Question

Briefly explain the process of charging a parallel plate capacitor, when it is connected across a dc source.

Hint:

The battery moves electrons from one plate to the other. Charging stops when the capacitor voltage equals the battery emf, giving Q = CV.

Answer 1

Step 1: Understanding the Concept:
A parallel plate capacitor has two conducting plates separated by an insulator. When it is joined to a dc source, charge flows on to the plates until the capacitor is fully charged.

Step 2: Start of charging:
The plate joined to the positive terminal of the battery is plate 1. The plate joined to the negative terminal is plate 2. The battery pulls free electrons off plate 1 and pushes them on to plate 2 through the connecting wires.
Plate 1 becomes positively charged (+Q) and plate 2 becomes negatively charged (-Q).

Step 3: Build-up of potential difference:
As the charge on the plates grows, the potential difference \(V_C = Q/C\) between them also grows. This opposes the battery emf, so the charging current falls with time.

Step 4: End of charging:
The current stops when the potential difference across the plates is equal to the emf \(V\) of the battery. The capacitor then holds the final charge:
\[ Q = CV \]
The charge is stored on the inner faces of the plates, and an electric field is set up in the gap.

Final Answer:
The battery moves electrons from one plate to the other until the plate voltage equals the battery voltage. The capacitor is then charged with \(Q = CV\).