Discharging of a capacitor Consider a charged capacitor with its positive plate holding charge Q. Now I join the capacitor to an circuit with resistance R . So the capacitor starts to discharge. Small charge $q$ flows out of positive plate in a small time $dt$ . My textbook says that the instantaneous current that flows is equal to $d$(Q-$q$)/$dt$ $=$ $-dq/dt$ . But I feel that as $q$ amount of charge has flown out the instantaneous current should be only $dq/dt$
 A: Firstly, you show some confusion between $Q$ an $q$. They are effectively the same thing, the charge stored by the capacitor. Since it is changing, it is customary to use lower case.
$dq/dt$ is the rate of change of $q$. As the capacitor discharges, $q$ falls; thus its rate of change $dq/dt$ is negative.
However the current $i$ flowing out is a positive current. In the equation $i = dq/dt$, we therefore also require the $dq/dt$ side to be positive. But it is negative, so we add an extra minus sign to yield a minus times a minus, written as $i = -dq/dt$.
This may still seem difficult or confusing to justify, so let's look at it a different way.
Consider the opposite situation, with a charging current $i$ where $q$ is increasing and $dq/dt$ is positive. Now we can happily write $i = dq/dt$ with no complications.
But that charging current is flowing the opposite way to the discharge current, so for the discharge situation we would write the current flowing in as $-i$. And, since $di/dt$ is then negative too, we can happily write $-i = dq/dt$.
But we do not want to express it as a negatively flowing charge current, we want a positive flowing discharge current going the other way. So we move the minus sign across and write $i = -dq/dt$.
