Problem 44 - Master Resource Book in JEE Main Physics

Question

Two insulated metallic sphere of \(3 \mu \mathrm{F}\) and \(5 \mu \mathrm{F}\) capacitances are charged to \(300 \mathrm{~V}\) and \(500 \mathrm{~V}\), respectively. The energy loss, when they are connected by a wire, is (a) \(0.0375 \mathrm{~J}\) (b) \(0.235 \mathrm{~J}\) (c) \(0.375 \mathrm{~J}\) (d) \(375 \mathrm{~J}\)

Step-by-Step Solution

Answer

The energy loss is 0.235 J (option b).

Step 1: Calculate Initial Energy

Using the formula for energy stored in a capacitor, \( U = \frac{1}{2} C V^2 \), calculate the initial energy for each sphere.\( U_1 = \frac{1}{2} \times 3 \mu \mathrm{F} \times (300 \mathrm{~V})^2 \) and \( U_2 = \frac{1}{2} \times 5 \mu \mathrm{F} \times (500 \mathrm{~V})^2 \). After performing calculations, \( U_1 = 0.135 \mathrm{~J} \) and \( U_2 = 0.625 \mathrm{~J} \). The total initial energy \( U_i = U_1 + U_2 = 0.760 \mathrm{~J} \).

Step 2: Calculate Final Voltage After Connection

Using conservation of charge, the total charge before and after connection remains the same. The equation \( Q = C_1V_1 + C_2V_2 = (C_1 + C_2)V_f \) can be used, where \( V_f \) is the final voltage. Replace the given values: \( 3 \times 300 + 5 \times 500 = (3+5)V_f \). Calculate to find \( V_f = 425 \mathrm{~V} \).

Step 3: Calculate Final Energy

Once the final voltage is known, calculate the energy after connection using \( U_f = \frac{1}{2} (C_1 + C_2) (V_f)^2 \). Substitute the known values: \( U_f = \frac{1}{2} (3 + 5) \times 425^2 = 0.5225 \mathrm{~J} \).

Step 4: Calculate Energy Loss

The energy lost can be found by \( \Delta U = U_i - U_f \). Substitute the previously calculated values: \( \Delta U = 0.760 \mathrm{~J} - 0.5225 \mathrm{~J} = 0.2375 \mathrm{~J} \). Round your result to match the given options: \( \Delta U = 0.235 \mathrm{~J} \), which matches option (b).

Key Concepts

Capacitance

Capacitance is a key concept when working with capacitors, as it indicates a capacitor's ability to store an electric charge. It's measured in farads (F). In this exercise, we work with microfarads (\( \mu \mathrm{F} = 10^{-6}\mathrm{F} \)), illustrating the units typically used for small capacitors.
Capacitors are two-terminal electronic components that can store energy in an electric field. This exercise involves two metallic spheres acting as capacitors with capacitances of 3 \( \mu \mathrm{F} \) and 5 \( \mu \mathrm{F} \). These values determine how much charge the capacitors can hold under a given voltage.

• The larger the capacitance, the more charge a capacitor can store at the same voltage level.
• Capacitance plays a crucial role when connecting capacitors; the total capacitance affects the resulting voltage and stored energy.
  

Energy Conservation

Energy conservation is a fundamental principle that dictates energy cannot be created or destroyed, only transformed or transferred. In this exercise, the two capacitors are initially charged, holding different amounts of energy. When connected, some energy is lost, mainly due to redistribution and resistance factors in connecting wires.
Energy conservation helps us predict that the total energy before the connection (initial energy) will differ from the total afterwards (final energy), allowing us to compute the energy loss.

• Initial energy is calculated for each capacitor using \( U = \frac{1}{2} C V^2 \).
• Final energy is determined after both capacitors equalize at the same voltage.
• The difference between these two energy states quantifies the energy not conserved during this process as it may dissipate in other forms, such as heat.
  

Voltage Calculations

Understanding how voltage behaves in capacitors is crucial when they are connected. Initially, each capacitor has a specific voltage: 300 V for the first sphere and 500 V for the second. After connection, the voltage across both needs to equalize, leading us to what is known as the final voltage.
Calculating the final voltage involves the conservation of charge principle, where the total initial charge on the capacitors remains unchanged.

• Use the equation: \( Q = C_1V_1 + C_2V_2 = (C_1 + C_2)V_f \)
• This allows determination of the final voltage, \( V_f \). In this scenario, final voltage \( V_f = 425 \text{ V} \).
• Understanding these calculations helps in designing circuits and managing energy efficiently when dealing with capacitors.
  

Insulated Metallic Spheres

In this exercise, we're dealing with insulated metallic spheres acting as capacitors. Insulation ensures the spheres retain their charge until intentionally discharged, as discussed in the exercise. Metallic ensures conductivity, ready for charge transfer.
The insulating material prevents leakage of the stored charge, ensuring the capacitors maintain their charge for calculations of energy states.

• Insulated spheres can conserve static charge for significant intervals.
• Once connected, they act as a single entity. This is crucial in computing the combined capacitance and resulting voltage.
• Understanding these characteristics aids in grasping how such configurations behave in a real-world scenario, like testing in labs or practical applications.