Problem 115 - Master Resource Book in JEE Main Physics

Question

A \(600 \mathrm{pF}\) capacitor is charged by a \(200 \mathrm{~V}\) supply. Then, it is disconnected from the supply and is connected to another uncharged \(600 \mathrm{pF}\) capacitor. How much electrostatic energy is lost in the process? [NCERT] (a) \(4 \times 10^{-6} \mathrm{~J}\) (b) \(6 \times 10^{-6} \mathrm{~J}\) (c) \(5 \times 10^{-6} \mathrm{~J}\) (d) \(8 \times 10^{-6} \mathrm{~J}\)

Step-by-Step Solution

Answer

The energy lost is \(6 \times 10^{-6} \text{ J}\); correct answer is (b).

Step 1: Calculate Initial Energy

The capacitance of the first capacitor is \( C_1 = 600 \text{ pF} = 600 \times 10^{-12} \text{ F} \), and the initial voltage is \( V = 200 \text{ V} \). The initial electrostatic energy stored in the capacitor can be calculated using the formula:\[ U_1 = \frac{1}{2} C_1 V^2 \]Substituting the values, we get:\[ U_1 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 \]\[ U_1 = 12 \times 10^{-6} \text{ J} \]

Step 2: Find New Capacitance and Charge

When the charged capacitor is connected to an uncharged capacitor, both capacitors will share the charge. The combined system has a capacity of:\[ C_{new} = C_1 + C_2 = 600 \text{ pF} + 600 \text{ pF} = 1200 \text{ pF} \]Since the capacitors are connected together, the total charge \( Q \) remains the same:\[ Q = C_1 \times V = 600 \times 10^{-12} \times 200 = 120 \times 10^{-9} \text{ C} \]

Step 3: Calculate Final Voltage After Redistribution

The final voltage \( V_f \) across both capacitors can be found using:\[ V_f = \frac{Q}{C_{new}} = \frac{120 \times 10^{-9}}{1200 \times 10^{-12}} \]\[ V_f = 100 \text{ V} \]

Step 4: Calculate Final Energy After Redistribution

The energy after redistribution of charge across both capacitors is given by:\[ U_2 = \frac{1}{2} C_{new} V_f^2 \]Substituting the known values, we find:\[ U_2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 \]\[ U_2 = 6 \times 10^{-6} \text{ J} \]

Step 5: Calculate Energy Lost

The energy lost is the difference between the initial energy and the final energy:\[ \text{Energy lost} = U_1 - U_2 = 12 \times 10^{-6} \text{ J} - 6 \times 10^{-6} \text{ J} \]\[ \text{Energy lost} = 6 \times 10^{-6} \text{ J} \]

Key Concepts

Capacitor

A capacitor is a fundamental component in the field of electrostatics. Imagine it as a tiny energy storage device, often used to store and release electrical energy as needed. Each capacitor consists of two conductive plates separated by an insulating material called a dielectric. When connected to a power source, one plate accumulates positive charges, while the other accumulates negative charges, creating an electric field. The ability of a capacitor to store charge is quantified by its capacitance, denoted as \( C \), measured in farads (F). In our exercise, we've examined a capacitor with a capacitance of \( 600 \text{ pF} \), or picofarads, which equals \( 600 \times 10^{-12} \text{ F} \). This high capacitance enables the capacitor to store a substantial amount of electrical energy.

• A capacitor's primary function is to store and release electrical energy.
• The capacitance value determines how much charge a capacitor can hold.
• An electric field is formed between the plates when it is charged.

By understanding capacitors, we gain insights into many electric circuits' operations, including power supplies, signal processing, and more.

Electrostatic Energy

Electrostatic energy in a capacitor refers to the energy stored when the capacitor is charged. This energy is a result of the electric field created between the capacitor plates. We calculate it using the formula \( U = \frac{1}{2} C V^2 \), where \( U \) is the electrostatic energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. The formula shows that the energy stored depends significantly on both the capacitance and the square of the voltage.
Initially, the charged capacitor held an energy of \( 12 \times 10^{-6} \text{ J} \). This energy reflects the ability of the capacitor to do work, like powering a circuit. However, when the charged capacitor is disconnected from the power supply and connected to another uncharged capacitor, the system's energy changes, as some energy is lost in the process.

• Electrostatic energy is stored due to the electric field between capacitor plates.
• The energy is influenced by both capacitance and voltage.
• Understanding energy calculations helps in predicting how a capacitor can power a system.

Charge Redistribution

Charge redistribution is a phenomenon that occurs when a charged capacitor is connected to an uncharged one. In this scenario, charge flows from the charged capacitor to the uncharged one until equilibrium is reached. The overall charge is conserved, but it's distributed across two capacitors. This process results in a voltage across both capacitors, which is lower than the initial voltage of the charged capacitor.
Floating charge means the total charge does not vanish; it only spreads out differently.
Initially, our charged capacitor had \( 200 \text{ V} \), but after connecting to an uncharged identical capacitor, the voltage dropped to \( 100 \text{ V} \).
Redistribution leads to a reduction in electrostatic energy due to changes in the electric field configuration. Calculating the new system's energy reveals a loss of energy, which in our case turned out to be \( 6 \times 10^{-6} \text{ J} \) during redistribution.

• Charge redistribution occurs between two capacitors until they reach equilibrium.
• Energy is lost to the environment as charges spread out.
• This process results in changes in voltage and stored energy.

Understanding how charges redistribute is critical when designing circuits involving multiple capacitors.