Spherically Symmetric Charge Distribution

Spherically Symmetric Charge Distribution

A spherically symmetric charge distribution refers to a distribution of electric charge that is symmetric about a central point (typically the origin) such that the charge density depends only on the radial distance $r$ from the center. This type of distribution is common in many physical systems, such as charged spherical shells, uniformly charged spheres, and point charges.

The primary advantage of dealing with spherically symmetric charge distributions is that it allows the use of Gauss’s Law to easily compute the electric field, as the symmetry simplifies the problem significantly. The electric field at any point depends only on the radial distance from the center and is directed radially outward (or inward, depending on the charge).

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1. General Formulation of Gauss’s Law for Spherically Symmetric Systems

For a charge distribution that is spherically symmetric, the electric field at any point is purely radial (i.e., it points directly away from or towards the center of the distribution). The magnitude of the electric field depends only on the radial distance $r$ from the center.

To apply Gauss’s Law in this case, consider a spherical Gaussian surface of radius $r$ centered at the charge distribution. The electric field is radial and has the same magnitude at all points on the surface due to the spherical symmetry of the system.

Gauss’s Law states that:

$$\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$$

Where:

• $\oint_S$ denotes a surface integral over a closed surface $S$,
• $\vec{E}$ is the electric field vector,
• $d\vec{A}$ is an infinitesimal vector area element on the spherical surface,
• $Q_{\text{enc}}$ is the total charge enclosed by the surface,
• $\epsilon_0$ is the permittivity of free space.

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2. Electric Field Outside a Spherically Symmetric Charge Distribution

Scenario:

Consider a spherically symmetric charge distribution with total charge $Q$ and radius $R$ (such as a uniformly charged sphere). We want to find the electric field outside the sphere (at a distance $r$ from the center, where $r > R$).

Solution:

Since the system is spherically symmetric, we choose a Gaussian surface that is a sphere of radius $r$ concentric with the charge distribution. The electric field $\vec{E}$ is radial and has the same magnitude at all points on the surface.

By Gauss’s Law:

$$\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$$

Since $\vec{E}$ is radial and constant over the spherical Gaussian surface, we can take the magnitude $E$ outside the integral:

$$E \oint_S dA = \frac{Q_{\text{enc}}}{\epsilon_0}$$

The surface area of the Gaussian sphere is $4\pi r^2$, so the equation becomes:

$$E \times 4\pi r^2 = \frac{Q}{\epsilon_0}$$

Solving for the electric field:

$$E = \frac{Q}{4\pi \epsilon_0 r^2}$$

Thus, the electric field outside a spherically symmetric charge distribution behaves as if all the charge were concentrated at the center of the sphere, and it follows the inverse-square law.

This result is exactly the same as the electric field due to a point charge. Therefore, for $r > R$, the electric field outside a spherically symmetric charge distribution is the same as if the entire charge were concentrated at the center of the distribution.

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3. Electric Field Inside a Spherically Symmetric Charge Distribution

Scenario:

Now, consider the case where the charge distribution is inside a sphere of radius $R$, and we want to find the electric field at a distance $r$ from the center where $r < R$.

Solution:

The approach is similar to the previous case, but now we have to account for the fact that only the charge enclosed within a sphere of radius $r$ contributes to the electric field at that point.

For a uniformly charged sphere, the charge density $\rho$ is:

$$\rho = \frac{Q}{\frac{4}{3} \pi R^3}$$

The total charge enclosed by a Gaussian surface of radius $r$ is:

$$Q_{\text{enc}} = \rho \times \frac{4}{3} \pi r^3 = \frac{Q}{R^3} \times r^3 = Q \times \frac{r^3}{R^3}$$

Now, using Gauss’s Law:

$$\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$$

The surface area of the Gaussian sphere is $4\pi r^2$, so:

$$E \times 4\pi r^2 = \frac{Q \times r^3}{R^3 \epsilon_0}$$

Solving for $E$:

$$E = \frac{Q r}{4\pi \epsilon_0 R^3}$$

Thus, the electric field inside a uniformly charged sphere increases linearly with the radial distance $r$ from the center. This result is similar to the electric field due to a point charge at the center of the sphere, but it only takes into account the charge inside the Gaussian surface.

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4. Example 1: Electric Field Due to a Uniformly Charged Sphere (Outside)

Consider a uniformly charged sphere with a total charge $Q = 5 \, \mu C$ and radius $R = 10 \, \text{cm}$. We want to find the electric field at a point located $r = 20 \, \text{cm}$ from the center of the sphere.

• From the previous formula for the electric field outside the sphere:

$$E = \frac{Q}{4\pi \epsilon_0 r^2}$$

Substitute the known values:

$$E = \frac{5 \times 10^{-6}}{4 \pi (8.85 \times 10^{-12}) (0.20)^2}$$

Calculating the electric field:

$$E \approx 1.13 \times 10^6 \, \text{N/C}$$

Thus, the electric field at $r = 20 \, \text{cm}$ is approximately $1.13 \times 10^6 \, \text{N/C}$.

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5. Example 2: Electric Field Inside a Uniformly Charged Sphere

Now, consider the same sphere with $Q = 5 \, \mu C$ and $R = 10 \, \text{cm}$. We want to find the electric field at a point located $r = 5 \, \text{cm}$ inside the sphere.

• From the formula for the electric field inside the sphere:

$$E = \frac{Q r}{4\pi \epsilon_0 R^3}$$

Substitute the known values:

$$E = \frac{5 \times 10^{-6} \times 0.05}{4 \pi (8.85 \times 10^{-12}) (0.10)^3}$$

Calculating the electric field:

$$E \approx 1.42 \times 10^5 \, \text{N/C}$$

Thus, the electric field at $r = 5 \, \text{cm}$ inside the uniformly charged sphere is approximately $1.42 \times 10^5 \, \text{N/C}$.

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6. Conclusion

• Gauss’s Law simplifies the calculation of the electric field for spherically symmetric charge distributions.
• Outside a spherically symmetric charge distribution (for $r > R$), the electric field behaves as if all the charge were concentrated at the center, following Coulomb's law $E = \frac{Q}{4\pi \epsilon_0 r^2}$.
• Inside a uniformly charged sphere (for $r < R$), the electric field increases linearly with distance, given by $E = \frac{Q r}{4\pi \epsilon_0 R^3}$.