Question

The capacitance of a spherical capacitor consisting of two concentric conducting spheres with radii \(r_{1}\) and \(r_{2}\left(r_{2}>r_{1}\right)\) is given by \(C=4 \pi \epsilon_{0} r_{1} r_{2} /\left(r_{2}-r_{1}\right) .\) Suppose that the space between the spheres, from \(r_{1}\) up to a radius \(R\left(r_{1}

Short answer

Question: Find the expression for the capacitance of a spherical capacitor when it's partially filled with a dielectric material with permittivity \(\epsilon = 10 \epsilon_0\), considering the capacitance without the dielectric material is given by the formula \(C = 4\pi \epsilon_0 r_1 r_2 / (r_2 - r_1)\). Answer: The expression for the overall capacitance of the spherical capacitor when partially filled with a dielectric material is: \(C = 4\pi (10 \epsilon_0)r_1 R / (R - r_1) + 4\pi \epsilon_0 r_1 r_2 / (r_2 - R)\).

Step-by-step solution

Finding the capacitance of the part with the dielectric

We'll first find the capacitance for the part with the dielectric material, ranging from \(r_1\) to \(R\). The formula for the capacitance with the dielectric material is \(C_\text{dielectric} = 4\pi \epsilon r_1 R / (R - r_1)\). We know that \(\epsilon = 10 \epsilon_0\), so we can substitute this into our formula and obtain the capacitance for the part with the dielectric material. $$C_\text{dielectric} = 4\pi \cdot (10 \epsilon_0)r_1 R / (R - r_1)$$

Finding the capacitance of the part without the dielectric

Next, we'll find the capacitance for the part without the dielectric material, ranging from \(R\) to \(r_2\). The formula for this capacitance is \(C_\text{empty} = 4\pi \epsilon_0 r_1 r_2 / (r_2 - R)\).

Combining the capacitances in parallel

Now, since these two capacitances are in parallel, the overall capacitance of the system can be calculated by adding the two capacitances together: $$C = C_\text{dielectric} + C_\text{empty}$$ $$C = 4\pi \cdot (10 \epsilon_0)r_1 R / (R - r_1) + 4\pi \epsilon_0 r_1 r_2 / (r_2 - R)$$

Checking the limits when \(R = r_1\) and \(R = r_2\)

Now, let's check the capacitance when \(R = r_1\): $$C = 4\pi \cdot (10 \epsilon_0)r_1 r_1 / (r_1 - r_1) + 4\pi \epsilon_0 r_1 r_2 / (r_2 - r_1)$$ $$C = 4\pi \epsilon_0 r_1 r_2 / (r_2 - r_1)$$ This matches the given formula for the capacitance without any dielectric material. Next, let's check the capacitance when \(R = r_2\): $$C = 4\pi \cdot (10 \epsilon_0)r_1 r_2 / (r_2 - r_1) + 4\pi \epsilon_0 r_1 r_2 / (r_2 - r_2)$$ $$C = 4\pi \cdot (10 \epsilon_0)r_1 r_2 / (r_2 - r_1)$$ This matches the formula for the capacitance when the entire space is filled with the dielectric material. So, the formula for the overall capacitance is: $$C = 4\pi \cdot (10 \epsilon_0)r_1 R / (R - r_1) + 4\pi \epsilon_0 r_1 r_2 / (r_2 - R)$$

Key concepts

Dielectric Material

Dielectric materials are insulating substances that are introduced between the plates of capacitors to enhance their capacitance. When placed in an electric field, these materials become polarized, meaning their molecular structure aligns in such a way that it reduces the overall field inside the material.

The presence of a dielectric generally increases the capacitance of a capacitor because it reduces the effective electric field within the capacitor, allowing it to store more charge for a given potential difference. The dielectric constant, also known as the relative permittivity (\(epsilon_r\)), is a measure of how much the dielectric material can increase the capacitance compared to the capacitance in a vacuum.

\(\text{If the dielectric constant is } epsilon_r, \text{ and the permittivity of free space is }epsilon_0, \text{ then the effective permittivity is } epsilon = epsilon_r \times epsilon_0\). The ratio of the capacitance with the dielectric (\(C_{dielectric}\)) to the capacitance without the dielectric (\(C_{vacuum}\)) gives the dielectric constant, \(\text{i.e., } epsilon_r = frac{C_{dielectric}}{C_{vacuum}}\). In our exercise, filling the space with a dielectric increases the capacitance by a factor of the dielectric constant (in this case, 10) within the range from \(r_1\) to \(R\).

Capacitance in Parallel

When dealing with circuits, capacitors can be arranged in two principal ways: in series or in parallel. The arrangement significantly affects the total capacitance of the system. For capacitors in parallel, the effective capacitance is the sum of the individual capacitances.

This is because the potential difference across each capacitor is the same, and the total charge stored on the capacitors is the sum of the charges on each. Mathematically, for capacitors in parallel: \(C_{total} = C_{1} + C_{2} + C_{3} + ... + C_{n}\). In our exercise, the two sections of the spherical capacitor—one with and one without the dielectric—are considered to be in parallel. You effectively add their individual capacitances, \(C_{dielectric}\) and \(C_{empty}\) to find the overall capacitance \(C\) of the capacitor.

Understanding Parallel Capacitance

Imagine having two storage tanks connected such that they can both be filled through a single valve. When you open the valve, the water flows into both tanks simultaneously, and the total volume of water stored is the sum of water in both tanks. Capacitors in parallel work similarly; they sum up their ability to store charge, analogous to how the tanks sum up their storage capacity.

Electric Constant (Epsilon)

The electric constant, also known as the permittivity of free space (\(\text{signified by }epsilon_0\)), is a fundamental physical constant which represents the capability of the vacuum to permit the passage of electric field lines.

Its numerical value depends on the units used, but in the International System of Units (SI), it is approximately \(8.854 \times 10^{-12}\) farads per meter (F/m). The role of the electric constant is pivotal in calculating the force between electric charges (as per Coulomb's law) and in defining the capacitance of a capacitor without any dielectric material.

In the context of our exercise, \(\text{the electric constant is used to determine the capacitance both with and without the dielectric material and to understand how much the presence of the dielectric amplifies the capacitance due to its electric properties}\). The increase in capacitance is not merely a factor of the dielectric constant but also a result of the fundamental interaction of electric fields with the space between the charges – an interaction characterized by \(\text{the electric constant}\).