A charge Q1 = -32 μC is fixed on the y axis at y = 4 m, and a charge Q2 = +18 μC is fixed on the x axis at x = 3 m.
a) Calculate the magnitude of the electric field E1 at the origin due to charge Q1. b) Calculate the magnitude of the electric field E2 at the origin due to charge Q2. c) On the diagram below, draw and label the electric fields E1, E2 and the net electric field at the origin.
d) Calculate the net electric field at the origin due to two charges Q1 and Q2.
Answer (1)
The net field is 25,456 N/C, directed southeast (diagonal between x and y axes).Given:[tex]Q_{1} =[/tex] -32μC at y = 4m[tex]Q_{2} =[/tex] +18μC at x = 4mFind electric field at origin from each charge and net fieldElectric field formula: [tex]E = \frac{k- /Q/ }{r^{2} }[/tex]Where:[tex]k = (9) (10^{9})(Nm^{2}) / C^{2}[/tex]Q is chargein Coulombs (μC = [tex]10^{-6} C[/tex] is distance from charge to origina) [tex]E_{1}[/tex] (due to Q₁ at y = 4 m)[tex]E_{1}= \frac{(9)(10^{9})(10^{-6} )}{4^{2} } \\ E_{1}= \frac{288,000}{16 }\\ E_{1}= 18,000 N/C[/tex]b) [tex]E_{2}[/tex] (due to Q₂ at x = 3 m)[tex]E_{2} = \frac{(9)(10^{-9}-18)(10x^{-6} )}{3^{2} }\\ E_{2} = \frac{162,000}{9}\\ E_{2} = 18,000 N/C[/tex]c) Net Electric Field at originE₁ and E₂ are perpendicular, so use Pythagorean theorem.[tex]E_{net} = \sqrt{(E^{2}_{1} )+(E^{2}_{2} )} \\E_{net} = \sqrt{(18,000^{2} ) + (18,000^{2} )} \\E_{net} = \sqrt{2(18,000^{2})} \\E_{net} = 18000\sqrt{2} \\E_{net} = 25,456 N/C[/tex]
