Parallel plates: 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

Question

Quizplus · Accepted Answer

The electric field strength between two parallel plates is given by:E = V/dwhere E is the electric field strength, V is the potential difference between the plates, and d is the distance between the plates.In this problem, we are given the electric field strength (E = 1.0 × 10^5 N/C) and the potential difference between the plates (V = 3.9 × 10^9 electrons * 1.60 × 10^-19 C/electron = 6.24 × 10^-10 V). We can use this to solve for the distance between the plates:d = V/E = 6.24 × 10^-10 V / 1.0 × 10^5 N/C = 6.24 × 10^-15 mThe diameter of the disks is twice the distance between the plates, so:diameter = 2d = 2 * 6.24 × 10^-15 m = 1.25 × 10^-14 m = 1.25 cmTherefore, the diameter of the disks is 1.25 cm, which is closest to choice A (3.0 cm).

Parallel Plates: the Electric Field Strength in the Space Between