Question

Let $S=100{\mathrm{mm}}^2,d=3\mathrm{mm}$, and ${ϵ}_r=12$ for a parallel-plate capacitor. (a) Calculate the capacitance. (b) After connecting a 6-V battery across the capacitor, calculate $E,D,Q$, and the total stored electrostatic energy. (c) With the source still connected, the dielectric is carefully withdrawn from between the plates. With the dielectric gone, recalculate $E,D,Q$, and the energy stored in the capacitor. $(d)$ If the charge and energy found in part $(c)$ are less than the values found in part $(b)$ (which you should have discovered), what became of the missing charge and energy?

Short answer

Answer: The differences in values when the dielectric is removed are as follows: - The electric field (E) remains the same because the voltage across the capacitor is unchanged. - The electric displacement field (D) decreases from $2.13\times {10}^{-7}\mathrm{C}/{\mathrm{m}}^2$ to $1.77\times {10}^{-8}\mathrm{C}/{\mathrm{m}}^2$. - The charge (Q) stored in the capacitor decreases from $2.12\times {10}^{-10}\mathrm{C}$ to $1.77\times {10}^{-11}\mathrm{C}$. - The total stored electrostatic energy (W) decreases from $6.37\times {10}^{-10}\mathrm{J}$ to $5.31\times {10}^{-11}\mathrm{J}$. The missing charge and energy are accounted for by the work done on removing the dielectric from between the plates.

Step-by-step solution

(a) Calculate the capacitance

To calculate the capacitance of the parallel-plate capacitor, we can use the formula: $$C={ϵ}_0{ϵ}_r\frac{A}{d}$$ where ${ϵ}_0=8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}$ is the vacuum permittivity, ${ϵ}_r$ is the relative permittivity of the dielectric, $A$ is the area of the plates, and $d$ is the separation between the plates. Plugging in the given values, we get: $$C=(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m})(12)\frac{100\times {10}^{-6}{\mathrm{m}}^2}{3\times {10}^{-3}\mathrm{m}}=3.54\times {10}^{-11}\mathrm{F}$$

(b) Calculate E, D, Q, and the total stored electrostatic energy

First, we'll calculate the electric field E using the formula: $$E=\frac{V}{d}$$ where V is the voltage across the capacitor (6V). Using this, we get: $$E=\frac{6\mathrm{V}}{3\times {10}^{-3}\mathrm{m}}=2000\mathrm{V}/\mathrm{m}$$ Next, we'll calculate the electric displacement field D using the formula: $$D={ϵ}_0{ϵ}_{r}E$$ Plugging in the calculated E and known constants, we get: $$D=(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m})(12)(2000\mathrm{V}/\mathrm{m})=2.13\times {10}^{-7}\mathrm{C}/{\mathrm{m}}^2$$ Now, we'll calculate the charge Q stored in the capacitor using the formula: $$Q=C\times V$$ Using the calculated capacitance and known voltage, we obtain: $$Q=(3.54\times {10}^{-11}\mathrm{F})(6\mathrm{V})=2.12\times {10}^{-10}\mathrm{C}$$ Finally, we will find the total stored electrostatic energy using the formula: $$W=\frac{1}{2}C{V}^2$$ Plugging in the known values, we get: $$W=\frac{1}{2}(3.54\times {10}^{-11}\mathrm{F})(6\mathrm{V}{)}^2=6.37\times {10}^{-10}\mathrm{J}$$

(c) Calculate E, D, Q, and the energy stored in the capacitor with the dielectric removed

With the dielectric removed, the capacitance changes. We can calculate the new capacitance ($C^{\prime }$) using the formula: $$C^{\prime }={ϵ}_0\frac{A}{d}$$ Plugging in the given values, we get: $$C^{\prime }=(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m})\frac{100\times {10}^{-6}{\mathrm{m}}^2}{3\times {10}^{-3}\mathrm{m}}=2.95\times {10}^{-12}\mathrm{F}$$ Since the voltage across the capacitor, V = 6V, remains unchanged, the new electric field E' would be the same as before: $$E^{\prime }=2000\mathrm{V}/\mathrm{m}$$ For the new electric displacement field D', we can calculate it using the formula: $$D^{\prime }={ϵ}_0{E}^{\prime }$$ Plugging in the constants, we get: $$D^{\prime }=(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m})(2000\mathrm{V}/\mathrm{m})=1.77\times {10}^{-8}\mathrm{C}/{\mathrm{m}}^2$$ We can now calculate the new charge Q' stored in the capacitor using the formula: $$Q^{\prime }=C^{\prime }\times V$$ Using the new capacitance and known voltage, we obtain: $$Q^{\prime }=(2.95\times {10}^{-12}\mathrm{F})(6\mathrm{V})=1.77\times {10}^{-11}\mathrm{C}$$ Finally, we will find the new stored electrostatic energy using the formula: $$W^{\prime }=\frac{1}{2}{C}^{\prime }{V}^2$$ Plugging in the values, we get: $$W^{\prime }=\frac{1}{2}(2.95\times {10}^{-12}\mathrm{F})(6\mathrm{V}{)}^2=5.31\times {10}^{-11}\mathrm{J}$$

