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I understand the need for current density to be vector, but I am now curious if the definition of current is changed and made to be the flow of charge from a perpendicular surface / charge passing through a perpendicular surface then in the accepted answer, we will not have a confusion of when to use $A$ and when to use $A cos \theta$ for area, and so then current density would be a scalar?, is this right, or would something else be a problem here now?

Brendan Darrer
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Paracetamol
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  • You're asking if the $\vec{j}$ in $\nabla\cdot\vec{j}+\dot{\rho}=0$ admits a scalar counterpart. Why would it? You may be confusing the respective roles of $\vec{j},,\rho$, the latter being a scalar. – J.G. Sep 27 '22 at 20:46

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