There are two conducting concentric hollow spheres of outer radii R2 and R1 ( R2>R1 ). The thickness of the material of both spheres is d .

The inner sphere is negatively charged with charge density -σ1, . The larger sphere is positively charged with charge density +σ2.

(a) What is the electric field (magnitude and direction) inside the inner sphere?

Direction: use units vectors in spherical coordinates ( r , Ѳ^ , ø^)

Magnitude: (Express your answer in terms of the following variables, if necessary R1, R2 , σ1 , σ2 , d , r and the constant epsilon_0 .
unanswered

What is the electric field (magnitude and direction) at distance from the center of the spheres for the following values of r :

(b) r<R1 but larger than R1-d

Direction: use units vectors in spherical coordinates ( r , Ѳ^ ,ø^)

Magnitude: (Express your answer in terms of the following variables, if necessary R1 , R2 , σ1 , σ2 , d , r and the constant .

unanswered

(c) r>R1 but smaller than R2-d

Direction: use units vectors in spherical coordinates ( r ,Ѳ^ , ø^ )

Magnitude: (Express your answer in terms of the following variables, if necessary , R1 , R2 ,σ1,σ2, d,r and the constant epsilon_0 .

unanswered

(d) r<R2 but larger than R2-d

Direction: use units vectors in spherical coordinates ( r , Ѳ^, ø^ )

Magnitude: (Express your answer in terms of the following variables, if necessary R1 , R2 , σ1 , σ2 , d , r and the constant epsilon_0 .

To solve this problem, we can use Gauss's law for electric fields. Gauss's law states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space (ε₀).

For part (a), we need to find the electric field inside the inner sphere. Since the inner sphere is negatively charged and the charge is distributed uniformly, the total charge enclosed by any surface within the inner sphere will be -σ₁ multiplied by the volume enclosed by that surface.

We can choose a spherical Gaussian surface with radius r (where r < R₁) inside the inner sphere. The electric field will be radial and its magnitude will be the same at every point on this surface (due to symmetry).

Using Gauss's law, the electric flux through this Gaussian surface is given by:

Flux = E * 4πr² = (-σ₁ * (4/3)πr³) / ε₀

Simplifying, we find:

E = (-σ₁ * (4/3)πr) / (ε₀ * r²)

The negative sign indicates that the electric field points radially inward towards the center of the spherical shell.

For parts (b), (c), and (d), we need to find the electric field at positions outside the inner sphere but within or outside the outer sphere.

For these regions, the electric field is influenced by both the charge distribution in the inner sphere and the charge distribution in the outer sphere.

To find the electric field, we can consider the equivalent charge distribution that arises from the inner and outer spheres combined. We do this by calculating the total charge Q enclosed by a Gaussian surface of radius r.

For part (b), we consider a spherical Gaussian surface with radius r, where R₁ < r < (R₁ + d). The total charge enclosed by this surface is:

Q = (4/3)π((R₁ + d)³ - R₁³) * (-σ₁) + (4/3)π(R₂³ - (R₁ + d)³) * σ₂

Using Gauss's law, the electric field is given by:

Flux = E * 4πr² = (Q) / ε₀

Simplifying, we find:

E = ((4/3)π((R₁ + d)³ - R₁³) * (-σ₁) + (4/3)π(R₂³ - (R₁ + d)³) * σ₂) / (4πr² * ε₀)

For parts (c) and (d), the calculations are similar but we need to consider different charge distributions.

For part (c), we consider a spherical Gaussian surface with radius r, where (R₁ + d) < r < (R₂ - d). The total charge enclosed by this surface is:

Q = (4/3)π(R₂³ - (R₁ + d)³) * σ₂

Using Gauss's law, the electric field is given by:

Flux = E * 4πr² = (Q) / ε₀

Simplifying, we find the same expression as in part (b) for the electric field magnitude.

For part (d), we consider a spherical Gaussian surface with radius r, where (R₂ - d) < r < R₂. The total charge enclosed by this surface is:

Q = (4/3)πr³ * σ₂

Using Gauss's law, the electric field is given by:

Flux = E * 4πr² = (Q) / ε₀

Simplifying, we find:

E = (4/3)πr * σ₂ / (ε₀ * r²)

In all these cases, the direction of the electric field will be radial, pointing outward from the center of the spheres.

Please note that these equations assume that the thickness of the material in both spheres is small compared to their radii (i.e., d << R₂ and d << R₁).