(d) Analyze the differences in charge and energy

Comparing the calculated values from parts (b) and (c), we notice that the charge (Q) and energy (W) stored in the capacitor decrease when the dielectric is removed: - Q without dielectric (Q') = $1.77\times {10}^{-11}\mathrm{C}$ < Q with dielectric = $2.12\times {10}^{-10}\mathrm{C}$ - W without dielectric (W') = $5.31\times {10}^{-11}\mathrm{J}$ < W with dielectric = $6.37\times {10}^{-10}\mathrm{J}$ The missing charge and energy are used to do work on removing the dielectric from between the plates. When the dielectric is removed from the capacitor, the voltage remains constant, but the capacitance decreases. As a consequence, the charge stored on the capacitor's plates and the stored electrostatic energy decrease as well. The difference in charge and energy between the cases with and without the dielectric is used to overcome the force applied by the electric field on the dielectric as it is withdrawn from between the plates. This work is ultimately dissipated as heat or other forms of energy.

Key concepts

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two large, flat plates placed close to each other. The space between the plates is filled with a dielectric material or vacuum. The primary purpose of a capacitor is to store electrical charge. The ability to store charge is measured by its capacitance, which is influenced by factors such as the area of the plates and the distance between them.

• The formula for capacitance is given by: $$C=\frac{{ϵ}_0{ϵ}_{r}A}{d}$$ where ${ϵ}_0$ represents the permittivity of free space, ${ϵ}_r$ is the relative permittivity, $A$ is the area of the plates, and $d$ is the separation distance.
• Larger plate areas and shorter distances between plates yield higher capacitance, meaning more charge can be stored.

Capacitors are widely used in electronic circuits for various applications, such as energy storage, filtering, and voltage regulation.

Dielectric Materials

Dielectric materials, when introduced between the plates of a capacitor, increase the capacitor's ability to store charge. These materials are insulators that become polarized in an electric field, reducing the field within the capacitor and allowing more charge to accumulate.

• The relative permittivity (${ϵ}_r$) of a dielectric material quantifies its effectiveness in increasing a capacitor's capacitance compared to a vacuum.
• When a dielectric is introduced, the capacitance of the capacitor increases by a factor equal to ${ϵ}_r$.

The removal of a dielectric while the capacitor is charged results in a decrease in capacitance, and consequently, the charge stored and energy also reduce. Understanding dielectric behavior is crucial for designing capacitors for specific electronic functions.

Electrostatic Energy

Electrostatic energy represents the energy stored within a capacitor. The energy is related to the charge accumulated on the plates and the voltage across them. The formula for electrostatic energy in a capacitor is: $$W=\frac{1}{2}C{V}^2$$ where $W$ is the energy, $C$ is the capacitance, and $V$ is the voltage.

• A higher capacitance or voltage results in more stored energy.
• When the dielectric is removed, the stored energy decreases because the capacitance is reduced.

Electrostatic energy can play significant roles in various applications, such as providing power in bursts, as seen in camera flashes or as part of the energy storage systems in electronic devices.

Electric Field

The electric field ($E$) within a capacitor is an essential factor that defines the potential difference created between the plates. It can be calculated using the formula: $$E=\frac{V}{d}$$ where $V$ is the voltage and $d$ is the distance between the plates.

• The electric field is directly proportional to the voltage and inversely proportional to the distance.
• Introducing a dielectric decreases the electric field within the capacitor, facilitating additional charge storage without increasing voltage.

Understanding the behavior of the electric field helps in designing capacitors that can efficiently manage voltage and charge, which is crucial in various electronic applications. By controlling the electric field through adjustments in plate separation or dielectric material use, engineers can tailor capacitors for specific needs.