Find the electric field of a uniformly charged solid sphere both inside and outside
the sphere if the total charge is Q0 and the radius of the sphere is d.

To find the electric field of a uniformly charged solid sphere both inside and outside the sphere, we use **Gauss's Law**, which states:

\[
\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}
\]

Where:
- \(\mathbf{E}\) is the electric field,
- \(d\mathbf{A}\) is the differential area element,
- \(Q_{\text{enc}}\) is the charge enclosed by the Gaussian surface, and
- \(\varepsilon_0\) is the permittivity of free space.

We’ll treat two cases: one for the electric field **inside** the sphere and one for the electric field **outside** the sphere.

### 1. Electric Field Inside the Sphere (\(r < d\)):

- Consider a spherical Gaussian surface of radius \(r\), where \(r < d\) (inside the charged sphere).
- The charge enclosed by this Gaussian surface, \(Q_{\text{enc}}\), is proportional to the volume enclosed, as the charge distribution is uniform. The charge density \(\rho\) is:

\[
\rho = \frac{Q_0}{\frac{4}{3} \pi d^3}
\]

The charge enclosed within radius \(r\) is:

\[
Q_{\text{enc}} = \rho \cdot \frac{4}{3} \pi r^3 = \frac{Q_0 r^3}{d^3}
\]

Using Gauss’s law:

\[
E \cdot 4 \pi r^2 = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{Q_0 r^3}{\varepsilon_0 d^3}
\]

Thus, the electric field inside the sphere is:

\[
E_{\text{inside}}(r) = \frac{Q_0 r}{4 \pi \varepsilon_0 d^3}
\]

### 2. Electric Field Outside the Sphere (\(r \geq d\)):

- For points outside the sphere, we consider a Gaussian surface of radius \(r\), where \(r \geq d\).
- The entire charge \(Q_0\) is enclosed within this surface, so \(Q_{\text{enc}} = Q_0\).

Using Gauss’s law:

\[
E \cdot 4 \pi r^2 = \frac{Q_0}{\varepsilon_0}
\]

The electric field outside the sphere is:

\[
E_{\text{outside}}(r) = \frac{Q_0}{4 \pi \varepsilon_0 r^2}
\]

### Summary of Results:
- **Inside the sphere** (\(r < d\)):

\[
E_{\text{inside}}(r) = \frac{Q_0 r}{4 \pi \varepsilon_0 d^3}
\]

- **Outside the sphere** (\(r \geq d\)):

\[
E_{\text{outside}}(r) = \frac{Q_0}{4 \pi \varepsilon_0 r^2}
\]

These are the expressions for the electric field both inside and outside a uniformly charged solid sphere.

Would you like a detailed explanation of the steps or have any questions?

Here are some related questions:
1. What happens to the electric field at the surface of the sphere?
2. How does the electric field inside the sphere compare to that of a conducting sphere?
3. What is the potential associated with this electric field inside and outside the sphere?
4. How would the electric field change if the charge distribution was not uniform?
5. How does this result compare to the electric field of a spherical shell?

**Tip**: The electric field inside a uniformly charged sphere increases linearly with distance from the center, while outside, it behaves like a point charge field.

Find the electric field of a uniformly charged solid sphere both inside and outside the sphere if the total charge is Q0 and the radius of the sphere is d.

Explore how to calculate the electric field of a uniformly charged solid sphere inside and outside the sphere using Gauss's Law. This detailed solution explains key concepts in electromagnetism, including spherical symmetry and electric field distribution for both the interior and exterior of the sphere. Perfect for college-level physics students.

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