In the figure below, Q₁ (-6 µC) by Q₂ (2 μC) are shown. Use k=9x10⁹ Nm²/c². Q1 O x=0 Q2 x=3cm x=4cm Using Part 1, the electric field due to Q₁ and Q₂ at x = 3 cm should have looked like this. Q₂ is closer to the point x=3 cm compared to Q₁1. Therefore, |E₂|> |E₁|. Both fields point in the negative x- direction. E₁ X E2 x=3cm (a) Calculate the electric field E, due to Q₁ at x = 3 cm. E₁: 6x107N/C (b) Calculate the electric field E₂ due to Q₂ at x = 3 cm. E₂: 18x107N/C (c) Calculate the net electric field at x = 3 cm due to Q₁ and Q₂. Enet 24x10' N/C Hint: Don't forget the sign of the electric fields.
In the figure below, Q₁ (-6 µC) by Q₂ (2 μC) are shown. Use k=9x10⁹ Nm²/c². Q1 O x=0 Q2 x=3cm x=4cm Using Part 1, the electric field due to Q₁ and Q₂ at x = 3 cm should have looked like this. Q₂ is closer to the point x=3 cm compared to Q₁1. Therefore, |E₂|> |E₁|. Both fields point in the negative x- direction. E₁ X E2 x=3cm (a) Calculate the electric field E, due to Q₁ at x = 3 cm. E₁: 6x107N/C (b) Calculate the electric field E₂ due to Q₂ at x = 3 cm. E₂: 18x107N/C (c) Calculate the net electric field at x = 3 cm due to Q₁ and Q₂. Enet 24x10' N/C Hint: Don't forget the sign of the electric fields.
College Physics
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Chapter1: Units, Trigonometry. And Vectors
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Transcribed Image Text:In the figure below, Q₁ (-6 µC) by Q₂ (2 μC) are shown.
Use k=9x10⁹ Nm²/C².
Q1
x=0
Q2
x=3cm x 4cm
Using Part 1, the electric field due to Q₁ and Q₂ at x = 3 cm should have looked like this. Q₂ is closer to
the point x=3 cm compared to Q₁. Therefore, |E2| > |E₁. Both fields point in the negative x-
direction.
E₁
X
E2
x=3cm
(a) Calculate the electric field E₁ due to Q₁ at x = 3 cm.
N/C
E₁: 6x107
(b) Calculate the electric field E₂ due to Q₂ at x = 3 cm.
E₂: 18x107 N/C
(c) Calculate the net electric field at x = 3 cm due to Q₁ and Q₂.
E.
N/C
net 24x107x
Hint: Don't forget the sign of the electric fields.
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