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Adding a capacitor A \(100 \mathrm{pF}\) capacitor is charged to 100 volts. After the charging battery is disconnected, the capacitor is connected in parallel with another capacitor. If the final voltage is 30 volts, what is the capacitance of the second capacitor? How much energy was lost, and what happened to it?

Short Answer

Expert verified
The capacitance of the second capacitor is calculated to be \(233.3pF\). The energy lost during the connection of the capacitors is \(0.4J\), which is converted into thermal energy.

Step by step solution

01

Determine the initial charge of the first capacitor

The charge \(Q\) in a capacitor can be calculated using the formula \(Q = CV\), where \(C\) is the capacitance and \(V\) is the voltage. For the first capacitor, the charge \(Q_1\) is equal to \(100pF * 100V = 10^{-2}C\).
02

Apply the charge conservation law

By using the charge conservation law, we can determine that the total initial charge is equal to the total final charge. Therefore, \(Q_1 = Q'_1 + Q'_2\), where \(Q'_1\) is the charge of the first capacitor and \(Q'_2\) is the charge of the second capacitor after they are connected in parallel. Since they are connected in parallel, the final voltage \(V'\) across them is the same and given as 30 volts, so \(Q'_1 = C_1*V'\) and \(Q'_2 = C_2*V'\), where \(C_1\) represents the capacitance of the first capacitor and \(C_2\) represents the capacitance of the second capacitor. Substituting for \(Q'_1\) and \(Q'_2\), we get \(10^{-2}C = 100pF * 30V + C_2 * 30V\). Solving this equation for \(C_2\) gives us the capacitance of the second capacitor.
03

Calculate the initial and final energies

The energy \(E\) stored in a capacitor is given by \(1/2*CV^2\). Therefore, the initial energy \(E_1\) is \(1/2*100pF*100V^2 = 0.5J\). The final energy \(E_2\) is the sum of the energy in the two capacitors when connected in parallel, \(E_2 = 1/2*100pF*30V^2 + 1/2*C_2*30V^2\).
04

Determine the energy lost

The energy lost \(E_{loss}\) is the difference between the initial and final energies, \(E_{loss} = E_1 - E_2\). This energy is converted into thermal energy due to resistance in the wires and other components.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance Calculation
Understanding how to calculate the capacitance of a system is critical for comprehending the behavior of circuits involving capacitors. Capacitance represents a capacitor's ability to store an electric charge per unit voltage increase. The formula to calculate the charge (\(Q\)) on a capacitor is \(Q = CV\), where \(C\) is the capacitance and \(V\) is the voltage.

In our exercise, we used this formula to determine the initial charge on the first capacitor. With a given capacitance of \(100 \text{pF}\) and a voltage of 100 volts, the charge amounted to \(10^{-2}C\). Knowing this, we can further explore the system of capacitors connected in parallel and find the unknown capacitance of the second capacitor.
Charge Conservation Law
The charge conservation law is a fundamental principle in physics that states the total electric charge in an isolated system remains constant. This means that the charge can neither be created nor destroyed, but it can be transferred from one part of the system to another.

When applying this to capacitors connected in parallel, we understand that the initial charge on the first capacitor must equal the total charge on both capacitors once they're connected. The parallel connection keeps the voltage the same across both capacitors, so using the charge equation \(Q = CV\), we determine the shared voltage and calculate the unknown capacitance of the second capacitor.
Energy in Capacitors
Capacitors not only store charge but also energy. The energy \(E\) stored in a capacitor is calculated using the formula \(E = \frac{1}{2}CV^2\). This equation reveals how the energy stored in a capacitor is directly proportional to the square of the voltage across it and the capacitance of the capacitor.

In the context of the exercise, we calculated the initial energy in the first capacitor when charged to 100 volts. After connecting another capacitor in parallel and equalizing the voltage to 30 volts, we then computed the final energy stored in both capacitors. These calculations are essential to understand how energy is distributed within a system of capacitors.
Electrical Energy Loss
During the process of connecting capacitors or any circuit manipulation, it's common to observe an energy loss. The energy is not lost in the sense of disappearing but is transformed due to the law of conservation of energy. In electrical circuits, this lost energy is usually converted into heat because of the resistance present in the wires and components.

Referring back to our problem, we determined the energy lost when the two capacitors were connected in parallel by finding the difference between the initial and final energies. This lost energy manifests as a subtle warmth in the system, and recognizing this transformation is pivotal for understanding real-world electrical systems and their efficiencies.

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Most popular questions from this chapter

If a point charge is located outside a hollow conducting shell, there is an electric field outside, but no electric field inside. On the other hand, if a point charge is located inside a hollow conducting shell, there is an electric field both inside and outside (although the external field would be zero in the special case where the shell happened to have charge exactly equal and opposite to the point charge). The situation is therefore not symmetric with respect to inside and

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