Question

In: Physics

A parallel-plate vacuum capacitor has 7.64 J of energy stored in it.

A parallel-plate vacuum capacitor has 7.64 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.30 mm,
a) what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?
b) What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?

Solutions

Expert Solution

Concept and reason

The concepts that are used to solve this problem are energy stored in the capacitor. Use the formula for the capacitance of a parallel plate capacitor and the energy stored in a capacitor. The relation between the separations between the two plates is obtained by using the formula for the capacitance of a parallel plate capacitor. The final potential energy is calculated using the formula for energy stored in a capacitor.

Fundamentals

Potential energy is defined as the energy possessed by an object by its position. Capacitance is defined as the ability of a capacitor to store energy. The capacitance is defined as the ratio of charge to potential difference and is given by, \(C=\frac{Q}{V}\)

Here, \(Q\) is the charge, and \(V\) is the voltage. The capacitance of a parallel plate capacitor is, \(C=\frac{\varepsilon_{0} A}{d}\)

Here, \(\varepsilon_{0}\) is the permittivity, \(\mathrm{A}\) is the area, and \(\mathrm{d}\) is the separation between the two plates. The potential energy stored in the capacitor is, \(U=\frac{Q^{2}}{2 C}\)

Here, \(Q\) is the charge, and \(C\) is the capacitance. The potential energy stored in the capacitor is, \(U=\frac{1}{2} C V^{2}\)

Here, \(V\) is the potential difference, and \(C\) is the capacitance.

(a) The capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates. So, the ratio of initial capacitance to final capacitance is, \(\frac{C_{1}}{C_{2}}=\frac{d_{2}}{d_{1}}\)

Explanation | Hint for next step

The initial capacitance of the capacitor is,

\(C_{1}=\frac{\varepsilon_{0} A_{1}}{d_{1}}\)

Here, \(\varepsilon_{0}\) is the permittivity, \(A_{1}\) is the plate area, and \(d_{1}\) is the initial separation between the two plates. The final capacitance of the capacitor is,

\(C_{2}=\frac{\varepsilon_{0} A_{2}}{d_{2}}\)

Here, \(\varepsilon_{0}\) is the permittivity, \(A_{2}\) is the plate area, and \(d_{2}\) is the final separation between the two plates. The ratio of the capacitance of the first capacitor to the capacitance of the second capacitor is \(\frac{C_{1}}{C_{2}}=\frac{\left(\frac{\varepsilon_{0} A_{1}}{d_{1}}\right)}{\left(\frac{\varepsilon_{0} A_{2}}{d_{2}}\right)}\)

\(=\left(\frac{A_{1}}{A_{2}}\right)\left(\frac{d_{2}}{d_{1}}\right)\)

The distance between the plates is decreased. The plates of the capacitors are not changed. So, the area of the plates remains the same.

Substitute A for \(A_{1}\) and \(A\) for \(A_{2}\). \(\begin{aligned} \frac{C_{1}}{C_{2}} &=\left(\frac{A_{1}}{A_{2}}\right)\left(\frac{d_{2}}{d_{1}}\right) \\ &=\left(\frac{A}{A}\right)\left(\frac{d_{2}}{d_{1}}\right) \\ &=\frac{d_{2}}{d_{1}} \end{aligned}\)

The potential energy stored in the first capacitor is, \(U_{1}=\frac{Q_{1}^{2}}{2 C_{1}}\)

Here, \(Q_{1}\) is the charge, and \(C_{1}\) is the capacitance. The potential energy stored in the second capacitor is, \(U_{2}=\frac{Q_{2}^{2}}{2 C_{2}}\)

Here, \(Q_{2}\) is the charge, and \(C_{2}\) is the capacitance. The ratio of potential energies is. \(\frac{U_{2}}{U_{1}}=\frac{\left(\frac{Q_{2}^{2}}{2 C_{2}}\right)}{\left(\frac{Q_{1}^{2}}{2 C_{1}}\right)}\)

\(=\left(\frac{Q_{2}}{Q_{1}}\right)^{2}\left(\frac{C_{1}}{C_{2}}\right)\)

The battery is disconnected, and the distance between the plates is decreased. Hence, the charge remains the same.

Substitute \(Q\) for \(Q_{1}\) and \(Q\) for \(Q_{2}\) \(\begin{aligned} \frac{U_{2}}{U_{1}} &=\left(\frac{Q}{Q}\right)^{2}\left(\frac{C_{1}}{C_{2}}\right) \\ &=\frac{C_{1}}{C_{2}} \\ \text { Substitute } \frac{d_{2}}{d_{1}} & \frac{C_{1}}{C_{2}} \end{aligned}\)

\(\frac{U_{2}}{U_{1}}=\frac{d_{2}}{d_{1}}\)

Rearrange the equation for the final potential energy. \(U_{2}=\left(\frac{d_{2}}{d_{1}}\right) U_{1}\)

Substitute \(1.30 \mathrm{~mm}\) for \(d_{2}, 2.30 \mathrm{~mm}\) for \(d_{1}\), and \(7.64 \mathrm{~J}\) for \(U_{1}\).

\(\begin{aligned} U_{2} &=\left(\frac{1.30 \mathrm{~mm}}{2.30 \mathrm{~mm}}\right)(7.64 \mathrm{~J}) \\ &=4.318 \mathrm{~J} \end{aligned}\)

Round off to three significant figures. So, the final potential energy is \(4.32 \mathrm{~J}\).

