Question

A parallel plate capacitor of capacitance \(C\) has plates of area \(A\) with distance \(d\) between them. When the capacitor is connected to a battery supplying potential difference \(V\), it has a charge of magnitude \(Q\) on its plates. While the capacitor is connected to the battery, the distance between the plates is decreased by a factor of \(3 .\) The magnitude of the charge on the plates and the capacitance will then be a) \(\frac{1}{3} Q\) and \(\frac{1}{3} C\). c) \(3 Q\) and \(3 C\). b) \(\frac{1}{3} Q\) and \(3 C\). d) \(3 Q\) and \(\frac{1}{3} C\).

Short answer

Answer: The new magnitude of the charge on the plates is 3 times the original charge (3Q) and the new capacitance is 3 times the original capacitance (3C).

Step-by-step solution

Find the original capacitance and charge using given formulae

Given: - Capacitance: C - Plate area: A - Distance between plates: d - Potential difference: V We know that the capacitance formula for a parallel plate capacitor is, C = ε₀A/d and the formula for the charge in a capacitor is, Q = CV. Using the above formulae, we have, Original Capacitance: C Original charge: Q = CV

Determine the new distance between the plates

The problem states that the distance between the plates is reduced by a factor of 3. So, New distance, d_new = d/3

Calculate the new capacitance

Since the potential difference V is constant, we can use the capacitance formula to find the new capacitance as the distance between the plates has changed. Keep in mind that the plate area A remains the same. New capacitance: C_new = ε₀A/d_new = ε₀A/(d/3) = 3ε₀A/d = 3C So, the new capacitance is 3 times the original capacitance (3C).

Calculate the new charge

Now that we have the new capacitance and the potential difference V remains constant, we can use the charge formula to find the new charge on the plates. New charge: Q_new = C_new × V = 3C × V Since Q = CV, then Q_new = 3Q. So, the new magnitude of the charge on the plates is 3 times the original charge (3Q). The correct answer is: d) 3Q and ⅓C.

Key concepts

Capacitance

Understanding capacitance is fundamental when studying electric circuits and systems. Capacitance is the ability of a system to store an electric charge. The unit of capacitance is the farad (F), which represents the charge in coulombs that can be stored per volt of potential difference. A capacitor is a two-terminal passive electronic component that consists of two conductive plates separated by an insulating material called a dielectric. The more area the plates have (A) and the closer they are together (smaller d), the higher the capacitance.

When the plates are brought closer without changing their area, as seen in the exercise where the distance was decreased by a factor of 3, the capacitance increases proportionally. In our example, the new capacitance (\(C_{new}\)) is equal to the original capacitance (\(C\)) multiplied by 3. That's because the formula for capacitance of parallel plates is \(C = \frac{\epsilon_0 A}{d}\), where \(\epsilon_0\) is the permittivity of free space.

Electric Charge

Electric charge is a fundamental property of matter, carried by certain particles that cause them to experience a force within electromagnetic fields. There are two types of electric charges: positive and negative. Like charges repel each other, whereas opposite charges attract. The SI unit for electric charge is the coulomb (C).

In a capacitor, charge is stored on the plates due to the potential difference supplied by a battery or any power source. The magnitude of charge (\(Q\)) on a capacitor's plates is directly proportional to both the capacitance (\(C\)) and the applied potential difference (\(V\)), as given by \(Q = CV\). During the exercise, when the distance between the plates was reduced, the capacitor remained connected to the battery, which means the potential difference was held constant. Therefore, the charge on the plates increased to three times the original amount (\(Q_{new} = 3Q\)) when the capacitance tripled.

Potential Difference

Potential difference, also known as voltage, is the work done per unit charge to move a small positive test charge between two points in an electric field. It's the driving force that pushes electric charge around a circuit, and its unit is the volt (V). A higher potential difference across a component in a circuit means more energy is transferred for a given amount of charge.

In the context of a parallel plate capacitor, the potential difference between the plates creates an electric field that allows the capacitor to store energy in the form of an electric charge. When connected to a battery, the potential difference across the plates is determined by the battery's voltage and remains constant as long as the capacitor is connected. In the example exercise, the fact that the voltage stays constant is the key reason the charge on the capacitor plates increases when the capacitance increases - the product of the constant voltage and increasing capacitance leads to an increased stored charge.