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 .

Please give answer.

To answer these questions, we need to apply Gauss's law for electrostatic fields, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the electric constant (ε₀).

(a) To find the electric field inside the inner sphere, we will consider a Gaussian surface in the shape of a sphere with radius r, where r is less than R1 (radius of the inner sphere). Since the inner sphere is negatively charged, the net charge enclosed by the Gaussian surface is -σ1 multiplied by the volume of the inner sphere (4/3πR1³), which gives -4/3πR1³σ1. Using Gauss's law, the electric flux through the Gaussian surface is given by E * 4πr², where E is the electric field.
According to Gauss's law, the electric flux is equal to the charge enclosed divided by the electric constant: E * 4πr² = (-4/3πR1³σ1) / ε₀.
Simplifying, we find the electric field magnitude inside the sphere:
|E| = (-R1³σ1 / (3ε₀r²))

(b) To find the electric field at a distance r where r is between R1 and (R1 - d), we need to consider two Gaussian surfaces.
First, let's consider a Gaussian surface located inside the hollow region between the two spheres, with a radius r. Since there is no charge enclosed by this surface, the electric flux through this Gaussian surface is zero. Therefore, the electric field inside this region is zero.

Next, let's consider a Gaussian surface in the shape of a sphere, with a radius r, located just inside the inner sphere (between R1 and (R1 - d)). The net charge enclosed by this Gaussian surface is -σ1 multiplied by the volume of the hollow region (4/3π(r³ - R1³)), which gives -4/3π(σ1(r³ - R1³)). Applying Gauss's law, we have E * 4πr² = (-4/3π(σ1(r³ - R1³))) / ε₀. Simplifying, we find the electric field magnitude inside this region:
|E| = (σ1(r³ - R1³)) / (3ε₀r²)

(c) To find the electric field at a distance r where r is between (R1 - d) and (R2 - d), we need to consider two Gaussian surfaces.
First, let's consider a Gaussian surface located just outside the larger sphere (between R2 and (R2 - d)). The net charge enclosed by this Gaussian surface is σ2 multiplied by the volume of the hollow region (4/3π((R2 - d)³ - R2³)), which gives 4/3π(σ2((R2 - d)³ - R2³)). Applying Gauss's law, we have E * 4πr² = (4/3π(σ2((R2 - d)³ - R2³))) / ε₀. Simplifying, we find the electric field magnitude inside this region:
|E| = (σ2((R2 - d)³ - R2³)) / (3ε₀r²)

Next, let's consider a Gaussian surface located just inside the larger sphere (between R1 and (R1 - d)). The net charge enclosed by this Gaussian surface is -σ1 multiplied by the volume of the hollow region (4/3π((R2-d)³ - R1³)), which gives -4/3π(σ1((R2-d)³ - R1³)). Applying Gauss's law, we have E * 4πr² = (-4/3π(σ1((R2-d)³ - R1³))) / ε₀. Simplifying, we find the electric field magnitude inside this region:
|E| = (σ1((R2-d)³ - R1³)) / (3ε₀r²)

(d) To find the electric field at a distance r where r is between (R2 - d) and R2, we need to consider two Gaussian surfaces.
First, let's consider a Gaussian surface located just outside the larger sphere (between R2 and r). The net charge enclosed by this Gaussian surface is σ2 multiplied by the volume of the larger sphere (4/3πr³ - R2³), which gives 4/3πσ2(r³ - R2³). Applying Gauss's law, we have E * 4πr² = (4/3πσ2(r³ - R2³)) / ε₀. Simplifying, we find the electric field magnitude inside this region:
|E| = σ2(r³ - R2³) / (3ε₀r²)

Next, let's consider a Gaussian surface located just inside the larger sphere (between R2 - d and r). The net charge enclosed by this Gaussian surface is -σ1 multiplied by the volume of the hollow region (4/3π(r³ - (R2 - d)³)), which gives -4/3πσ1(r³ - (R2 - d)³). Applying Gauss's law, we have E * 4πr² = (-4/3πσ1(r³ - (R2 - d)³)) / ε₀. Simplifying, we find the electric field magnitude inside this region:
|E| = σ1(r³ - (R2 - d)³) / (3ε₀r²)

Note: The direction of the electric field can be determined using the unit vectors in spherical coordinates (r, θ^, ø^), where r is the radial direction, θ^ is the azimuthal direction, and ø^ is the polar direction. The direction of the electric field vector will be determined by the net charge enclosed within the Gaussian surface.