The integral form: $\frac{d q}{d t}+\oint_{S} \vec{j} \cdot d \vec{S}=0$
The differential form: ${\frac {\partial \rho }{\partial t}}+\nabla \cdot \vec{j} =0$
How to intuitively understand the change from total derivative to partial?
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1$\begingroup$ Welcome to PSE! I have removed your second question because there should only be 1 question per post. You are more than welcome to make a new post asking that question though :) $\endgroup$– BioPhysicistCommented Jan 21, 2022 at 4:22
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1 Answer
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In answer to first question:
The quantity $q$ is the total charge of the system, so it’s not going to vary across space (because it’s the sum charge), so it can only vary across time.
However, the quantity $\rho$ is a charge density. The distribution of charge can vary across space (some areas have more charge/volume than others). It can also vary across time. Because it has multiple parameters, we have to take a partial derivative.
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$\begingroup$ Thank you for the answer, good one! $\endgroup$ Commented Jan 21, 2022 at 3:42
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1$\begingroup$ I think maybe I should make a separate post on the second question. I will wait for days for good answers here, if not make it I will make a new post. $\endgroup$ Commented Jan 21, 2022 at 3:43
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$\begingroup$ I think that’d be a good idea. $\endgroup$ Commented Jan 21, 2022 at 3:54