Problem 25 - University Physics with Modern Physics

Question

A thin spherical shell with radius \(R_1 =\) 3.00 cm is concentric with a larger thin spherical shell with radius \(R_2 =\) 5.00 cm. Both shells are made of insulating material. The smaller shell has charge \(q_1 = +\)6.00 nC distributed uniformly over its surface, and the larger shell has charge \(q_2 = -\)9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) \(r =\) 0; (ii) \(r =\) 4.00 cm; (iii) \(r =\) 6.00 cm? (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

Step-by-Step Solution

Answer

(i) 1798 V, (ii) -269.7 V, (iii) -449.5 V; The potential difference is 2247.5 V, the inner shell is at higher potential.

Step 1: Understand the Setup

We have two concentric spheres with radii \(R_1 = 3.00\) cm and \(R_2 = 5.00\) cm. The inner sphere has charge \(q_1 = +6.00\) nC, and the outer sphere has charge \(q_2 = -9.00\) nC. The potential at an infinite distance is zero.

Step 2: Electric Potential Inside a Charged Spherical Shell

The electric potential at a distance \(r\) from the center due to a spherical shell with a charge \(q\) is given by \(V(r) = \frac{kq}{R}\) when \(r < R\), where \(R\) is the radius of the shell. For \(r = 0\), the potential is only due to the inner shell since \(r < R_1\).

Step 3: Calculate Potential at \(r = 0\)

For \(r = 0\), the potential \(V_0\) is due to the inner shell only. Thus, \(V_0 = \frac{kq_1}{R_1}\), where \(k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\). Substitute \(q_1 = 6.00 \times 10^{-9} \text{ C}\) and \(R_1 = 0.03 \text{ m}\) to find \(V_0 = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-9}}{0.03} = 1798 \text{ V}\).

Step 4: Electric Potential Between Two Spherical Shells

Between the two shells (\(R_1 < r < R_2\)), the potential is the sum of potentials due to both shells: \(V(r) = \frac{kq_1}{r} + \frac{kq_2}{R_2}\).

Step 5: Calculate Potential at \(r = 4.00\) cm

At \(r = 0.04\) m, substitute \(r = 0.04\) m, \(q_1 = 6.00 \times 10^{-9}\) C, and \(q_2 = -9.00 \times 10^{-9}\) C in \(V(r) = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-9}}{0.04} + \frac{8.99 \times 10^9 \times (-9.00 \times 10^{-9})}{0.05}\). Calculate \(V(0.04) = 1348.5 \text{ V} - 1618.2 \text{ V} = -269.7 \text{ V}\).

Step 6: Electric Potential Outside Both Spherical Shells

Outside both shells (\(r > R_2\)), total potential is \(V(r) = \frac{k(q_1 + q_2)}{r}\).

Step 7: Calculate Potential at \(r = 6.00\) cm

At \(r = 0.06\) m, since \(q_1 + q_2 = -3.00 \times 10^{-9}\) C, \(V(0.06) = \frac{8.99 \times 10^9 \times (-3.00 \times 10^{-9})}{0.06} = -449.5 \text{ V}\).

Step 8: Calculate Potential Difference Between Shells

The potential difference \(\Delta V = V_{inner} - V_{outer}\). From the steps, \(V_{inner} = 1798 \text{ V} (\text{at } R_1) \) and \(V_{outer} = -449.5 \text{ V} (\text{at } R_2)\). Thus, \(\Delta V = 1798 + 449.5 = 2247.5 \text{ V}\). The inner shell is at a higher potential.

Key Concepts

Spherical Shells

Spherical shells are three-dimensional surfaces which are perfectly symmetric around a central point. Think of them as the outer layer of an onion or the shell of a soccer ball, but completely smooth and uniform. In physics, when these shells are made up of insulating materials and carry electric charge, they exhibit unique properties.

These spherical shells affect their surroundings with the electric field they create. Since the shells in this context are thin, all the charge is concentrated on their surfaces. This is a key aspect when calculating electric potential and fields as opposed to how a solid sphere would behave.

• The shells consider the distance from their center – the inner shell influences closer points, and the outer shell affects points further away.
• As the shells are concentric, their centers align, allowing shared analysis of their combined effects on charges at various distances.

Understanding these shells aids in grasping how they affect each other and their environment, particularly when evaluating factors like electric potential and field strength.

Coulomb's Law

Coulomb's Law is fundamental in understanding electric forces between two charged objects. It tells us how the force and potential energy change based on distance and charge magnitude.

According to this law, the electric potential (\(V\)) at a distance from a charged object will be directly proportional to the charge (\(q\)) and inversely proportional to that distance. Thus, for a spherical shell with radius \( R\), the potential at any point \( r\) outside the shell is calculated using \( V(r) = \frac{kq}{r}\), where \( k\) is Coulomb's constant.

• Inside a spherical shell, the potential remains constant no matter how close one gets to the center, as long as they stay within the boundary of that shell.
• Between the shells, the potential is a sum of the potentials from each shell calculated separately.

This law simplifies complex systems into manageable calculations, as seen in finding potentials at various distances from two concentric shells. It is crucial to understand this relationship to analyze systems with multiple charges and varying distances.

Electric Fields

The concept of electric fields revolves around the idea of a force field produced by charges around them. Imagine an invisible aura that affects how other charges move in its presence. Though the electric field itself does no work, it dictates how the potential energy of charges changes.

Electric fields (\(E\)) from spherical shells can be visualized as lines that start perpendicular from the shell's surface and move outwards, becoming sparser further away; this indicates weaker fields.

• Inside a charged spherical shell, the electric field is zero – the charges on the surface cancel their mutual effect.
• Outside the shells, however, the field behaves as if all the shell's charge were concentrated at the center, simplifying into a point-like source.

This visualization helps predict the movement and behavior of charges in and around such setups. Decisions involving potential differences and the strength of the fields between these shells rely heavily on these features of the electric field.