Question

A \(5 \mu \mathrm{F}\) capacitor is charged by a \(220 \mathrm{v}\) supply. It is then disconnected from the supply and is connected to another uncharged \(2.5 \mu \mathrm{F}\) capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ? (A) \(0.02 \mathrm{~J}\) (B) \(0.121 \mathrm{~J}\) (C) \(0.04 \mathrm{~J}\) (D) \(0.081 \mathrm{~J}\)

Short answer

Final Voltage = 146.67 V #tag_title#Step 3: Calculate the final energy stored in the connected capacitors#tag_content#Now that we have the final voltage across the connected capacitors, we can calculate the final energy stored in them using the energy stored in a capacitor formula: Final Energy = \(\frac{1}{2}\) × Total Capacitance × Final Voltage^2 Final Energy = \(\frac{1}{2}\) × 7.5 × 10^{-6} F × (146.67 V)^2 #tag_title#Step 4: Determine the energy lost#tag_content#Finally, we can find the energy lost in the process by subtracting the final energy stored in the connected capacitors from the initial energy stored in the charged capacitor: Energy Lost = Initial Energy - Final Energy Energy Lost = \(\frac{1}{2} \times 5 \times 10^{-6} \text{F} \times (220 \text{V})^2 - \frac{1}{2} \times 7.5 \times 10^{-6} \text{F} \times (146.67 \text{V})^2\) Energy Lost = 0.121 J The energy lost in the form of heat and electromagnetic radiation during the process is (B) 0.121 J.

Step-by-step solution

Calculate the initial energy stored in the charged capacitor

The initial energy stored in the 5 μF capacitor charged by a 220 V supply can be calculated using the formula for the energy stored in a capacitor: Energy = \(\frac{1}{2}\) × Capacitance × Voltage^2 Let's plug in the given values and calculate the initial energy. Energy = \(\frac{1}{2}\) × 5 × 10^{-6} F × (220 V)^2

Determine the final voltage across the connected capacitors

When the capacitors are connected, the capacitor's charge will be shared, and the voltage across both capacitors will be the same. The total charge will remain constant, but the total capacitance will change. We can use the formula for capacitors connected in parallel to find the total capacitance: Total Capacitance = Capacitance1 + Capacitance2 Total Capacitance = 5 × 10^{-6} F + 2.5 × 10^{-6} F = 7.5 × 10^{-6} F Now, we can use the constant charge formula (Q = C × V) to find the final voltage across the connected capacitors: Initial Charge = Initial Capacitance × Initial Voltage Final Charge = Total Capacitance × Final Voltage As the charge is constant: Initial Charge = Final Charge Initial Capacitance × Initial Voltage = Total Capacitance × Final Voltage Final Voltage = \(\frac{\text{Initial Capacitance} \times \text{Initial Voltage}}{\text{Total Capacitance}}\) Final Voltage = \(\frac{5 \times 10^{-6} \text{F} \times 220 \text{V}}{7.5 \times 10^{-6} \text{F}}\)

Key concepts

Capacitance

Capacitance is the ability of a system to store an electric charge. It is a critical concept in understanding how capacitors work, which are devices that hold and store electrical energy. The capacitance of a capacitor is determined by the formula:\[ C = \frac{Q}{V} \]where \( C \) is the capacitance measured in farads, \( Q \) is the charge in coulombs, and \( V \) is the voltage across the capacitor in volts.

A larger capacitance means that the capacitor can store more charge for the same voltage. Practical capacitors range in size depending on their use, often described in microfarads (\( \mu F \)), where \( 1 \mu F = 10^{-6} \) farads. The relationship between capacitance, voltage, and stored charge forms the basis for many electronic components and systems, helping us to manage and manipulate electrical energy effectively.

Energy stored in a capacitor

A capacitor stores energy in the electric field created between its plates when connected to a voltage source. The energy stored in a capacitor depends on its capacitance and the voltage applied to it. The formula to calculate this energy is:\[ \text{Energy} = \frac{1}{2} C V^2 \]where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.

In the problem given, a capacitor with a capacitance of \( 5 \mu F \) is charged with a voltage of \( 220 V \). Plugging these values into the formula helps in determining the initial energy stored in that capacitor. This concept is vital in understanding how capacitors function in circuits, providing a means to retain energy and release it when needed, much like a battery but faster and in pulses.

Voltage

Voltage, also known as electric potential difference, is the force that pushes electric charges to move in a circuit. It's a measure of how much energy is available to push electrons from one point to another and is typically measured in volts (V).

In capacitors, voltage determines the potential for energy storage. When two capacitors are connected, as in the exercise, they equalize at a new common voltage that dictates the final energy distribution. This new voltage is calculated using the principle of charge conservation, which ensures that the initial total charge in the system remains constant. Consequently, understanding the relationship between voltage and how it contributes to energy distribution helps clarify many electrical circuit operations.

Charge conservation

Charge conservation is a fundamental principle in physics stating that the total electric charge in an isolated system remains constant over time, regardless of any changes within the system. This is crucial when dealing with capacitors that are charged and then connected to other capacitors.

When a charged capacitor is connected to an uncharged one, the total charge initially present in the system gets redistributed between the two after reaching equilibrium. Mathematically, this is expressed as:\[ Q_\text{initial} = Q_\text{final} \]In practice, this means:\[ C_1V_1 = (C_1 + C_2)V_\text{final} \]where \( C_1 \) and \( C_2 \) are the capacitances and \( V_1 \) is the initial voltage. The goal is to find \( V_\text{final} \) once equilibrium is reached. This principle helps in calculating how voltage adjusts when capacitors with different capacitances interact, ensuring that no charge is lost in the process.