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 (rˆ,θˆ,ϕˆ)
b)Magnitude: (Express your answer in terms of the following variables, if necessary R1, R2, σ1, σ2, d, r and the constant ϵ0. To enter σ1 type sigma_1, for σ2 type sigma_2, for R1 type R_1, etc..)
((d) r<R2 but larger than R2−d
Direction: use units vectors in spherical coordinates (rˆ,θˆ,ϕˆ)
To determine the electric field inside the inner sphere, we can use Gauss's Law. Gauss's Law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.
(a) Electric Field inside the Inner Sphere:
1. Choose a Gaussian surface that encloses the inner sphere.
2. Since the inner sphere is negatively charged, the electric field will point radially inward towards the center of the sphere. Thus, we choose a Gaussian surface in the shape of a concentric sphere with radius r within the inner sphere.
3. According to Gauss's Law, the electric flux through this Gaussian sphere is given by the formula φ = E * 4πr^2, where E is the magnitude of the electric field inside the sphere.
4. The enclosed charge within the Gaussian sphere is -σ1 * volume of inner sphere = -σ1 * (4/3)π(r^3).
5. By substituting these values into Gauss's Law, we have E * 4πr^2 = -σ1 * (4/3)π(r^3) / ε0, where ε0 is the permittivity of free space.
6. Solving for E, we get E = -σ1 * (r/3ε0).
(b) Magnitude of Electric Field inside the Inner Sphere:
To express the magnitude of the electric field in terms of the given variables, we simply take the absolute value of the expression derived in step 6.
So, the magnitude of the electric field inside the inner sphere is E = |σ1 * (r/3ε0)|.
(d) Electric Field inside Larger Sphere, r < R2 but larger than R2−d:
1. Similar to the previous case, we choose a Gaussian surface within the larger sphere, but outside the inner sphere. This Gaussian surface will be in the shape of a concentric sphere with radius r.
2. Since the larger sphere is positively charged, the electric field will point radially outward from the center of the sphere. Thus, we choose a Gaussian surface with radius r inside the larger sphere.
3. According to Gauss's Law, the electric flux through this Gaussian sphere is given by the formula φ = E * 4πr^2, where E is the magnitude of the electric field inside the sphere.
4. The enclosed charge within the Gaussian sphere is (σ2 * volume of larger sphere) - (-σ1 * volume of inner sphere) = (σ2 * (4/3)π(R2^3) - (-σ1 * (4/3)π(R1^3)).
5. By substituting these values into Gauss's Law, we have E * 4πr^2 = (σ2 * (4/3)π(R2^3) - (-σ1 * (4/3)π(R1^3))) / ε0.
6. Solving for E, we get E = (σ2 * (R2^3) - (-σ1 * (R1^3))) / (4πε0 * r^2).
Note: In both cases, the direction of the electric field is already mentioned in the question. Inside the inner sphere, it points radially inward, and inside the larger sphere but outside the inner sphere, it points radially outward.