Problem 27 - University Physics with Modern Physics

Question

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{cm} .\) Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) 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) \(r=4.00 \mathrm{cm} ;\) (iii) \(r=6.00 \mathrm{cm} ?\) (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) 179.8 V; (ii) 1348.5 V; (iii) -449.5 V. Inner shell is at higher potential by 719.2 V.

Step 1: Electric Potential at the Center (r = 0 cm)

For a point inside a uniformly charged spherical shell, by Gauss's Law, the electric field is zero. Therefore, the electric potential at the center due to each shell is constant throughout the sphere. Since the potential at infinity is zero, the potential at the center is simply the sum of potentials due to each shell:- Potential due to first shell:\[ V_1 = k \frac{q_1}{R_1} = (8.99 \times 10^9) \frac{6 \times 10^{-9}}{0.03} = 1798 \text{ V}\]- Potential due to second shell:\[ V_2 = k \frac{q_2}{R_2} = (8.99 \times 10^9) \frac{-9 \times 10^{-9}}{0.05} = -1618.2 \text{ V}\]- Total potential at the center:\[ V = V_1 + V_2 = 1798 - 1618.2 = 179.8 \text{ V}\]

Step 2: Electric Potential at r = 4 cm

At point r = 4 cm, we are outside the smaller shell but inside the larger shell. The potential is constant as it depends only on the total charge within radius r.Since only the smaller shell contributes within this region:- The potential contribution from the smaller shell is:\[ V_1 = k \frac{q_1}{r} = (8.99 \times 10^9) \frac{6 \times 10^{-9}}{0.04} = 1348.5 \text{ V}\]- The larger shell contributes no potential because it's external to the point considered. The total potential at r = 4 cm is:\[ V = V_1 = 1348.5 \text{ V}\]

Step 3: Electric Potential at r = 6 cm

For r = 6 cm, which is outside both shells, the potential depends on the net charge of both shells treated as a point charge:- The net charge is:\[ q_{net} = q_1 + q_2 = 6 \times 10^{-9} - 9 \times 10^{-9} = -3 \times 10^{-9} \text{ C} \]- The potential at r = 6 cm is:\[ V = k \frac{q_{net}}{r} = (8.99 \times 10^9) \frac{-3 \times 10^{-9}}{0.06} = -449.5 \text{ V}\]

Step 4: Potential Difference Between Shell Surfaces

To find the potential difference between the surfaces of the two shells, we should calculate the potential at each shell's surface and then find their difference.- Potential at the surface of the inner shell (r = 3 cm):\[ V_{inner} = k \frac{q_1}{3 \times 10^{-2}} + k \frac{q_2}{5 \times 10^{-2}} = 1798 - 1618.2 = 179.8 \text{ V}\]- Potential at the surface of the outer shell (r = 5 cm):\[ V_{outer} = k \left( \frac{6 \times 10^{-9}}{5 \times 10^{-2}} + \frac{-9 \times 10^{-9}}{5 \times 10^{-2}} \right) = -539.4 \text{ V} \]- Potential difference: \[ \Delta V = V_{inner} - V_{outer} = 179.8 - (-539.4) = 719.2 \text{ V} \]The inner shell is at a higher potential than the outer shell by 719.2 V.

Key Concepts

Spherical Shells

Spherical shells are three-dimensional objects that feature concentric layers or surfaces, shaped like a sphere, surrounding a common center. In physics, when dealing with charges, these shells are often considered in problems concerning electrostatics and electric fields. A critical factor about spherical shells is that they can be conductive or insulating. In the case of insulating shells, charges are strictly confined to the surface and don't move across it. This exercise demonstrates a system composed of two thin concentric spherical shells, each possessing a distinct uniform charge distribution over their surface.
Key characteristics of these shells include:

• Each shell has its respective radius (like the smaller shell with radius 3.00 cm and the larger one with radius 5.00 cm).
• The charge distribution could be positive or negative, as indicated by the charges of the shells.
• The interactions of these charges and their effects on electric potential are computed using fundamental electrostatic principles.

Gauss's Law

Gauss's Law is a pivotal principle in electromagnetism that defines the relationship between the electric field and the charge distribution within a given volume. The law states that the total electric flux through a closed surface is directly proportional to the enclosed charge. Mathematically, it's expressed as:\[ \Phi = \int \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where \(\Phi\) is the electric flux, \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is a differential area on the closed surface, \(Q_{enc}\) is the enclosed charge, and \(\varepsilon_0\) is the permittivity of free space.

For spherical shells, Gauss’s Law helps simplify calculations by indicating that the electric field inside a charged spherical shell is zero when observed from a point within the material. This principle is used to determine potentials for different regions within and around the shells. For instance, at a point inside the smaller shell but outside the larger shell, only the smaller shell contributes to the electric potential since its electric field is uniform.

Potential Difference

Potential difference, also known as voltage, is the difference in electric potential between two points in space. It indicates how much work is required to move a charge from one point to another. For spherical shells, potential difference is significant as it determines the relative energy states between the charges on the shells.

In the exercise, we calculated the potential difference between the surfaces of two concentric spherical shells by subtracting the potentials calculated at each shell's surface. The formula for potential due to a charge \(q\) at a distance \(r\) from it is:\[ V = k \frac{q}{r} \]where \(k\) is Coulomb's constant. The potential difference is found as:\[ \Delta V = V_{inner} - V_{outer} \]This allows us to observe that the inner shell exhibited a higher potential than the outer shell, a result of their specific charge values and separation.

Insulating Material

Insulating materials are substances that do not allow the free flow of electric charge. In the context of spherical shells, insulating materials act as barriers that hold charge on their surfaces without allowing it to move through the shell or interact with internal charges.

The significance of using insulating materials in these shells is reducing interaction and charge redistribution within the material, sticking the charges strictly to the surfaces. This makes calculations of electric potential and fields more straightforward as each surface charge distribution maintains its potential defined by its own geometry and the distance to any point of interest. Understanding insulation helps clarify why electric fields and potential can be considered as originating primarily from surface charges, particularly in such electrostatic conditions.