Question

A conducting sphere of radius r1 = 0.45 m has a total charge of Q = 1.1 μC. A separate uncharged conducting sphere of radius r2 = 0.21 m connects to the first by a slim conducting wire. The spheres are separated by a huge distance compared to their size

. Part (b)  What is the surface charge density of the second sphere, σ2, after they are connected in coulombs per square meter?

Answer 1

Answer 2

When the two conducting spheres are connected by a conducting wire, they will come to the same potential due to the transfer of charge between them. Since the total charge on the first sphere (Q1) is 1.1 μC and the second sphere is initially uncharged, the final charge on each sphere will be Q1/2 = 0.55 μC.

The surface charge density (σ) of a conducting sphere is defined as the charge per unit area on its surface. The surface charge density of the second sphere (σ2) can be calculated as follows:

σ2 = Q2 / A2

where: Q2 is the final charge on the second sphere (0.55 μC) A2 is the surface area of the second sphere.

The surface area of a sphere is given by the formula: A = 4πr^2

Let's calculate σ2:

Given data: Q2 = 0.55 μC (1.1 μC / 2) r2 = 0.21 m

Calculate the surface area of the second sphere:

A2 = 4πr2^2 A2 = 4π * (0.21 m)^2

Now, we can calculate σ2:

σ2 = Q2 / A2 σ2 = (0.55 μC) / [4π * (0.21 m)^2]

Now, let's convert the charge to coulombs and solve for σ2:

σ2 = (0.55 × 10^-6 C) / [4π * (0.21 m)^2]

σ2 ≈ 8.39 × 10^-6 C/m^2

So, the surface charge density of the second sphere after they are connected is approximately 8.39 × 10^-6 coulombs per square meter (C/m²).