Problem 94 - Master Resource Book in JEE Main Physics

Question

A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles, then which of the following is correct? (a) The charge on the capacitor increases (b) The voltage across the plates increases (c) The capacitance increases (d) The electrostatic energy stored in the capacitor increases

Step-by-Step Solution

Answer

The voltage across the plates and the electrostatic energy increase.

Step 1: Understanding the Scenario

We have a parallel plate capacitor that is initially charged with a battery, and then the battery is disconnected. Next, the plates of the capacitor are moved farther apart.

Step 2: Charge Conservation

Since the battery is disconnected, the charge on the capacitor remains constant. The correct option about charge is: the charge does not increase.

Step 3: Effect on Capacitance

The capacitance of a parallel plate capacitor is given by the formula: \( C = \frac{\varepsilon_0 A}{d} \), where \( A \) is the area of the plates, and \( d \) is the distance between them. As the plates are moved farther apart, \( d \) increases, which means the capacitance \( C \) decreases.

Step 4: Determining Voltage Change

The voltage \( V \) across a capacitor is given by: \( V = \frac{Q}{C} \), where \( Q \) is the constant charge. Since \( C \) decreases, \( V \) must increase.

Step 5: Analyzing Energy Storage

The energy \( U \) stored in a capacitor is \( U = \frac{1}{2} \frac{Q^2}{C} \). As \( C \) decreases with the increased separation, \( U \) increases due to the squared term \( \frac{Q^2}{C} \).

Key Concepts

Charge Conservation

When a parallel plate capacitor is initially charged, electrons are moved from one plate to another, creating an electric charge. If a battery is used to charge the capacitor and then disconnected, the charge remains fixed on the plates, no matter how we alter the system. This is known as charge conservation.

In simple words, the amount of positive charge on one plate and negative charge on the other plate remains the same unless an external circuit is created to alter the distribution of electrons. This concept is crucial because it tells us that the charge, denoted by \( Q \), does not change simply by moving the plates apart. Instead, it stays constant, leading us to better understand how the system's voltage and energy respond when we alter other physical conditions.

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. For parallel plate capacitors, it can be calculated using the formula \( C = \frac{\varepsilon_0 A}{d} \), where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of one of the plates, and \( d \) is the distance between the plates.

When the plates of a capacitor are moved further apart, the distance \( d \) increases, causing the capacitance \( C \) to decrease. Since capacitance directly depends on the geometric factors of the capacitor, any change in the distance between the plates will inversely affect the capacitance. A larger separation means less capacitance, indicating a reduced ability to store charge at a given voltage. Understanding how these factors affect capacitance is key to predicting the behavior of capacitors in circuits.

Voltage Change

The voltage across a capacitor is related to its charge and capacitance by the formula \( V = \frac{Q}{C} \). Since we know from charge conservation that the charge \( Q \) remains constant when we disconnect the battery, any changes in capacitance will directly affect the voltage.

With increased separation of the plates and consequently decreased capacitance, the formula tells us that the voltage \( V \) must increase. It's like squeezing a partially filled sponge: as the firm squeeze reduces the space (or equivalently, as the space is stretched which is opposite here), the pressure (or voltage) intensifies. This concept demonstrates the intrinsic balance between charge, voltage, and capacitance in the performance of the capacitor.

Energy Storage in Capacitors

The energy stored in a capacitor is given by the equation \( U = \frac{1}{2} \frac{Q^2}{C} \). This relationship illustrates how energy storage is affected by both the charge and capacitance. When the capacitance \( C \) decreases due to the increased separation of the plates, the energy \( U \) stored in the capacitor increases.

This occurs because the energy stored in a capacitor is inversely proportional to its capacitance, and directly proportional to the square of the charge. So, with a constant charge \( Q \) and a reducing capacitance \( C \), the energy must rise: much like stretching a rubber band, more energy gets stored as it is pulled further apart. Understanding the relation between energy storage, charge, and capacitance helps in designing circuits with efficient energy usage.