Respuesta :
Answer:
The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 8.20 cm from the origin O . For each of the situations in the drawing, determine the magnitude of the net electric field at the origin.
Given, all charges in the figure a and b are in the units of μ C .
The answer to the question is
[tex]E_{Net} =[/tex] 7.796×10⁶ N / C at 45.0 ° above the x axis
Explanation:
q₁ = 2.0 μ C
q₂ = -3.0 μ C
q₃ = -5.0 μ C
The principle of superposition states that the net electric field is equal to the sum of the individual electric fields
[tex]E_{Net} =[/tex] ∑ [tex]E_{i}[/tex]
Therefore [tex]E_{Net0} =[/tex] ∑ [tex]E_{i}[/tex] = [tex]E_{2->0}[/tex] + [tex]E_{-5->0}[/tex] + [tex]E_{-3->0}[/tex]
[tex]E_{2->0}[/tex] = k ×[tex]\frac{q_{+2} ^{2} }{d^{2} }[/tex] = 8.99×10⁹ N m²/ C₂× (2.0×10⁻⁶)/(8.2×10⁻² m)² = 2.674×10⁶ N / C + x
[tex]E_{-5->0}[/tex] = k ×[tex]\frac{q_{+2} ^{2} }{d^{2} }[/tex] = 8.99×10⁹ N m²/ C₂× (5.0×10⁻⁶)/(8.2×10⁻² m)² = 6.685×10⁶ N / C + y
[tex]E_{-3->0}[/tex] =k ×[tex]\frac{q_{+2} ^{2} }{d^{2} }[/tex] = 8.99×10⁹ N m²/ C₂× (3.0×10⁻⁶)/(8.2×10⁻² m)² = 4.011×10⁶ N / C + x
[tex]E_{Net} =[/tex] 2.674×10⁶ N / C + x + 6.685×10⁶ N / C + y + 4.011×10⁶ N / C + x =
= 6.685×10⁶ N / C + x+ 6.685×10⁶ N / C + y
Hence [tex]E_{Net}[/tex] = √(( 6.685×10⁶)²+(6.685×10⁶)²) = 7.796×10⁶ N / C
θ = tan ⁻¹ (6685008.923/6685000) = 45.0 °
Therefore [tex]E_{Net} =[/tex] 7.796×10⁶ N / C at 45.0 ° above the x axis
