⚡ المجال الكهربائي لكرة مصمتة باستخدام قانون جاوس ⚡ Electric Field of a Solid Sphere Using Gauss's Law

Calculating Electric Field Inside and Outside a Charged Solid Sphere

To calculate the electric field at a point inside or outside a charged solid sphere, we use Gauss's law which states:

Where:

• E: Electric field intensity
• dA: Differential area element of the closed surface
• Q_inside: Charge enclosed within the closed surface
• ε₀: Permittivity of free space

Calculation Steps:

1. We choose an imaginary spherical surface (Gaussian surface) passing through the point where we want to calculate the field
2. We calculate the electric flux through this surface
3. We determine the charge enclosed within the Gaussian surface
4. We apply Gauss's law to calculate the field intensity

Case of Point Inside the Sphere \( r < R \)

When the point is inside the sphere \( r < R \), the Gaussian surface will be a sphere of radius \( r \), and the charge enclosed within this surface is part of the total charge.

Since the charge is uniformly distributed, we can use the proportionality between volumes and charges:

Where \( K = \frac{1}{4\pi\epsilon_0} \) is Coulomb's constant.

Note: The field inside the sphere is directly proportional to the distance from the center \( E \propto r \).

Case of Point Outside the Sphere \( r > R \)

When the point is outside the sphere \( r > R \), the Gaussian surface will be a sphere of radius \( r \), and the charge enclosed within this surface is the total charge of the sphere.

By applying Gauss's law:

Note: The field outside the sphere is inversely proportional to the square of the distance from the center \( E \propto \frac{1}{r^2} \).

Important Notes:

• The electric field inside a charged solid sphere is directly proportional to the distance from the center
• The electric field outside a charged solid sphere is inversely proportional to the square of the distance from the center
• At the surface of the sphere \( r = R \), the field values from both internal and external relationships are equal
• The graph of field versus distance shows a maximum value at the surface of the sphere

Interactive Simulation

\[1 \star \star \star\]

The relationship between the field and distance from the center of a charged
non-conducting solid sphere was plotted, resulting in the following graph.
The magnitude of the field at a point at a distance of \( 0.15 \, \text{m} \) from the center
equals

Choose the correct answer

---

A

\[E = 2800 \, \text{N/C}\]

B

\[E = 3000 \, \text{N/C}\]

C

\[E = 2700 \, \text{N/C}\]

D

\[E = 2900 \, \text{N/C}\]

Click here to show solution method