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Chapter 3: Problem 66

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

Conductor in a capacitor (a) The plates of a capacitor have area \(A\) and separation \(s\) (assumed to be small). The plates are isolated, so the charges on them remain constant; the charge densities are \(\pm \sigma .\) A neutral conducting slab with the same area \(A\) but thickness \(s / 2\) is initially held outside the capacitor, see Fig. \(3.40 .\) The slab is released. What is its kinetic energy at the moment it is completely inside the capacitor? (The slab will indeed get drawn into the capacitor, as evidenced by the fact that the kinetic energy you calculate will be positive.) (b) Same question, but now let the plates be connected to a battery that maintains a constant potential difference. The charge densities are initially \(\pm \sigma\). (Don't forget to include the work done by the battery, which you will find to be nonzero.)

Distribution of charge on a capacitor Consider a parallel-plate capacitor with different magnitudes of charge on the two plates. Let the charges be \(Q_{1}\) and \(Q_{2}\) (which we normally set equal to \(Q\) and \(-Q\) ). Find the four amounts of charge on the inner and outer surfaces of the two plates.

Tetrahedron resistance \(* *\) A tetrahedron has equal resistors \(R\) along each of its six edges. Find the equivalent resistance between any two vertices. Do this by: (a) using the symmetry of the tetrahedron to reduce it to an equivalent resistor; (b) laying the tetrahedron flat on a table, hooking up a battery with an emf \(\mathcal{E}\) to two vertices, and writing down the four loop equations. It's easy enough to solve this system of equations by hand, but it's even easier if you use a computer.

Principal radii of curvature * Consider a point on the surface of a conductor. The principal radii of curvature of the surface at that point are defined to be the largest and smallest radii of curvature there. To find the radii of curvature, consider a plane that contains the normal to the surface at the given point. Rotate this plane around the normal, and look at the curve representing the intersection of the plane and the surface. The radius of curvature is defined to be the radius of the circle that locally matches up with the curve. For example, a sphere has its principal radii everywhere equal to the radius \(R\). A cylinder has tone principal radius equal to the cross-sectional radius \(R\), and the other equal to infinity. It turns out that the spatial derivative (in the direction of the\\} normal) of the electric field just outside a conductor can be written in terms of the principal radii, \(R_{1}\) and \(R_{2}\), as follows: $$ \frac{d E}{d x}=-\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) E $$ (a) Verify this expression for a sphere, a cylinder, and a plane. (b) Prove this expression. Use Gauss's law with a wisely chosen pillbox just outside the surface. Remember that near the surface, the electric field is normal to it.

Tapered rod Two graphite rods are of equal length. One is a cylinder of radius \(a\). The other is conical, tapering (or widening) linearly from radius \(a\) at one end to radius \(b\) at the other. Show that the end-to-end electrical resistance of the conical rod is \(a / b\) times that of the cylindrical rod. Hint: Consider the rod to be made up of thin, disk-like slices, all in series. (This result is actually only an approximate one, valid in the limit where the taper is slow. See Problem \(4.6\) for a discussion of this.)
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