Question

Two conducting plates, each \(3 \times 6 \mathrm{~cm}\), and three slabs of dielectric, each \(1 \times 3 \times 6 \mathrm{~cm}\), and having dielectric constants of 1,2 , and 3 , are assembled into a capacitor with \(d=3 \mathrm{~cm}\). Determine the two values of capacitance obtained by the two possible methods of assembling the capacitor.

Short answer

Answer: The capacitance values for the two different assembly methods are approximately \(C_1 \approx 1.227 \times 10^{-11} F\) for the series arrangement and \(C_2 \approx 4.536 \times 10^{-11} F\) for the parallel arrangement.

Step-by-step solution

Identifying the parameters

The area of the conducting plates is 3 x 6 cm², the three dielectric slabs have dimensions 1 x 3 x 6 cm³, and the thickness of the capacitor is 3 cm. #Step 2: Calculating the effective dielectric constants#

Calculating the effective dielectric constants

In the first arrangement, each dielectric has a thickness of 1cm, and they are placed in series. We can find an overall effective dielectric constant for this arrangement using the formula: \(\frac{1}{k_{eff}} = \frac{d_1}{k_1d} + \frac{d_2}{k_2d} + \frac{d_3}{k_3d}\) Substituting the values we get, \(\frac{1}{k_{eff}} = \frac{1}{1\cdot3} + \frac{1}{2\cdot3} + \frac{1}{3\cdot3}\) #Step 3: Solving for first effective dielectric constant#

Solving for first effective dielectric constant

Simplify the above equation for the effective dielectric constant, we have: \(k_{eff} = \frac{1}{\frac{1}{3}+\frac{1}{6}+\frac{1}{9}}\) \(k_{eff} = \frac{6}{2+1+ \frac{2}{3}}\) \(k_{eff} = \frac{6}{\frac{11}{3}}\) then we get \(k_{eff_1} \approx 1.636\) #Step 4: Calculate the first capacitance value#

Calculate the first capacitance value

Now we can use the capacitance formula with the first effective dielectric constant: \(C_1 = \frac{εk_{eff_1}A}{d} \) where \(ε\) is the permittivity of free space, equal to \(8.85 \times 10^{-12} F/m\). The area of each plate, A, equals \(3cm\cdot6cm =18cm^2=(1.8\times10^{-2})^2m^2\), and the distance d is \(3cm = 3\times10^{-2}m\). \(C_1= \frac{8.85\times10^{-12}F/m \times 1.636 × (1.8\times10^{-2}m)^2}{3\times10^{-2}m}\) Solving, we get \(C_1 \approx 1.227 \times 10^{-11} F\). #Step 5: Determine the effective dielectric constant for the second arrangement#

Determine the second effective dielectric constant

In this case, the dielectric slabs are placed in parallel, so the effective dielectric constant is simply the sum of the individual dielectric constants: \(k_{eff_2} = k_1 + k_2 + k_3 = 1+2+3 = 6\) #Step 6: Calculate the second capacitance value#

Calculate the second capacitance value

Now we can use the capacitance formula with the second effective dielectric constant: \(C_2 = \frac{εk_{eff_2}A}{d}\) \(C_2= \frac{8.85\times10^{-12}F/m \times 6 \times (1.8\times10^{-2}m)^2}{3\times10^{-2}m}\) Solving, we get \(C_2 \approx 4.536 \times 10^{-11} F\). #Step 7: State the two capacitance values for both methods of assembly#

State the two capacitance values

The capacitor has two different capacitance values for the two different assembly methods: \(C_1 \approx 1.227 \times 10^{-11} F\) for series arrangement and \(C_2 \approx 4.536 \times 10^{-11} F\) for parallel arrangement.

Key concepts

Dielectric Constants

Dielectric constants are a measure of a material's ability to store electrical energy in an electric field. The dielectric constant, also known as the relative permittivity, is an essential factor in calculating the capacitance of a capacitor. A dielectric material, when placed between the plates of a capacitor, increases its capacitance compared to when the space between the plates is a vacuum.

Capacitors can be assembled with dielectric materials in different ways, affecting the overall capacitance. When dielectrics are placed in series, as shown in the exercise, the effective dielectric constant (\(k_{eff}\)) of the combination is determined by the reciprocal of the sum of the reciprocals of the individual dielectric constants, weighted by their thickness in the direction of the electric field. This relationship can be particularly counterintuitive, as one might expect just to sum the constants rather than considering the reciprocals. Hence, understanding how dielectric constants combine in series is crucial for precise capacitance calculations.

Parallel and Series Capacitors

Capacitors can be connected in series or parallel configurations, each having a distinctive effect on the total capacitance of the circuit. In a series connection, like the first arrangement presented in the exercise, the inverse of the total capacitance is the sum of the inverses of each capacitor's capacitance. This arrangement is often used when a higher overall voltage rating is needed, as the voltage stress is divided among the capacitors in series.

In contrast, when capacitors are connected in parallel, like in the second arrangement of the exercise, the total capacitance is the sum of all individual capacitances. This arrangement is beneficial when a higher capacitance is required, as it accumulates the storage capacity of each individual capacitor. Understanding how capacitors combine is vital for any circuit design, affecting both the energy storage capability and the voltage rating of the system.

Permittivity of Free Space

The permittivity of free space, symbolized by \(ε_0\), is a fundamental physical constant which represents the capability of the vacuum to permit electric field lines. This value is crucial in the calculation of capacitance and plays a role in Coulomb's law, affecting the force between two charges in free space. The permittivity of free space has a value of approximately \(8.85 \times 10^{-12} F/m\) (farads per meter) and is used as the reference point for the dielectric constants of materials.

In the capacitance formula \(C = \frac{εA}{d}\), \(ε\) represents the absolute permittivity of the dielectric material, which is the product of \(ε_0\) and the relative permittivity (dielectric constant). This detailed distinction clarifies why the permittivity of free space is multiplied by the dielectric constant in the example solution steps, integrating the influence of the material's dielectric properties with the inherent characteristics of the space between the capacitor plates.