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A parallel-plate air capacitor has a capacitance of \(920 \mathrm{pF}\). The charge on each plate is \(3.90 \mu \mathrm{C}\). (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

Short Answer

Expert verified
The potential difference between the plates is 4239.13 V. If the plate separation is doubled, keeping charge constant, the potential difference will be 8478.26 V. The work required to double the separation is 8493.36 J.

Step by step solution

01

Find the Potential Difference (Voltage)

By using the formula for voltage \( V = Q / C \), where C is the capacitance and Q is the charge on the capacitor, we find the voltage. Here, we use \( C = 920 \, pF = 920 \times 10^{-12} \, F \) and \( Q = 3.90 \, \mu C = 3.90 \times 10^{-6} \, C \). Substituting these values, we find \( V = (3.90 \times 10^{-6} \, C) / (920 \times 10^{-12} \, F) = 4239.13 \, V \).
02

Find the New Voltage After Doubling the Plate Separation

If the plate separation in a parallel plate capacitor is doubled, while the charge Q remains constant, the voltage V should also double, as the capacitance C becomes half. So, the new voltage will be \( V' = 2 \times V = 2 \times 4239.13 \, V = 8478.26 \, V \).
03

Calculate the Work Done to Double the Separation

To find the work done, we use the formula \( W = (1/2) \times CV^2 - (1/2) \times C' V'^2 \), where C and V are the initial capacitance and voltage, C' and V' are the new capacitance and voltage after doubling the separation. Note that the capacitance is inversely proportional to separation distance, thus if the separation is doubled, capacitance will be halved. Hence, C' = 460 pF. Substituting these values, \( W = [(1/2) \times (920 \times 10^{-12} \, F) \times (4239.13 \, V)^2] - [(1/2) \times (460 \times 10^{-12} \, F) \times (8478.26 \, V)^2] = 8493.36 \, J \).

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Key Concepts

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

Capacitance
Capacitance is a measure of a capacitor's ability to store charge per unit of electric potential difference between its plates. It is defined by the equation, \( C = \frac{Q}{V} \) where \( C \) is the capacitance, \( Q \) is the charge stored, and \( V \) is the potential difference across the plates. Capacitors are commonly used in electronic circuits to store energy and filter signals.

In the case of a parallel-plate capacitor, the capacitance is also affected by the area of the plates and the distance between them, given by \( C = \frac{\varepsilon_0 A}{d} \) where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the separation between the plates. This is crucial as it means that if we change the separation while keeping the charge constant, the capacitance will change. In our example, doubling the separation of the plates halves the capacitance, which in turn affects the potential difference and the energy stored in the capacitor.
Electric Potential Difference

Understanding Potential Difference

Electric potential difference, often called voltage, is the amount of electric potential energy per unit charge at one point in space compared to another point. In a capacitor, it is the energy required to move a charge from one plate to another.

Using the exercise as an example, the initial potential difference can be found by dividing the charge on the plates by the capacitance. When the plate separation is doubled, the capacitance is halved, so the potential difference is doubled under the condition that the charge remains the same, as shown in the step-by-step solution. It is essential to note that this relationship holds due to the inverse proportionality between capacitance and the separation between the plates in a parallel-plate capacitor.
Work Done in Capacitors

Calculating Work in a Capacitor

When discussing work done on capacitors, we are generally referring to the energy required to move charge within the electric field between the plates, which in turn changes the electric potential energy of the system. The formula to calculate the work done to change the capacitance (or the separation of the plates) is \( W = \frac{1}{2} C V^2 - \frac{1}{2} C' V'^2 \) where \( W \) is the work done, \( C \) and \( C' \) are the initial and final capacitance values, and \( V \) and \( V' \) are the initial and final voltages. This equation shows that the work done is equal to the change in stored energy.

In our scenario, the work required to double the separation is calculated using the change in capacitance and voltage due to the increased distance. This reflects the energy needed to increase the separation against the attractive force between the oppositely charged plates. Knowing how to calculate this work is critical in designing circuits and understanding the energy dynamics within them.

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Most popular questions from this chapter

A \(5.00 \mathrm{pF},\) parallel-plate, air-filled capacitor with circular to a \(12.0 \mathrm{~V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the \(12.0 \mathrm{~V}\) battery after the radius of each plate was doubled without changing their separation?

A fuel gauge uses a capacitor to determine the height of the fuel in a tank. The effective dielectric con- stant \(K_{\text {eff }}\) changes from a value of 1 when the tank is empty to a value of \(K\), the dielectric constant of the fuel, when the tank is full. The appropriate electronic circuitry can determine the effective dielectric constant of the combined air and fuel between the capacitor plates. Each of the two rectangular plates has a width \(w\) and a length \(L\) (Fig. \(\mathbf{P 2 4 . 6 6}\) ). The height of the fuel between the plates is \(h\). You can ignore any fringing effects. (a) Derive an expression for \(K_{\text {eff }}\) as a function of \(h\). (b) What is the effective dielectric constant for a tank \(\frac{1}{4}\) full, \(\frac{1}{2}\) full, and \(\frac{3}{4}\) full if the fuel is gasoline \((K=1.95) ?\) (c) Repeat part (b) for methanol \((K=33.0)\). (d) For which fuel is this fuel gauge more practical?

A parallel-plate capacitor has capacitance \(C_{0}=8.00 \mathrm{pF}\) when there is air between the plates. The separation between the plates is \(1.50 \mathrm{~mm}\). (a) What is the maximum magnitude of charge \(Q\) that can be placed on each plate if the electric field in the region between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{~V} / \mathrm{m} ?\) (b) A dielectric with \(K=2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{~V} / \mathrm{m} ?\)

Two air-filled parallel-plate capacitors with capacitances \(C_{1}\) and \(C_{2}\) are connected in parallel to a battery that has a voltage of \(36.0 \mathrm{~V}\) \(C_{1}=4.00 \mu \mathrm{F}\) and \(C_{2}=6.00 \mu \mathrm{F}\). (a) What is the total positive charge stored in the two capacitors? (b) While the capacitors remain connected to the battery, a dielectric with dielectric constant 5.00 is inserted between the plates of capacitor \(C_{1}\), completely filling the space between them. Then what is the total positive charge stored on the two capacitors? Does the insertion of the dielectric cause total charge stored to increase or decrease?

A \(10.0 \mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a \(12.0 \mathrm{~V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the \(12.0 \mathrm{~V}\) battery after the radius of each plate was doubled without changing their separation?

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