Question

A large capacitor has a charge +q on one plate and - q on the other. At time t=0, the capacitor is connected in series to two ammeters and a light bulb. Immediately after the circuit is closed, the ammeterconnected to the positive plate of the capacitor reads $I^{p}$ and the ammeter connected to the $I^{N}$
Each ammeter reads positive if current flows through thecircuit in a clockwise direction (from the + to the -terminal of the meter).


Immediately after time t=0, what happens to the charge on the capacitor plates?
Check all that apply.

Individualcharges flow through the circuit from the positive to the negativeplate of the capacitor.
Individualcharges flow through the circuit from the negative to the positiveplate of the capacitor.
The positiveand negative charges attract each other, so they stay in thecapacitor.
Currentflows clockwise through the circuit.
Currentflows counterclockwise through the circuit.

At any given instant after t=0, what is the relationship between the currentflowing through the two ammeters, $I^{P}$ and $I^{N}$, and the current through the bulb, $I^{B}$


$I_{\rm P} /\ \textgreater \ I_{\rm B} \ \textgreater \ I_{\rm N} I_{\rm P} = I_{\rm B} /\ \textgreater \ I_{\rm N} I_{\rm P} /\ \textgreater \ I_{\rm B} = I_{\rm N} I_{\rm P} = I_{\rm B} = I_{\rm N}$

What is the relationship between the current and charge? As the charge q(t) on the positive plate of the capacitor decreases, what happens to the value of the current?

The current

increases.
decreases.
does not change.

Answer 1

The individual charges flow from negative to positive in the circuit
(conventional) current flows clockwise (positive to negative)
Ip = Ib = In
Q (charge) = I*t (current*time)
less charge = less current