(a) Calculate the electric field E, due to Q₁ at x = 3 cm. E₁: N/C (b) Calculate the electric field E₂ due to Q₂ at x = 3 cm. E₂: N/C (c) Calculate the net electric field at x = 3 cm due to Q₁ and Q₂. Enet ✔N/C
(a) Calculate the electric field E, due to Q₁ at x = 3 cm. E₁: N/C (b) Calculate the electric field E₂ due to Q₂ at x = 3 cm. E₂: N/C (c) Calculate the net electric field at x = 3 cm due to Q₁ and Q₂. Enet ✔N/C
College Physics
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Chapter1: Units, Trigonometry. And Vectors
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Transcribed Image Text:1. In the figure below, Q₁ (-6 µC) by Q₂ (2 µC) are shown.
Use k =9x10⁹ Nm²/C².
Q1
x=0
x-3cm x 4cm
X
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
E₂ x=3cm
(a) Calculate the electric field E, due to Q₁ at x = 3 cm.
1
E₁:
✔N/C
(b) Calculate the electric field E₂ due to Q₂ at x = 3 cm.
E₂:
|N/C
(c) Calculate the net electric field at x = 3 cm due to Q₁ and Q₂.
N/C
E net
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Transcribed Image Text:1. In the figure below, Q₁ (-6 μC) by Q₂ (2 μC) are shown.
Use k=9x10° 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.
EL
E₂ x 3cm
(a) Calculate the electric field E, due to Q₁ at x = 3 cm.
E₁: 60x10 N/C
(b) Calculate the electric field E₂ due to Q₂ at x = 3 cm.
E₂: 180x10N/C
(c) Calculate the net electric field at x = 3 cm due to Q₁ and Q₂.
Enet 240x10N/C
!
No, that's not the correct answer.
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