A uniform line of charge with linear charge density λ is at rest on th

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A uniform line of charge with linear charge density λ is at rest on the x-axis. At the dot, we can calculate a small contribution to the the y-component of the electric field due to an infinitesimally small amount of charge at the shaded region. What is the y-component of the electric field due to the shaded region of charge? A. dEy = λdx4πϵ0(x2+y2) (B. dEy = λdx4πϵ0(x2+y2)yx C. dEy = λdx4πϵ0(x2+y2)xy D. dEy = λdx4πϵ0(x2+y2)yx2+y2 E. dEy = λdx4πϵ0(x2+y2)xx2+y2

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A uniform line of charge with linear charge density $λ$ is at rest on the $x$ -axis. At the dot, we can calculate a small contribution to the the y-component of the electric field due to an infinitesimally small amount of charge at the shaded region. What is the $y$ -component of the electric field due to the shaded region of charge? A. $d E_{y} = \frac{λ d x}{4 π ϵ_{0} (x^{2} + y^{2})}$ (B. $d E_{y} = \frac{λ d x}{4 π ϵ_{0} (x^{2} + y^{2})} \frac{y}{x}$ C. $d E_{y} = \frac{λ d x}{4 π ϵ_{0} (x^{2} + y^{2})} \frac{x}{y}$ D. $d E_{y} = \frac{λ d x}{4 π ϵ_{0} (x^{2} + y^{2})} \frac{y}{\sqrt{x^{2} + y^{2}}}$ E. $d E_{y} = \frac{λ d x}{4 π ϵ_{0} (x^{2} + y^{2})} \frac{x}{\sqrt{x^{2} + y^{2}}}$