A metal sphere with radius $r_a$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_b$. There is charge $+q$ on the inner sphere and charge $-q$ on the outer spherical shell. Use $E_r=\left[1/\right(4\pi {ϵ}_0\left)\right]\left(q/r^2\right)$ and the result from part (a) to find the electric field at a point outside the larger sphere at a distance $r$ from the center, where $r>r_b$. Note: Part (a) asked to calculate the potential $V(r)$ for (i) $r < r_a$; (ii) $r_a < r < r_b$; (iii) $r>r_b$. (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite..

Verified step by step guidance

1. Step 1: To calculate the potential V(r) for different regions, start by considering the potential due to a point charge. The potential at a distance r from a point charge q is given by V(r) = q / (4πϵ₀r).
2. Step 2: For region (i) r < r_a, the potential inside a conductor is constant and equal to the potential at its surface. Therefore, V(r) = V(r_a) = q / (4πϵ₀r_a).
3. Step 3: For region (ii) r_a < r < r_b, the potential is the sum of the potentials due to the inner sphere and the outer shell. The potential due to the inner sphere is V_inner = q / (4πϵ₀r), and the potential due to the outer shell is constant because the shell does not affect the potential inside it. Thus, V(r) = q / (4πϵ₀r) + constant.
4. Step 4: For region (iii) r > r_b, the potential is the sum of the potentials due to both spheres, but since the outer shell has charge -q, it cancels the effect of the inner sphere at distances greater than r_b. Therefore, V(r) = 0.
5. Step 5: To show the potential difference V_ab = q / (4πϵ₀) (1/r_a - 1/r_b), use the expression for the potential at the surface of each sphere and subtract the potential at r_b from the potential at r_a. This gives V_ab = V(r_a) - V(r_b) = q / (4πϵ₀) (1/r_a - 1/r_b).

Key Concepts

Electric Potential

Electric potential, V, is the work done per unit charge in bringing a positive test charge from infinity to a point in space. It is a scalar quantity and is measured in volts. In this problem, the potential at a point is the sum of potentials due to the inner and outer spheres, and it is zero at infinity.

Gauss's Law

Gauss's Law relates the electric field E to the charge distribution. It states that the electric flux through a closed surface is proportional to the enclosed charge. For spherical symmetry, it simplifies the calculation of electric fields, allowing us to determine E(r) using the symmetry of the charge distribution on the spheres.

Electric Field

The electric field, E, is a vector field representing the force per unit charge exerted on a test charge at any point in space. It is derived from the gradient of the electric potential, E = -∂V/∂r. In this problem, E is calculated for regions inside, between, and outside the spheres, considering the charge distribution.