A disk with a uniform positive surface charge density lies in the x-y plane, centered on the origin.The disk contains 2.5 x 10^-6 C/m² of charge, and is 7.5 cm in radius.What is the electric field at z = 15 cm? A)30 N/C B)300 N/C C)3000 N/C D)3.0 x 10⁴ N/C E)3.0 x 10⁷ N/C

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The electric field of a disk with uniform surface charge density is given by:$$E = \frac{\sigma}{2\epsilon_0} \left ( 1 - \frac{z}{\sqrt{z^2 + R^2}} ight )$$where:$$\sigma$$ is the surface charge density$$\epsilon_0$$ is the permittivity of free space$$z$$ is the distance from the center of the disk to the point where the electric field is being calculated$$R$$ is the radius of the diskPlugging in the given values, we get:$$E = \frac{2.5 	imes 10^{-6} C/m^2}{2(8.85 	imes 10^{-12} C^2/Nm^2)} \left ( 1 - \frac{0.15 m}{\sqrt{0.15^2 m^2 + 0.075^2 m^2}} ight )$$$$E = 3.0 	imes 10^4 N/C$$

A Disk with a Uniform Positive Surface Charge Density Lies