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3 years ago

In the diagram, q1 = +6.6010^-9 C and q2 = +3.1010^-9 C. Find the magnitude of the total electric field at point P. pls help?

Physics

2 answers:

PilotLPTM [1.2K]3 years ago

8 0

A I just answer that question

kirza4 [7]3 years ago

3 0

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰ (angle with respect x-axis)

Explanation:

E(t) Electric Field is a vector, so we need to determine module and direction

E(t) = E(q₁) + E (q₂) Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂ respectively.

E(q₁) = K * q₁/ (d₁)² K = 9 *10⁹ [N*m²/C²] d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225 [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) = 9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9 tanα = 2,395 α = 67,5⁰

The angle of the vector electric field with the x-axis

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