A thin,circular disk of radius 30.0 cm is oriented in the yz-plane with its center at the origin.The disk carries a total charge of +3.00 μC distributed uniformly over its surface.Calculate the magnitude of the electric field due to the disk at the point x = 15.0 cm along the x-axis. (ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A)9.95 × 10⁵ N/C B)4.98 × 10⁵ N/C C)3.31 × 10⁵ N/C D)2.49 × 10⁵ N/C E)1.99 × 10⁵ N/C

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The electric field due to a uniformly charged disk at a point along its axis can be calculated using the formula derived from integrating Coulomb's law over the disk's surface. The formula for the electric field $E$ at a distance $x$ from the center of the disk along its axis is given by:$$E = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)$$where $\sigma$ is the surface charge density ($\sigma = \frac{Q}{\pi R^2}$), $Q$ is the total charge on the disk, $R$ is the radius of the disk, $x$ is the distance from the disk along its axis, and $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} C^2/N \cdot m^2$).Given:- $Q = 3.00 \mu C = 3.00 \times 10^{-6} C$- $R = 30.0 cm = 0.30 m$- $x = 15.0 cm = 0.15 m$- $\epsilon_0 = 8.85 \times 10^{-12} C^2/N \cdot m^2$First, calculate the surface charge density $\sigma$:$\sigma = \frac{Q}{\pi R^2} = \frac{3.00 \times 10^{-6} C}{\pi (0.30 m)^2}$Then, substitute $\sigma$, $x$, $R$, and $\epsilon_0$ into the formula for $E$:$$E = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{0.15 m}{\sqrt{(0.15 m)^2 + (0.30 m)^2}}\right)$$After performing the calculations, the result matches the value given in choice C, which is $3.31 \times 10^5 N/C$. This calculation involves integrating Coulomb's law over the disk's surface to account for the contributions of all the infinitesimal charge elements on the disk to the electric field at the point of interest.

A Thin,circular Disk of Radius 30