If an insulated non conducting sphere of radius R has charge density (rho). The electric field at a distance r from the centre of sphere(r

When dealing with an insulated non-conducting sphere that has a uniform charge density, it’s essential to understand how the electric field behaves both inside and outside the sphere. Let’s break this down step by step.

Understanding Charge Distribution

In this scenario, we have a non-conducting sphere with radius R and a uniform charge density, denoted by ρ (rho). The charge density tells us how much charge is distributed throughout the volume of the sphere. Since the sphere is insulated, the charge remains fixed in place and does not move, which affects the electric field around it.

Electric Field Inside the Sphere

For points inside the sphere (where r < R), we can use Gauss's Law to find the electric field. Gauss's Law states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀). For a Gaussian surface (a sphere) of radius r (where r < R), the charge enclosed can be calculated as follows:

• Volume of the Gaussian sphere: \( V = \frac{4}{3} \pi r^3 \)
• Charge enclosed: \( Q_{\text{enc}} = \rho V = \rho \left(\frac{4}{3} \pi r^3\right) \)

According to Gauss's Law:

Φ = \( \frac{Q_{\text{enc}}}{\varepsilon_0} \)

Since the electric field (E) is uniform over the Gaussian surface, we can express the electric flux (Φ) as:

Φ = \( E \cdot 4 \pi r^2 \)

Equating these expressions gives us:

\( E \cdot 4 \pi r^2 = \frac{\rho \left(\frac{4}{3} \pi r^3\right)}{\varepsilon_0} \)

Solving for E, we find:

\( E = \frac{\rho r}{3\varepsilon_0} \)

Electric Field Outside the Sphere

Now, for points outside the sphere (where r ≥ R), the entire charge of the sphere can be treated as if it were concentrated at the center. The total charge (Q) can be calculated as:

• Total charge: \( Q = \rho \left(\frac{4}{3} \pi R^3\right) \)

Using Gauss's Law again for a Gaussian surface of radius r (where r ≥ R), we have:

Φ = \( E \cdot 4 \pi r^2 = \frac{Q}{\varepsilon_0} \)

Substituting for Q gives:

\( E \cdot 4 \pi r^2 = \frac{\rho \left(\frac{4}{3} \pi R^3\right)}{\varepsilon_0} \)

Solving for E in this case results in:

\( E = \frac{Q}{4\pi \varepsilon_0 r^2} = \frac{\rho \left(\frac{4}{3} \pi R^3\right)}{4\pi \varepsilon_0 r^2} = \frac{\rho R^3}{3 \varepsilon_0 r^2} \)

Summary of Electric Fields

To summarize, the electric field at a distance r from the center of the insulated non-conducting sphere is:

• For r < R: \( E = \frac{\rho r}{3\varepsilon_0} \)
• For r ≥ R: \( E = \frac{\rho R^3}{3 \varepsilon_0 r^2} \)

This behavior illustrates how the electric field varies with distance both inside and outside the sphere, reflecting the principles of electrostatics and Gauss's Law effectively. Understanding these concepts is crucial for analyzing electric fields in various configurations. If you have any more questions or need clarification, feel free to ask!