6.7k views
0 votes
Two 2.0 cm by 2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires of the terminals of a 9.0 V battery.

a. What are the charge on each electrode and the potential difference between them?
b. If the wires are disconnected, and insulating handles are used to pull the plates apart to a new spacing of 2.0 mm, what are the charge on each electrode and the potential difference between them?
c. If instead, the plates from part a. remained connected to the battery while the insulating handles pull them apart to their new spacing of 2.0 mm, what are the charge on each electrode and the potential difference between them?

User Vek
by
4.1k points

1 Answer

4 votes

Final answer:

The charge on metal electrodes and the potential difference between them depends on whether they are connected to a power source and their separation. When disconnected, the charge remains constant but the potential changes with separation distance. When connected, the potential remains constant but the capacitors' charge changes as separation is altered.

Step-by-step explanation:

For a pair of parallel plates with a potential difference (PD) applied across them, the charge on each electrode is determined by the capacitance and the PD between the plates. The capacitance C is given by C = ε0A / d, where ε0 is the vacuum permittivity, A is the area of one plate, and d is the separation between the plates.

a. Initial charge and potential difference

The initial charge Q on each electrode can be calculated using the formula Q = C × V, where V is the potential difference across the plates. Since V is 9.0 V and the area A is 2.0 cm by 2.0 cm, and the plates are separated by 1.0 mm, we get a certain capacitance C and thus can find Q. The potential difference remains 9.0 V.

b. Charge and potential difference after disconnection

After disconnection, the charge on the plates remains the same because there's no path for charge to leave the plates, so Q remains unchanged. However, the potential difference V increases as the plates are separated further to 2.0 mm because V = Q / C and capacitance decreases.

c. Charge and potential difference while connected to the battery

When the plates remain connected to the battery and are pulled apart, the battery maintains a constant potential difference of 9.0 V across the plates. However, the capacitance C decreases as they're pulled apart, which causes the charge Q on each plate to change to maintain the constant PD of 9.0 V (since Q = C × V).

User Gaurav Saluja
by
4.1k points