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Suppose we have a capacitor which is charging with applied voltage $(V(t)).$ I want to know what the surface current is which is shown in figure.

enter image description here

I know the continuity equation says the current of wires is equal to surface current i.e. $$ic=c~\dot V$$ but it doesn't say anything about the direction of current. (Assumptions: You can neglect spilling of the electric field outside of the capacitor and and the slope of changing V(t) is little, you can suppose it is V=kt), Also I know the current is a function of x axis, Could anyone help me?

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  • $\begingroup$ Without any approximations, this is not a trivial problem! Do you know something about the shape of the plates? Are the plates so close to each other that one can neglect spilling of the electric field outside of the capacitor? How is the cable soldered to the capacitor? $\endgroup$ – Martin Ueding Oct 2 '16 at 19:51
  • $\begingroup$ Tnx, I've edited question. $\endgroup$ – Panda Oct 2 '16 at 19:54
  • $\begingroup$ You seem to have the answer already : the continuity equation says the current of wires is equal to surface current. The direction is the same as the direction of the current in the wires. What exactly is the difficulty in your mind? $\endgroup$ – sammy gerbil Oct 2 '16 at 20:57
  • $\begingroup$ When current comes in capacitor we should have also normal direction and I don't know how much of current goes through the tangential and how much goes through normal. However, I know sum of them are equal to total current. $\endgroup$ – Panda Oct 2 '16 at 21:12
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Perhaps I may make the following approximation? The voltage is turned on so slowly that the electrons find their way close to the surface in an ordered manner and then slowly expose themselves on the surface. This way the current density is the same over the whole surface and only in the surface-normal direction.

I think “surface current” is somewhat misleading. I would think of a current along the surface. This will happen when you have the wire connected to one side of the capacitor and let the charges flow in. Here I will assume that there the change in voltage is so slow that the tangential currents can be neglected and we only want the current that is normal to the surface.

The charge $Q$ that a capacitor can hold on one of the plates depends on the capacity $C$ and voltage $V$ like $Q = CV$. The new charges that get to the surface when the voltage is increased by $\Delta V$ is $\Delta Q = C \, \Delta V$. These new charges are evenly distributed on the whole surface.

For a current $I$ we also have $Q = I T$, after a time $T$ the charge $Q$ will be transferred by the current. A current density $i$ can be obtained by dividing through the area $A$, so $i = I/A$.

From this you should be able to compute the current density at the surface. If you are stuck or finished, you can continue reading.

We divide the above relation by some short time interval $\Delta t$, take the limit $\Delta t \to 0$ and end up with a time derivative. So the expression is $\dot Q = C \dot V$. But $\dot Q = I$ already. We divide by the surface area $A$ of the capacitor and obtain $i = C \dot V / A$.

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  • $\begingroup$ Tnx but you didn't consider the direction in here. I mean if this is the current in tangential direction what is current in normal direction? $\endgroup$ – Panda Oct 2 '16 at 20:12
  • $\begingroup$ Sorry, I compute the current that is normal to the surface; that is made up from charges which come from inside the conductor and expose themselves on the surface. $\endgroup$ – Martin Ueding Oct 2 '16 at 20:15

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