The electric field strength in the space between two closely spaced parallel disks is .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 $E$ between two parallel plates is given by $E = \frac{\sigma}{\varepsilon_0}$, where $\sigma$ is the surface charge density ($\frac{Q}{A}$, with $Q$ being the total charge and $A$ the area of the plate), and $\varepsilon_0$ is the permittivity of free space. Given $E = 1.0 \times 10^{5} \, \text{N/C}$, $e = 1.60 \times 10^{-19} \, \text{C}$, and $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2$, and the number of electrons transferred $N = 3.9 \times 10^{19}$, we first find the total charge $Q = N \cdot e = 3.9 \times 10^{19} \cdot 1.60 \times 10^{-19} \, \text{C} = 6.24 \times 10^{-0} \, \text{C}$. Using the equation for $E$, we solve for $A$, the area of the disk: $A = \frac{Q}{E \cdot \varepsilon_0}$. Substituting the given values, $A = \frac{6.24 \times 10^{-0}}{1.0 \times 10^{5} \cdot 8.85 \times 10^{-12}} \, \text{m}^2$, which calculates to approximately $7.05 \times 10^{-4} \, \text{m}^2$. The area of a disk is given by $A = \pi r^2$, where $r$ is the radius of the disk. Solving for $r$, we get $r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{7.05 \times 10^{-4}}{\pi}} \, \text{m}$, which is approximately $0.015 \, \text{m}$ or $1.5 \, \text{cm}$. Since the diameter is twice the radius, the diameter is $3.0 \, \text{cm}$.

The Electric Field Strength in the Space Between Two Closely