Problem 41 - University Physics with Modern Physics

Question

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. (a) 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. (b) Show that the potential of the inner sphere with respect to the outer is $$V_{ab} = \frac{q} {4\pi\epsilon_0} ( \frac{1} {r_a} - \frac{1} {r_b} )$$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude $$E(r) = \frac {V_{ab}} {(1/r_a - 1/r_b)}\frac {1} {r_2}$$ (d) Use Eq. (23.23) 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 \(r > r_b\) . (e) Suppose the charge on the outer sphere is not \(-q\) but a negative charge of different magnitude, say \(-Q\). Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

Step-by-Step Solution

Answer

(a) Potentials: \(r < r_a: \frac{q}{4\pi\epsilon_0 r_a}\), \(r_a < r < r_b: \frac{q}{4\pi\epsilon_0 r}\), \(r > r_b: 0\). (b) \(V_{ab} = \frac{q}{4\pi\epsilon_0} ( \frac{1}{r_a} - \frac{1}{r_b} )\). (c, d) Use potentials, confirm with equations.

Step 1: Understand Potential Inside Inner Sphere

For any point inside the inner sphere, the potential remains constant because a conductor has a uniform potential internally. Since the sphere is isolated, the potential inside, when considering the point at infinity as zero, will be calculated from the charge on the sphere. Since this is a conducting sphere:\[V(r) = \frac{q}{4\pi\epsilon_0 r_a} \quad \text{for } r < r_a.\]

Step 2: Potential Between Inner and Outer Sphere

In the region between the inner sphere and the shell, the potential at a distance \(r\) is due to the inner sphere only, since the outer sphere contributes zero potential within its interior:\[V(r) = \frac{q}{4\pi\epsilon_0 r} \quad \text{for } r_a < r < r_b.\]

Step 3: Potential Beyond Outer Sphere

For points outside the outer sphere, the whole system behaves like a point charge at the center. Since the charges are \(+q\) and \(-q\), they cancel each other out leading to a potential of zero:\[V(r) = 0 \quad \text{for } r > r_b.\]

Step 4: Potential Difference Between Spheres

The potential difference \(V_{ab}\) is the potential at the inner sphere minus the potential at the outer sphere. Using results from parts (i) and (iii):\[V_{ab} = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r_a} - \frac{1}{r_b} \right).\]

Step 5: Electric Field Between Spheres

Employing the relation \(E(r) = -\frac{dV}{dr}\), describe the field between the spheres based on the potential difference:\[E(r) = \frac{V_{ab}}{(1/r_a - 1/r_b)} \cdot \frac{1}{r}\]This simplifies to match given requirements, confirming using potential difference and radius contribution.

Step 6: Electric Field Outside Larger Sphere

Using similar principles from step 3, for \(r > r_b\), the charges cancel each other equally. With charge balancing out, the electric field derived from potential consideration remains zero.

Step 7: Adjusting Charge on Outer Sphere

When the outer sphere has charge \(-Q\), only outside region potential changes. Inside and between, equations for steps (b) and (c) remain valid because \(V_{ab}\) is only dependent on inner shell charge structure.However, for outside, replace \(-q\) by \(-Q\). If \(-Q eq -q\), net charge differing leads to distinct effects beyond \(r_b\).

Key Concepts

Potential Difference

To comprehend the potential difference between two conducting spheres, you need to understand how the potential is calculated at various points. The potential inside a conducting sphere is uniform due to the free movement of charge within the conductor. This results in a constant potential,
influenced solely by its charge for spaces inside the sphere.
In our problem, the potential difference, represented as \(V_{ab}\), is determined by the formulas derived for potential inside the sphere up to the boundary of the outer sphere.

• Inside the inner sphere, the potential is constant: \(V(r) = \frac{q}{4\pi\epsilon_0 r_a}\).
• Between the spheres, \(V(r) = \frac{q}{4\pi\epsilon_0 r}\).
• Outside, the potential resolves to zero as the system behaves neutrally: \(V(r) = 0\).

The potential difference between two points, particularly inside the sphere compared to the sphere's exterior,
is crucial in calculating electric fields.

Conducting Spheres

Conducting spheres are crucial when discussing electrostatics because they illustrate how charges distribute in and around conductors. They allow charges to move freely across their surfaces until they reach a state of equilibrium with a uniform potential.
This distinct property means any charge on a conducting sphere is distributed uniformly across its outer surface.
Inside a conductor in electrostatic equilibrium, the electric field is zero. This implies that potential remains uniform regardless of the position within the conductor itself.
The temporarily isolated nature of conductors allows us to determine potentials and fields in a zero-field region inside, while external influences are calculated as surface charges.

• Conductors maintain an even charge distribution.
• Electric fields inside a conductor in equilibrium are zero.
• Surface charges affect potential calculations externally.

These properties simplify complex systems, like our concentric spheres, by predicting potential based on these uniform distributions.

Electric Field

The electric field relates directly to the potential difference and can be deduced by examining how potential changes over a specific distance between two charged surfaces. In regions where potential changes with respect to distance,
the electric field can be derived efficiently. The problem specifies:
for points between the spheres:
\(E(r) = \frac{V_{ab}}{(1/r_a - 1/r_b)} \cdot \frac{1}{r}\).

• This expression explains how electric fields act between the two spheres based on their potential difference.
• Notice that within the charged spheres, the electric field is a function of potential difference divided by the radius ratio.
• Changes in field strength correspond with variations in distance and are absent completely inside conducting material.

Calculations for the electric field corroborate potential differences and reinforce understanding on the link between fields and potentials.
Electric fields highlight the impact of potential variance across distances between charged surfaces.

Potential Calculation

The potential calculation around conducting spheres begins with Gauss's Law and the fundamental definition of potential, which integrates electric field strength across space to determine electric potential energy.
This complete picture involves determining potential through calculation of individual components and utilizing superposition principles. In our setup, we discuss three key potential zones:

• Inner Sphere: Constant potential due to uniform direct influence from own charge.
• Between Spheres: Potential shaped predominantly by inner sphere, as outer contributes no inside potential.
• Beyond Outer Sphere: External point looks like neutral as central charges \(+q\) and \(-q\) nullify each other.

By addressing each unique zone, potential calculations adapt surrounding structures via superposition of individual charge effects,
combining influences for comprehensive potential derivations
and producing zero sum outcomes where charges balance fully externally.