Question

An isolated solid spherical conductor of radius \(5.00 \mathrm{~cm}\) is surrounded by dry air. It is given a charge and acquires potential \(V\), with the potential at infinity assumed to be zero. a) Calculate the maximum magnitude \(V\) can have. b) Explain clearly and concisely why there is a maximum.

Short answer

Answer: The maximum potential a solid spherical conductor with a radius of 5 cm can have is 150,000 V. There is a maximum potential because of the breakdown voltage of the surrounding dry air. When the electric field strength at the surface of the sphere exceeds the breakdown voltage of the dry air, a discharge occurs, and the sphere loses some of its charge, preventing the potential from increasing further.

Step-by-step solution

Identify the Knowns and Unknowns

We know the following information: - The radius of the solid spherical conductor: \(r = 5.00\,\text{cm}\) - We assume the potential at infinity to be zero: \(V_{\infty}=0\) We need to find: a) The maximum magnitude of the potential \(V\) of the sphere. b) Explain why there is a maximum potential.

Recall the Formula for the Potential of a Sphere

The formula for the potential of a sphere with charge \(Q\), at distance \(r\) from its center is: \(V = \frac{1}{4π \varepsilon_0} \cdot \frac{Q}{r}\) Where, - \(V\) is the potential - \(Q\) is the charge of the sphere - \(r\) is the distance from the center of the sphere - \(\varepsilon_0\) is the permittivity of free space (\(8.85\times10^{-12}\,\text{C}^2\text{/N m}^2\))

Consider the Breakdown Voltage of Dry Air

The maximum potential a sphere in dry air can have is limited by the breakdown voltage of air. When the electric field strength at the surface is greater than the breakdown voltage of dry air, it will cause a discharge (like a spark) and the potential will not be able to increase further. The breakdown field strength of dry air is around \(E_{max} \approx 3\times10^6\,\text{V/m}\).

Calculate the Electric Field at the Surface of the Sphere

We need to find the electric field at the surface of the sphere when the potential is at its maximum. To calculate the electric field, we can use the formula \(E = \frac{V}{r}\). Now, we can set the electric field equal to the breakdown field strength of dry air and solve for the maximum potential \(V_{max}\). \(E_{max} = \frac{V_{max}}{r}\) \(V_{max} = E_{max} \cdot r\)

Find the Maximum Potential \(V_{max}\)

Plug in the values for \(E_{max}\) and \(r\) to find the maximum potential: \(V_{max} = (3\times10^6\,\text{V/m}) \cdot (5.00\,\text{cm})\) \(V_{max} = (3\times10^6\,\text{V/m}) \cdot (0.05\,\text{m})\) \(V_{max} = 150,000\,\text{V}\) a) The maximum magnitude of the potential \(V_{max}\) is \(150,000\,\text{V}\).

Explain Why There is a Maximum Potential

b) There is a maximum potential because of the breakdown voltage of the surrounding dry air. When the electric field strength at the surface of the sphere exceeds the breakdown voltage of the dry air, a discharge occurs (like a spark), and the sphere loses some of its charge, preventing the potential from increasing further. Therefore, there is a maximum potential that the solid spherical conductor can have.