Dielectric breakdown occurs in the air at an electric field strength of E_max = 3.0 × 10⁶ V/m. What is the maximum surface charge density that can be placed on a spherical conductor of radius 1.5 m before breakdown?

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The electric field strength at the surface of a spherical conductor is given by:$$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$$where:* E is the electric field strength (V/m)* Q is the charge on the conductor (C)* r is the radius of the conductor (m)* $\epsilon_0$ is the permittivity of free space (8.85 × 10^-12 C^2/Nm^2)We are given that the maximum electric field strength that can be placed on a spherical conductor before breakdown is Emax = 3.0 × 10^6 V/m. We are also given that the radius of the spherical conductor is r = 1.5 m. We can use this information to solve for the maximum surface charge density that can be placed on the conductor:$$E_{max} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$$$$Q = 4\pi\epsilon_0r^2E_{max}$$$$Q = 4\pi(8.85 × 10^{-12} C^2/Nm^2)(1.5 m)^2(3.0 × 10^6 V/m)$$$$Q = 2.7 × 10^{-5} C$$The maximum surface charge density is:$$\sigma = \frac{Q}{4\pi r^2}$$$$\sigma = \frac{2.7 × 10^{-5} C}{4\pi(1.5 m)^2}$$$$\sigma = 2.7 × 10^{-5} C/m^2$$Therefore, the correct answer is A.

Dielectric Breakdown Occurs in the Air at an Electric Field