(b) The potential energy stored in the first capacitor is,

\(U_{1}=\frac{1}{2} C_{1} V_{1}^{2}\)

Here, \(V_{1}\) is the electric potential, and \(C_{1}\) is the capacitance. The potential energy stored in the second capacitor is,

\(U_{2}=\frac{1}{2} C_{2} V_{2}^{2}\)

Here, \(V_{2}\) is the electric potential, and \(C_{2}\) is the capacitance. The ratio of potential energies is \(\frac{U_{2}}{U_{1}}=\frac{\left(\frac{1}{2} C_{2} V_{2}^{2}\right)}{\left(\frac{1}{2} C_{1} V_{1}^{2}\right)}\)

The battery is connected. Hence, the electric potential remains the same.

Substitute \(V\) for \(V_{1}\) and \(V\) for \(V_{2}\).

\(\begin{aligned} \frac{U_{2}}{U_{1}} &=\frac{\left(\frac{1}{2} C_{2} V^{2}\right)}{\left(\frac{1}{2} C_{1} V^{2}\right)} \\ &=\frac{C_{2}}{C_{1}} \\ &=\frac{1}{\left(\frac{C_{1}}{C_{2}}\right)} \\ \text { Substitute } \frac{d_{2}}{d_{1}} \text { for } \frac{C_{1}}{C_{2}} . \\ \frac{U_{2}}{U_{1}} &=\frac{1}{\left(\frac{d_{2}}{d_{1}}\right)} \\ &=\frac{d_{1}}{d_{2}} \end{aligned}\)

Rearrange the equation for the final potential energy. \(U_{2}=\left(\frac{d_{1}}{d_{2}}\right) U_{1}\)

Substitute \(1.30 \mathrm{~mm}\) for \(d_{2}, 2.30 \mathrm{~mm}\) for \(d_{1}\), and \(7.64 \mathrm{~J}\) for \(U_{1}\).

\(\begin{aligned} U_{2} &=\left(\frac{2.30 \mathrm{~mm}}{1.30 \mathrm{~mm}}\right)(7.64 \mathrm{~J}) \\ &=13.516 \mathrm{~J} \end{aligned}\)

Round off to three significant figures. So, the final potential energy is \(13.5 \mathrm{~J}\).

The capacitor is connected to the source of potential, and the separation between the plates is decreased. Since the capacitor is connected to the potential source, the electric potential does not change. Hence, the initial and final potentials of the capacitor are the same.


Part a

The potential energy of the capacitor when the battery is disconnected is \(4.32 \mathrm{~J}\).

Part b

The potential energy of the capacitor when the battery is disconnected is \(13.5 \mathrm{~J}\).

Related Solutions

A parallel-plate vacuum capacitor has 7.22J of energy stored in it. The separation between the plates...
A parallel-plate vacuum capacitor has 7.22J of energy stored in it. The separation between the plates is 2.70mm . If the separation is decreased to 1.85mm , a) what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed? b) What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?
A parallel-plate vacuum capacitor has 8.22 JJ of energy stored in it. The separation between the...
A parallel-plate vacuum capacitor has 8.22 JJ of energy stored in it. The separation between the plates is 3.50 mmmm. Part A If the separation is decreased to 1.80 mmmm, what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed? Express your answer in joules. U = ?? J Part B If the separation is decreased to 1.80 mmmm, what is the energy now stored if the...
Electric Potential (Parallel Plate Capacitor Potential Energy and Potential) A parallel plate capacitor has two terminals,...
Electric Potential (Parallel Plate Capacitor Potential Energy and Potential) A parallel plate capacitor has two terminals, one (+) and the other (-). When you move a test positive charge, q at uniform velocity from the negative terminal (Ui and Vi) to the positive terminal (Uf and Vf), work W = ΔU = qΔV is done on the charge, increasing the energy of the field by this amount. The work done by the field on the charge is – W. (V...
S-16) a) Design a parallel plate capacitor with a capacitance of 5 pF with vacuum between...
S-16) a) Design a parallel plate capacitor with a capacitance of 5 pF with vacuum between the plates. b) What will change if a dielectric is placed between the plates with dielectric constant of 2. Alternate problem 16 (your choice, do the above capacitor problem or the resistor problem below) S-16r) a) Given that the resistivity of carbon in graphite form is 5 x 10-5 Ω*m, design a 10 Ω resister. Hint: choose a cylindrical shape and specify the length...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V. (C is the capacitance and U is the stored energy.) Answer the following questions regarding the capacitor charged by a battery. For each statement below, select True or False. With the capacitor connected to the battery, decreasing d increases U. A: True B: False After being disconnected from the battery, inserting a dielectric with...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V. (C is the capacitance and U is the stored energy.) Answer the following questions regarding the capacitor charged by a battery. For each statement below, select True or False. True or False With the capacitor connected to the battery, increasing d decreases Q. True or False  After being disconnected from the battery, inserting a dielectric...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V. (C is the capacitance and U is the stored energy.) Answer the following questions regarding the capacitor charged by a battery. For each statement below, select True or False. 1. With the capacitor connected to the battery, inserting a dielectric with κ will increase C. 2. With the capacitor connected to the battery, decreasing...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V. (C is the capacitance and U is the stored energy.) Answer the following questions regarding the capacitor charged by a battery. For each statement below, select True or False. After being disconnected from the battery, inserting a dielectric with κ will increase C. With the capacitor connected to the battery, inserting a dielectric with...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V volts. Let C be the capacitance and U the stored energy. Select True or False for each statement. 1. After being disconnected from the battery, decreasing d increases C. 2. With the capacitor connected to the battery, decreasing d increases C. 3. With the capacitor connected to the battery, decreasing d decreases Q. 4....
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is...
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V. (C is the capacitance and U is the stored energy.) Answer the following questions regarding the capacitor charged by a battery. For each statement below, select True or False. and please explain answers Thanks! 1) With the capacitor connected to the battery, decreasing d increases U 2) After being disconnected from the battery, inserting...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT