Respuesta :
Answer:
A)Magnitude of the net electric field at the origin due q₁ and q₂
E₀= 131.6 N/C
B) Direction of the net electric field at the origin due q₁ and q₂
β=3.91°
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻⁹ C
Data
q₁ = -4 nC = -4*10⁻⁹ C
q₂ = +6nC =+ 6*10⁻⁹ C
k = 9*10⁹ N*m²/C²
[tex]d_{1} =\sqrt{0.6^{2}+0.8^{2} } =1 m[/tex]
d₂ = 0.6 m
Graphic attached
The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:
E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.
E₂: Total field at point x=0 , y=0 due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.
Calculation of the electric field at the origin of x-y coordinates due to the charge q₁
E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C
Components (x-y) of the field due to q1:
E₁x=E₁*cosα = 36*(0.6/1) = 36*(0.6) = 21.6 N/C
E₁y=E₁*sinα = 36*(0.8/1) = 36*(0.8) =28.8 N/C
Calculation of the electric field at the origin of x-y coordinates due to the charge q₂
E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C
Calculation of the electric field components at the origin of x-y coordinates
E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)
E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C
E₀y= E₁y = 28.8 N/C
A )Magnitude of the net electric field at the origin due q₁ and q₂
[tex]E_{o} =\sqrt{128.4^{2}+28.8^{2} }[/tex]
E₀= 131.6 N/C
B) Direction of the net electric field at the origin due q₁ and q₂
[tex]\beta =tanx{-1} \frac{E_{oy} }{Eox}[/tex]
[tex]\beta =tan^{-1} \frac{28.8}{128.4}[/tex]
β=3.91°
The magnitude and the direction of the net electric force due the point charge q1 and q2 is,
- Thus, the magnitude of the net electric field at the origin due to these two point charges is 878.8 N/C.
- b) The direction of the net electric field at the origin due to these two point charges is at the 55 degree of angle.
What is electric charge?
When a body is places inside a electric field, it fill a force due to the electric field. This force felt by body is called the electric charge. It can be either positive or negative.
The charge of point one is of magnitude -4.00 nC and the charge of point two is of magnitude +6.00 nC.
The point one is placed along the at x = 0.60 m, y = 0.80 m and point 2 is placed at the point x = 0.60 m , y = 0. The value of point one is x cm and the value of point two is 0.4 cm.
- A. The magnitude of the net electric field at the origin due to these two point charges.
The vector form the of the electric force, due to point A can be given as,
[tex]E_{A}=\dfrac{9\times10^9\times6\times10^{-9}}{0.6^2}\\E_{A}=900\rm N/C[/tex]
The vector form the of the electric force, due to point B can be given as,
[tex]E_B=\dfrac{9\times10^9\times4\times10^{-9}}{\sqrt{0.6^2+0.8^2}}\\E_B=36\rm N/C[/tex]
The angle is given by,
[tex]\theta=180-\tan^{-1}\dfrac{0.8}{0.6}\\\theta=127^o[/tex]
The net electric force due to charge 1 and 2 can be given as,'
[tex]E_{net}=\sqrt{900^2+36^2+2(900)(36)\cos(127^o)}\\E_{net}=878.8\rm N/C[/tex]
Thus, the magnitude of the net electric field at the origin due to these two point charges is 878.8 N/C.
- B. The direction of the net electric field at the origin due to these two point charges.
The angle for the direction can be given as,
[tex]\tan \alpha=\dfrac{b\sin \theta}{a+b\cos\theta}\\\tan \alpha=\dfrac{900\sin \dfrac{4}{5}}{36+900\cos\dfrac{-3}{5}}\\\alpha=55^o[/tex]
Thus, the direction of the net electric field at the origin due to these two point charges is at the 55 degree of angle.
The magnitude and the direction of the net electric force due the point charge q1 and q2 is,
- Thus, the magnitude of the net electric field at the origin due to these two point charges is 878.8 N/C.
- b) The direction of the net electric field at the origin due to these two point charges is at the 55 degree of angle.
Learn more about the electric charge here;
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