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 electric potential (V) for a spherical conductor with a radius of 5.00 cm is \(1.35 \times 10^6 \mathrm{~V}\). There is a maximum value for V because the dielectric strength of air restricts the electric field that can be sustained without causing a breakdown of the insulating properties of the air. If the electric field exceeds the dielectric strength of air, the air will ionize, and a spark will occur, discharging the sphere. Thus, there is a maximum potential difference that can be held by the conductor without causing a breakdown of the surrounding air.

Step-by-step solution

Formula for electric potential

The electric potential (V) for a charged sphere with charge Q and radius R at its surface is given by the formula: \[V = \frac{kQ}{R}\] k is the electrostatic constant: \( k = 8.99 \times 10^9 \frac{N \cdot m^2}{C^2} \)

Dielectric strength of air

To find the maximum electric potential, we need to take into account the dielectric strength of air. When the electric field is too strong, it can cause the air to ionize and create a spark. The dielectric strength of dry air is approximately \(3.0 \times 10^6 \mathrm{~V/m}\).

Maximum electric field at the surface

To calculate the maximum electric field allowed at the surface of the spherical conductor, we need to find the electric field (E) produced by the charged sphere at the surface. The expression for E is: \[E = \frac{kQ}{R^2}\] Since the dielectric strength of air is given, we can equate the electric field produced by the sphere (E) to the dielectric strength of air (E_air) to find the maximum electric field allowed at the surface of the spherical conductor. \[E = E_{\text{air}}\] \[\frac{kQ}{R^2} = 3.0 \times 10^6 \mathrm{~V/m}\]

Calculate the maximum charge (Q)

Now, we can calculate the maximum charge (Q) that can be held by the spherical conductor before the air breaks down. \[Q = 3.0 \times 10^6 \frac{\mathrm{V} \cdot \mathrm{m}}{\mathrm{C}} \times R^2\] Substitute R with the given radius of the spherical conductor (0.05m): \[Q = 3.0 \times 10^6 \frac{\mathrm{V} \cdot \mathrm{m}}{\mathrm{C}} \times (0.05\mathrm{~m})^2\] \[Q = 7.5 \times 10^{-6} \mathrm{~C}\]

Calculate the maximum electric potential (V)

Now that we have the maximum charge (Q) that can be held by the spherical conductor, we can calculate the maximum electric potential (V) using the formula: \[V = \frac{kQ}{R}\] Substitute the values for k, Q, and R: \[V = \frac{8.99 \times 10^9 \frac{\mathrm{N}\cdot \mathrm{m}^2}{\mathrm{C^2}} \times 7.5 \times 10^{-6} \mathrm{~C}}{0.05\mathrm{~m}}\] \[V = 1.35 \times 10^6 \mathrm{~V}\] a) The maximum magnitude V can have is \(1.35 \times 10^6 \mathrm{~V}\). b) There is a maximum value for V because the dielectric strength of air restricts the electric field that can be sustained without causing a breakdown of the insulating properties of the air. In other words, if the electric field exceeds the dielectric strength of air, the air will ionize, and a spark will occur, discharging the sphere. Therefore, there is a maximum potential difference that can be held by the conductor without causing a breakdown of the surrounding air.