Current in Capacitor Discharge Circuit - 5 μC per s

[sachin123]

Homework Statement

A fully charged capacitor is connected in a closed circuit with nothing else in the circuit.
So the capacitor starts discharging.At a certain instant of time the charge on the capacitor is
reducing at the rate of 5 micro coulomb per second(dq/dt =5 μC per s).
What is the current in the circuit at this instant of time?2. The attempt at a solution

My answer is that the current is twice dq/dt which is equal to 10 μC per s because ...

when the capacitor is getting discharged at some rate,that means that the charge on each plate of the capacitor reduces at that rate.But a capacitor has opposite charges on its plates.SO,the current due to +ve charges and -ve charges (in opposite directions)add up ,thus doubling the value.

However the value mentioned as answer is only 5 μC per s.
Can someone tell me why?
Thank You

 

[da_nang]

Current's defined as the number of positive charges that passes at a certain time. At that certain time, a charge equal to 5 μC passes through the capacitor. That's the positive current. Of course there's always a negative current as well, in this case it's the flow of the electrons. The positive current is the flow of the "holes" the electrons leave. So in one direction there are electrons, in the other positive "holes". They both have the same but opposite charge, so the currents are equal but are in different directions. Thus by convention, current's defined for the flow of the positive charges. So in this case, what they're asking for is the current in one of the directions, specifically the positive one.

 

[sachin123]

Oh yes.
Dumb of me.
It is the electrons that flow through the circuit and reach the positive plate and neutralize the positive charge on the plate.

Thanks !