Capacitance Using Calculus: Videos & Practice Problems

To find the capacitance of two concentric spherical shells with radii a and b (where a < b), consider the charge on each sphere as ±q. The capacitance (C) is defined as the charge (q) divided by the potential difference (V). The electric field (E) between the shells is given by Gauss's law as E = kq/r². The potential difference is calculated using the integral of the electric field, resulting in V = kq(1/b - 1/a). Thus, the capacitance is expressed as $=\frac{4\pi \epsilon (ab)}{(a-b)}$.

Capacitance of Spherical Capacitor

To determine the capacitance of two concentric spherical shells, one with radius \( a \) and the other with radius \( b \) (where \( a < b \)), we start by recalling that capacitance (\( C \)) is defined as the charge (\( Q \)) divided by the potential difference (\( V \)). In this scenario, the inner shell carries a charge of \( +Q \) and the outer shell carries a charge of \( -Q \).

To find the potential difference between the two shells, we utilize the electric field (\( E \)) in the region between them. According to Gauss's law, the electric field at a distance \( r \) from the center (where \( a < r < b \)) is given by:

\( E = \frac{kQ}{r^2} \hat{r} \

where \( k \) is Coulomb's constant. The potential difference (\( V \)) can be calculated by integrating the electric field from radius \( a \) to radius \( b \):

\( V = -\int_{a}^{b} E \cdot dr = -\int_{a}^{b} \frac{kQ}{r^2} dr \)

This integral simplifies to:

\( V = -\left[-\frac{kQ}{r}\right]_{a}^{b} = kQ\left(\frac{1}{a} - \frac{1}{b}\right) \)

To express this in a more standard form, we can find a common denominator:

\( V = kQ \left(\frac{b - a}{ab}\right) \)

Now, substituting this expression for \( V \) back into the capacitance formula, we have:

\( C = \frac{Q}{V} = \frac{Q}{kQ \left(\frac{b - a}{ab}\right)} \)

Here, the charge \( Q \) cancels out, leading to:

\( C = \frac{ab}{k(b - a)} \)

Recalling that \( k = \frac{1}{4\pi \epsilon_0} \), we can rewrite the capacitance as:

\( C = 4\pi \epsilon_0 \frac{ab}{b - a} \)

This formula represents the capacitance of the two concentric spherical shells, highlighting the relationship between the radii of the shells and the permittivity of free space.

Capacitance of Cylindrical Capacitor

To determine the capacitance per unit length of two concentric infinitely long cylindrical shells, one with radius \( a \) and the other with radius \( b \) (where \( a < b \)), we start by recognizing that these shells form a capacitor with charges \( +q \) and \( -q \) on the inner and outer shells, respectively.

The capacitance \( C \) is defined as the charge \( q \) per unit voltage \( V \). To find \( V \), we need to calculate the potential difference between the two cylinders, which can be expressed as:

\[ V = -\int_{a}^{b} \mathbf{E} \cdot d\mathbf{r} \]

Using Gauss's law, the electric field \( \mathbf{E} \) in the region between the two cylinders depends only on the inner cylinder and is given by:

\[ \mathbf{E} = \frac{2k\lambda}{r} \hat{r} \]

Here, \( \lambda \) is the charge per unit length, and \( k \) is the Coulomb's constant. The integral for the potential difference becomes:

\[ V = -\int_{a}^{b} \frac{2k\lambda}{r} dr \]

This integral simplifies to:

\[ V = -2k\lambda \left[ \ln(r) \right]_{a}^{b} = -2k\lambda (\ln(b) - \ln(a)) = -2k\lambda \ln\left(\frac{b}{a}\right) \]

We can further simplify this by bringing the negative sign inside the logarithm:

\[ V = 2k\lambda \ln\left(\frac{a}{b}\right) \]

Now, substituting this expression for \( V \) back into the capacitance formula gives:

\[ C = \frac{q}{V} = \frac{q}{2k\lambda \ln\left(\frac{a}{b}\right)} \]

Since \( \lambda = \frac{q}{L} \) (where \( L \) is the length of the cylinder), we can substitute \( \lambda \) into the capacitance equation:

\[ C = \frac{q}{2k \left(\frac{q}{L}\right) \ln\left(\frac{a}{b}\right)} = \frac{L}{2k \ln\left(\frac{a}{b}\right)} \]

To find the capacitance per unit length, we divide by \( L \):

\[ C' = \frac{1}{2k \ln\left(\frac{a}{b}\right)} \]

Substituting \( k = \frac{1}{4\pi\epsilon_0} \) into the equation yields:

\[ C' = \frac{2\pi\epsilon_0}{\ln\left(\frac{a}{b}\right)} \]

This result represents the capacitance per unit length of two infinitely long concentric cylindrical shells, highlighting the relationship between the geometry of the shells and their capacitance.

Here’s what students ask on this topic:

What is the capacitance of two concentric spherical shells with radii a and b?

The capacitance of two concentric spherical shells with radii a and b (where a < b) can be found using the formula:

$\frac{4\pi \epsilon ₀ab}{a-b}$

Here, ε₀ is the permittivity of free space. This formula is derived by calculating the potential difference between the shells and then using the definition of capacitance, which is the charge divided by the potential difference.

How do you derive the capacitance formula for concentric spherical shells using calculus?

To derive the capacitance formula for concentric spherical shells using calculus, follow these steps:

1. Calculate the electric field between the shells using Gauss's law: $E=\frac{kq}{r^2}$

2. Integrate the electric field to find the potential difference (V) between the shells: $-{\int }_{atob}\frac{kq}{r^2}dr$

3. Simplify the integral to get: $V=\frac{kq}{(b-a)}$

4. Use the definition of capacitance: $C=\frac{q}{V}$

5. Substitute V into the capacitance formula to get: $\frac{4\pi \epsilon ₀ab}{a-b}$

What is the role of Gauss's law in finding the capacitance of concentric spherical shells?

Gauss's law is crucial in finding the capacitance of concentric spherical shells because it helps determine the electric field between the shells. According to Gauss's law, the electric field (E) at a distance r from the center of a charged sphere is given by:

$E=\frac{kq}{r^2}$

where k is Coulomb's constant and q is the charge on the inner sphere. This electric field is then integrated over the distance between the two shells to find the potential difference (V). The capacitance (C) is then calculated using the formula:

$C=\frac{q}{V}$

How does the potential difference between two concentric spherical shells affect their capacitance?

The potential difference (V) between two concentric spherical shells directly affects their capacitance (C). Capacitance is defined as the charge (q) divided by the potential difference:

$C=\frac{q}{V}$

If the potential difference increases while the charge remains constant, the capacitance decreases. Conversely, if the potential difference decreases, the capacitance increases. The potential difference is determined by integrating the electric field between the shells, which depends on the radii of the shells and the charge on the inner shell. Therefore, the geometry and charge distribution of the shells play a significant role in determining the capacitance.

Why is the capacitance of concentric spherical shells expressed in terms of the permittivity of free space?

The capacitance of concentric spherical shells is expressed in terms of the permittivity of free space (ε₀) because ε₀ is a fundamental constant that characterizes the ability of the vacuum to permit electric field lines. The formula for the capacitance of concentric spherical shells is:

$\frac{4\pi \epsilon ₀ab}{a-b}$

Here, ε₀ appears because it relates the electric field to the charge density in a vacuum. Including ε₀ in the formula ensures that the capacitance is correctly scaled according to the properties of the surrounding medium, which in this case is free space.