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 of 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 are then a) $\frac{1}{3}Q$ and $\frac{1}{3}C$. c) $3Q$ and $3C$. b) $\frac{1}{3}Q$ and $3C$. d) $3Q$ and $\frac{1}{3}C$.

Short answer

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

Step-by-step solution

Calculate the initial capacitance

Using the given values of $A$ and $d$ and the formula for capacitance, we can compute the initial capacitance as $C=\frac{{ϵ}_{0}A}{d}.$

Determine the potential difference

The potential difference $V$ is given in the problem. We can use this and the charge magnitude $Q$ to compute the initial capacitance. As $Q=CV$, we have $C=\frac{Q}{V}$.

Calculate the new distance between the plates

The distance between the plates is decreased by a factor of 3, so the new distance between the plates is $\frac{d}{3}$.

Calculate the new capacitance

Using the new distance and the initial capacitance formula, we can compute the new capacitance as $C^{\prime }=\frac{{ϵ}_{0}A}{\frac{d}{3}}=\frac{3{ϵ}_{0}A}{d}=3C$.

Calculate the new charge magnitude

Since the capacitor is connected to the battery while the distance between the plates is being reduced, the potential difference $V$ remains constant. Using the new capacitance value, we can compute the new charge magnitude as $Q^{\prime }=C^{\prime }V=3CV=3Q$. Now, we can see that the answer is The new magnitude of the charge on the plates: $3Q$ The new capacitance: $3C$ So, the correct answer is (c) $3Q$ and $3C$.

Key concepts

Capacitance Calculation

Capacitance is a crucial concept in understanding how capacitors work. It represents the ability of a capacitor to store electric charge per unit potential difference across its plates. The capacitance of a parallel plate capacitor is given by the equation $C=\frac{{ϵ}_{0}A}{d}$, where ${ϵ}_0$ is the permittivity of free space, $A$ is the area of one plate, and $d$ is the distance between the plates.

When the distance between the plates is decreased while the capacitor is connected to a battery, the capacitance increases because the denominator of the capacitance formula gets smaller. In the example provided, reducing the distance by a factor of three triples the capacitance. This calculation is a vital practice for students learning about capacitors, as it shows how physical changes to the capacitor affect its ability to store charge.

Electric Charge

Electric charge is an essential concept in electromagnetism, defined as the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of charges, positive and negative. In the context of capacitors, the electric charge on the plates is the amount of excess electrons that have been moved from one plate to the other.

In our exercise, the magnitude of the charge on a capacitor's plates $Q$ is directly proportional to both the capacitance $C$ and the potential difference $V$ across the plates, which is shown by the equation $Q=CV$. Importantly, when a parallel plate capacitor's plate separation is altered while connected to a constant voltage source, such as a battery, the charge stored on the plates changes in direct proportion to the change in capacitance, since $V$ remains constant.

Potential Difference

The potential difference, often called voltage, is the work done per unit charge to move a charge between two points in an electric field. In a circuit, a battery provides a constant potential difference, which is what drives electrons to flow. For a capacitor, the potential difference is related to how much energy is stored in the electric field between the plates.

The potential difference across the plates of a capacitor does not change when the plates are moved closer together if the capacitor remains connected to the battery. In the given problem, since the battery remains connected and maintains a constant potential difference, the reduction in distance between the plates does not affect the potential difference. However, it significantly affects the stored charge and capacitance of the system, which increases in direct proportion to one another when the distance decreases.