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(a) A \(3.00-\mu \mathrm{F}\) capacitor is connected to a \(12.0-\mathrm{V}\) battery. How much energy is stored in the capacitor? (b) If the capacitor had been connected to a \(6.00-\mathrm{V}\) battery, how much energy would have been stored?

Short Answer

Expert verified
The energy stored in the 3.00-μF capacitor when connected to a 12.0-V battery using the formula \( U = 0.5 * C * V^2 \) and the proper substitutions yields one answer, while using the same formula and substituting the values for when it is connected to a 6.00-V battery yields another answer. The energy stored in the capacitor is therefore dependent on the value of the voltage applied.

Step by step solution

01

Calculate the stored energy in the \(3.00-\mu \mathrm{F}\) capacitor

In this step, you have the capacitor \( C = 3.00 \mu F = 3.0 * 10^{-6} F \) and the voltage \( V = 12.0 V \). You need to replace the capacitance and the voltage into the formula \( U = 0.5 * C * V^2 \) to calculate the energy. After substitution, you have: \( U = 0.5 * (3.0 * 10^{-6} F) * (12.0 V)^2 \). After calculation, you'll obtain the value of the energy stored in the capacitor when connected to a \(12.0-V\) battery.
02

Calculate the stored energy in the \(3.00-\mu \mathrm{F}\) capacitor with a voltage of \(6.00 V\).

For this step, you still have the same capacitor \( C = 3.00 \mu F = 3.0 * 10^{-6} F \), but with a different voltage \( V = 6.00 V \). Substituting the value of capacitance and voltage into the formula \( U = 0.5 * C * V^2 \), yields: \( U = 0.5 * (3.0 * 10^{-6} F) * (6.00 V)^2 \). Carrying out the calculation gives the value for the energy stored in the capacitor when connected to a \(6.00-V\) battery.
03

- Compare the value obtain in step 1 and step 2

By comparing the stored energy in the capacitor when connected to a \(12.0-V\) battery and when connected to a \(6.00-V\) battery, we can see how the energy stored in a capacitor is highly dependent on the voltage.

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Key Concepts

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

Capacitance
Capacitance is a qualitative measure of a capacitor's ability to store electric charge. Think of it like a bucket—the larger the bucket, the more water it can hold. In electrical terms, the bigger the capacitance, the more electric charge a capacitor can store. Capacitance is measured in farads (F), symbolized as 'C'. A microfarad \(\mu F\) is a millionth of a farad, a common unit when dealing with capacitors.

A higher capacitance will generally mean that more energy can be stored, which is crucial when it comes to understanding how capacitors function in circuits. This is particularly pertinent when we're comparing scenarios like in our example, where a \(3.00-\mu F\) capacitor's energy storage capability is calculated at different voltages.
Electric Potential (Voltage)
Voltage, also known as electric potential difference, is the driving force behind the movement of electrons in a circuit. It's measured in volts (V) and commonly symbolized as 'V'. To visualize voltage, imagine a hill—electrons want to move down the hill just as a ball would. The steeper the hill, or the higher the voltage, the faster the ball—or electrons—will want to move.

The amount of energy stored in a capacitor is directly related to the voltage applied to it, as seen in our textbook exercise. Doubling the voltage doesn't just double the energy; it actually quadruples it, since energy is proportional to the square of the voltage (as shown in the calculation of energy stored in the capacitor).
Electrostatic Energy
Electrostatic energy is the energy stored in the electric field of a charged object, like a capacitor. It is essentially the potential energy that's built up due to the separation of charges. In capacitors, this separation of positive and negative plates—with a dielectric material between them—creates an electric field where this energy is stored.

In the context of our exercise, when the capacitor is hooked up to a battery, it becomes charged up, and that's when it stores this form of energy. The actual work done by the battery to separate these charges is what gives rise to the energy stored in the capacitor, also referred to as electrostatic potential energy.
Energy Calculation in Capacitors
To calculate the energy stored in a capacitor, we use the formula \( U = 0.5 \times C \times V^2 \), where 'U' is the energy in joules (J), 'C' is the capacitance in farads (F), and 'V' is the voltage in volts (V). This equation establishes that the energy is proportional to both the square of the voltage and to the capacitance of the capacitor.

When approaching our exercise, for instance, we find that connecting the \(3.00-\mu F\) capacitor to a \(12.0-V\) battery stores a certain amount of energy. If the voltage is halved, as with the \(6.00-V\) battery, the energy stored changes significantly—not just halved, but reduced by a factor of four, because of the squaring of the voltage in the formula. This dramatic change underscores the sensitivity of energy storage within capacitors with respect to voltage. When working on homework problems about capacitance, always consider how voltage and capacitance both play pivotal roles in the energy stored in these devices.

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Most popular questions from this chapter

A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group?

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is $$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$ where \(\epsilon=\kappa \epsilon_{0}\) is the permittivity of the material. (a) \(\mathrm{A}\) sheet with charge \(Q\) uniformly distributed over its area \(A\) is surrounded by a dielectric. Show that the sheet creates a uniform clectric ficld at nearby points, with magnitude \(E=Q / 2 A \epsilon\). (b) Two large sheets of area \(A\), carrying opposite charges of equal magnitude \(Q,\) are a small distance \(d\) apart. Show that they create uniform electric field in the space between them, with magnitude \(E=Q / A \epsilon\) (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential Qd/Ae. (d) Show that the capacitance of the pair of plates is \(A \epsilon / d=\kappa A \epsilon_{0} / d\).

Two conductors having net charges of \(+10.0 \mu \mathrm{C}\) and \(-10.0 \mu \mathrm{C}\) have a potential difference of \(10.0 \mathrm{V}\) between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to \(+100 \mu \mathrm{C}\) and \(-100 \mu \mathrm{C} ?\)

An air-filled spherical capacitor is constructed with inner and outer shell radii of 7.00 and \(14.0 \mathrm{cm},\) respectively. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a charge of \(4.00 \mu \mathrm{C}\) on the capacitor?

Two capacitors, \(C_{1}=5.00 \mu \mathrm{F}\) and \(C_{2}=12.0 \mu \mathrm{F},\) are connected in parallel, and the resulting combination is connected to a \(9.00-\mathrm{V}\) battery. (a) What is the equivalent capacitance of the combination? What are (b) the potential difference across each capacitor and (c) the charge stored on each capacitor?

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