HURRY ITS TIMED
2 answers:
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Answer:
Ex= -17.1 N/C
Ey = +26.9 N/C
Explanation:
We apply formula of electric field:
Ep=k*q/d²
Ep: Electric field at point ( N/C)
q: Electric charge (C)
k: coulomb constant (N.m²/C²)
d: distance from charge q to point P (m)
In the attached graph we observe the directions of the electric field at P(0,0) due to q1 and q2
Calculation of the field at point P due to the load q₁
E₁=k*q₁/d₁² = 9*10⁹*4.25*10⁻⁹/1.080²= 32.8 N/C : Magnitude of E1
Direction of E₁ :Because the charge q₁ is negative the field enters the charge (+ y)
Calculation of the field at point P due to the load q₂
d₂=1.375 m
E₂=k*q₂/d₂² = 9*10⁹*3.80*10⁻⁹/ 1.375² = 18.09 N/C Magnitude of E₂
Direction of E₂ :Because the charge q₂ is positive the field leaves the charge in direction of angle β
, then,E₂ tiene componentes x-y en P.
E₂x=-E₂cos β= -18.09*(1.3/1.375)= -17.1 N/C
E₂y=-E₂sin β= -18.09*(0.45/1.375)= -5.9 N/C
Calculation of the electric field at point P located at the origin(0,0)
Ex=E₂x= -17.1 N/C
Ey=E₁y+E₂y =32.8 N/C -5.9 N/C = 26.9 N/C
