Question

$q=\frac{l}{\omega }sin\omega t$ Was derived by using the relationship $l=dp>dt$between the current and the charge on the capacitor. In Fig. 31.9a the positive counterclockwise current increases the charge on the capacitor. When the charge on the left plate is positive but decreasing in time, is$l=dp>dt$still correct or should it be$l=dp>dt$? Is $l=dp>dt$ still correct when the right-hand plate has positive charge that is increasing or decreasing in magnitude? Explain.

Short answer

When the charge on left plate is positive but decreasing in time so time rate of charges $\frac{dq}{dt}$will be negative value and current direction will be clockwise but current equation will be same $l=\frac{dq}{dt}$.

Step-by-step solution

Define convention of current equation

The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery.

Explanation

The current is positive when charge on left plate is increasing satisfying$l=dq>dt$.

When charge on left plate is positive decreasing with time so $l=-dq>dt$but since current is flowing in anticlockwise direction so $l=\frac{dq}{dt}$will be positive.

When the charge on right plate is positive which is increasing or decreasing with time, the current equation $l=\frac{dq}{dt}$will be same via same reasoning of left plate.

Thus, the current equation will always be same for any plate.