The electric field strength in the space between two closely spaced parallel disks is 1.0 × 10⁵ N/C. This field is the result of transferring 3.9 × 10⁹ electrons from one disk to the other. What is the diameter of the disks? (e= 1.60 × 10^-19C, ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A) 3.0 cm B) 1.5 cm C) 4.5 cm D) 6.0 cm

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The electric field strength between two closely spaced parallel disks is given by:$$E = \frac{\sigma}{\epsilon_0}$$where:* E is the electric field strength (N/C)* σ is the surface charge density (C/m^2)* ε0 is the permittivity of free space (8.85 × 10^-12 C^2/N ∙ m^2)The surface charge density is given by:$$\sigma = \frac{Q}{A}$$where:* Q is the charge on the disk (C)* A is the area of the disk (m^2)Substituting the given values into the above equations, we get:$$1.0 	imes 10^5 N/C = \frac{3.9 	imes 10^9 C}{\epsilon_0 A}$$Solving for A, we get:$$A = \frac{3.9 	imes 10^9 C}{(1.0 	imes 10^5 N/C) \epsilon_0} = 3.48 	imes 10^-5 m^2$$The diameter of the disks is given by:$$d = 2\sqrt{\frac{A}{\pi}} = 2\sqrt{\frac{3.48 	imes 10^-5 m^2}{\pi}} = 3.0 cm$$Therefore, the correct answer is A.

The Electric Field Strength in the Space Between Two Closely