∂\boldsymbol E}{∂t}.$$

Now what did Maxwell do to conclude that indeed $\frac{\partial \bf E}{\partial t}$ was required to solve the problem? Did he conduct experiment to conclude or just some of his obscure calculations?

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.