A capacitor is charged to a potential of 12V and then disconnected from the batt
Practice Questions
Q1
A capacitor is charged to a potential of 12V and then disconnected from the battery. If the plate area is doubled, what will be the new potential difference? (2022)
6V
12V
24V
It cannot be determined
Questions & Step-by-Step Solutions
A capacitor is charged to a potential of 12V and then disconnected from the battery. If the plate area is doubled, what will be the new potential difference? (2022)
Step 1: Understand that a capacitor stores electrical charge and has a certain voltage (potential difference) when connected to a battery.
Step 2: When the capacitor is charged to 12V and then disconnected from the battery, the charge on the capacitor plates remains constant.
Step 3: Know that the capacitance (C) of a capacitor is related to its plate area (A) and the distance between the plates (d) by the formula: C = ε * (A/d), where ε is the permittivity of the material between the plates.
Step 4: If the plate area (A) is doubled, the capacitance (C) increases because capacitance is directly proportional to the area.
Step 5: However, since the capacitor is disconnected, the total charge (Q) on the plates remains the same. The relationship between charge, capacitance, and voltage is given by the formula: Q = C * V.
Step 6: Since the charge (Q) is constant and capacitance (C) has increased, the voltage (V) must decrease to keep the equation balanced.
Step 7: Therefore, even though the capacitance increases, the potential difference (voltage) across the capacitor does not change from the original 12V because the charge is fixed.
Capacitance and Charge – The relationship between capacitance, charge, and potential difference in a capacitor.
Effect of Plate Area on Capacitance – Doubling the plate area of a capacitor increases its capacitance but does not affect the charge stored when disconnected from a battery.
