Problem 46 - College Physics

Question

A parallel-plate capacitor \(C\) is charged up to a potential \(V_{0}\) with a charge of magnitude \(Q_{0}\) on each plate. It is then disconnected from the battery, and the plates are pulled apart to twice their original separation. (a) What is the new capacitance in terms of \(C ?\) (b) How much charge is now on the plates in terms of \(Q_{0} ?(\mathrm{c})\) What is the potential difference across the plates in terms of \(V_{0} ?\)

Step-by-Step Solution

Answer

(a) \(\frac{C}{2}\); (b) \(Q_0\); (c) \(2V_0\).

Step 1: Understand the Initial Setup of the Capacitor

Initially, we have a parallel-plate capacitor with capacitance \(C\), charged to a potential \(V_0\), and with a charge \(Q_0\). The capacitance \(C\) is given by \(C = \frac{\varepsilon_0 A}{d}\), where \(A\) is the plate area, \(d\) is the separation between the plates, and \(\varepsilon_0\) is the permittivity of free space. The charge \(Q_0\) and the potential \(V_0\) relate to each other by \(Q_0 = C \cdot V_0\).

Step 2: Calculate the New Capacitance

The plates are pulled apart to twice their original separation, so the new separation \(d' = 2d\). The capacitance of a parallel-plate capacitor is inversely proportional to the separation distance: \(C' = \frac{\varepsilon_0 A}{d'} = \frac{\varepsilon_0 A}{2d} = \frac{C}{2}\). Thus, the new capacitance \(C'\) is \(\frac{C}{2}\).

Step 3: Determine the Charge After Disconnection

Since the battery is disconnected before the plates are moved, charge cannot enter or leave the plates. Thus, the charge remains the same: \(Q' = Q_0\).

Step 4: Calculate the New Potential Difference

Using the relation \(V' = \frac{Q'}{C'}\) where \(Q' = Q_0\) and \(C' = \frac{C}{2}\), we find the new potential \(V' = \frac{Q_0}{\frac{C}{2}} = \frac{2Q_0}{C} = 2V_0\). The new potential difference across the plates is \(2V_0\).

Key Concepts

Parallel-Plate Capacitor

A parallel-plate capacitor is a device that stores electrical energy by maintaining a separation of charges on two conductive plates. These plates are arranged with an insulating material, or dielectric, sandwiched between them. The basic setup involves two plates that are parallel to each other, which helps in achieving a uniform electric field between the plates. In this setup, the distance between the plates and the surface area of the plates are key factors that affect the storing capacity of the capacitor. By changing the distance between these plates, we can influence the capacitance, as seen in the problem above when the plates are pulled apart. The uniformity of the electric field makes parallel-plate capacitors popular in various electronic applications where predictable behavior is needed.

Capacitance

Capacitance is a measure of a capacitor's ability to store charge. It is defined by the capacitance equation: \[C = \frac{\varepsilon_0 A}{d}\] where:

• \(C\) is the capacitance,
• \(\varepsilon_0\) represents the permittivity of free space,
• \(A\) is the area of one of the plates,
• \(d\) is the separation between the plates.

Capacitance is directly proportional to the plate area and inversely proportional to the distance between the plates. This means if you increase the area while keeping everything else constant, the capacitance increases. However, if the separation distance is doubled, the capacitance is halved, as seen when the plates are pulled apart to twice their original separation, causing the capacitance to change from \(C\) to \(\frac{C}{2}\). This inversely proportional relationship between capacitance and distance is a critical factor in the design and application of capacitors.

Electric Potential Difference

The electric potential difference, also known as voltage, between two points in a circuit is a measure of the energy difference per charge between those points. For a capacitor, this is the voltage across its plates. Initially, the potential difference across the plates is given by \(V_0\). When the distance between the plates is altered, the capacitance changes, affecting the potential difference. After the battery is disconnected and the plates are pulled apart, the potential difference increases. The change in potential difference can be calculated using the formula:\[V' = \frac{Q'}{C'}\]With the charge remaining constant after disconnection \((Q' = Q_0)\) and the capacitance halved \((C' = \frac{C}{2})\), the new potential difference becomes twice the original, \(2V_0\). This illustrates how potential difference adjusts to maintain the relationship with charge and capacitance.

Charge Conservation

Charge conservation is a fundamental principle that states the total electric charge in an isolated system remains constant over time. This principle applies to the parallel-plate capacitor in our scenario. When the capacitor is disconnected from the battery, it becomes an isolated system with respect to its charge. Therefore, even though the plates are physically moved, resulting in a change in capacitance and potential difference, the charge \(Q\), originally \(Q_0\), remains unchanged. This demonstrates that while capacitors are capable of changing how charge is distributed or stored depending on physical alterations to the system, they strictly adhere to the law of conservation of charge, which is crucial for maintaining balance in electrical circuits.