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A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled?

Short Answer

Expert verified
The stored energy increases by 100% or doubles when the plate separation is doubled.

Step by step solution

01

Understand the Concepts

Recall the formula for the energy stored in a parallel-plate capacitor, which is given by \( U = \frac{1}{2}CV^2 \). Here, \( C \) is the capacitance and \( V \) is the potential difference across the plates. The capacitance of a parallel-plate capacitor is given by \( C = \frac{\varepsilon_0A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates and \( d \) is the separation between the plates.
02

Relate Capacitance to the Plate Separation

Since the capacitor is disconnected from the battery, the charge \( Q \) on the capacitor remains constant. The potential difference \( V \) will change if the plate separation changes, as \( V = \frac{Q}{C} \). We see that the capacitance \( C \) is inversely proportional to the plate separation \( d \), so when the separation is doubled, the capacitance is halved (\( C' = \frac{C}{2} \)).
03

Calculate the New Stored Energy

The new stored energy, when the plate separation is doubled, can be found using the same energy formula \( U' = \frac{1}{2}C'V'^2 \). Since \( Q \) is constant and \( C' = \frac{C}{2} \), then \( V' = 2V \), resulting in \( U' = \frac{1}{2}\frac{C}{2}(2V)^2 = 2U \). Thus, the stored energy is doubled.
04

Determine the Fractional Change in Stored Energy

The fractional change in stored energy is the new energy divided by the original energy, minus one: \( \frac{U'}{U} - 1 = \frac{2U}{U} - 1 = 1 \). The stored energy increases by a factor of 1, which means it doubles or increases by 100%.

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Key Concepts

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

Capacitance Calculation
Capacitance is a measure of a capacitor's ability to store electric charge. The capacitance of a parallel-plate capacitor is particularly easy to calculate based on its physical dimensions and the properties of the material between the plates—the dielectric. For such a capacitor, the formula is
\( C = \frac{\varepsilon_0 A}{d} \),
where
  • \( \varepsilon_0 \) represents the permittivity of free space,
  • \( A \) is the area of one plate, and
  • \( d \) is the separation distance between the two plates.
This equation shows that the capacitance is directly proportional to the plate area and inversely proportional to the plate separation. Understanding this relationship is crucial, as changes to the plate area or separation will impact the capacitance, which in turn affects other properties of the capacitor, such as the stored energy and the electric potential difference.
Stored Energy in Capacitors
A capacitor stores electrical energy when electric charges of equal magnitude, but opposite polarity, are separated across its plates. The energy stored in a capacitor can be expressed by the formula:
\( U = \frac{1}{2}CV^2 \),
where \( U \) is the stored energy, \( C \) is the capacitance, and \( V \) is the electric potential difference across the capacitor's plates. When the plate separation of a disconnected parallel-plate capacitor is doubled, the charge \( Q \) remains constant because it's disconnected from the battery that provided the charge. However, because capacitance depends on the plate separation, the act of doubling the separation will halve the capacitance (\( C' = \frac{C}{2} \)).

As a result, the potential difference across the plates must increase to maintain the constant charge (since \( Q = CV \)), specifically it will double (\( V' = 2V \)). When these new values are placed into the stored energy formula, it can be seen that the stored energy, surprisingly, doubles. This is a non-intuitive outcome that highlights the importance of considering all aspects of the capacitor's relationships—capacitance, charge, potential difference, and stored energy—to understand how they influence each other.
Electric Potential Difference
The electric potential difference, or voltage, between two points is a measure of the work done to move a unit charge from one point to the other. In the context of capacitors, the electric potential difference across the capacitor's plates is crucial because it determines how much charge is stored for a given capacitance.
The relationship between charge (\( Q \)), capacitance (\( C \)), and voltage (\( V \)) is given by the simple equation \( Q = CV \). Therefore, if the amount of stored charge on a capacitor remains constant while the capacitance is reduced (for example by increasing the distance between plates), the voltage across the capacitor must increase accordingly.

This effect is precisely what happens when the plate separation of a disconnected parallel-plate capacitor is doubled: the capacitance is halved, but since the charge cannot go anywhere (the capacitor is disconnected), the electric potential difference must double to maintain the equation balance. This is why the stored energy, contrary to what might be expected, increases when the plate separation increases in a charged, disconnected capacitor.

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Most popular questions from this chapter

A parallel-plate capacitor has a charge \(Q\) and plates of area \(A .\) What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is \(E=Q / A \epsilon_{0},\) you might think that the force is \(F=Q E=Q^{2} / A \epsilon_{0} .\) This is wrong, because the field \(E\) includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually \(F=Q^{2} / 2 \epsilon_{0} A\) . (Suggestion: Let \(C=\epsilon_{0} A / x\) for an arbitrary plate separation \(x ;\) then require that the work done in separating the two charged plates be \(W=\int F d x\) . The force exerted by one charged plate on another is sometimes used in a machine shop to hold a workpiece stationary.

Two capacitors, \(C_{1}=25.0 \mu \mathrm{F}\) and \(C_{2}=5.00 \mu \mathrm{F}\) , are connected in parallel and charged with a \(100-\mathrm{V}\) power supply. (a) Draw a circuit diagram and calculate the total energy stored in the two capacitors. (b) What If? What potential difference would be required across the same two capacitors connected in series in order that the combination stores the same amount of energy as in (a)? Draw a circuit diagram of this circuit.

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 identical parallel-plate capacitors, each with capacitance \(C,\) are charged to potential difference \(\Delta V\) and connected in parallel. Then the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy.

One conductor of an overhead electric transmission line is a long aluminum wire 2.40 \(\mathrm{cm}\) in radius. Suppose that at a particular moment it carries charge per length 1.40\(\mu \mathrm{C} / \mathrm{m}\) and is at potential 345 \(\mathrm{kV}\) . Find the potential 12.0 \(\mathrm{m}\) below the wire. Ignore the other conductors of the transmission line and assume the electric field is everywhere purely radial.

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