Respuesta :
Answer:
[tex]\vec{E} = (-0.383~\^x)~{\rm N/C}[/tex]
Explanation:
The electric field of a point charge can be calculated by the following formula:
[tex]\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r[/tex]
where r is the distance from the point source.
For the first particle, the magnitude of electric field is
[tex]E_1 = \frac{1}{4\pi\epsilon_0}\frac{8 \times 10^{-9}}{(16)^2 + (12\times 10^{-3})^2} = 0.282~N/C[/tex]
The x- and y-components of the E-field are
[tex]E_{1x} = E_1\cos(\theta_1) = E_1 \frac{16}{\sqrt{16^2 + (12\times 10^{-3})^2}} \sim 0.286~N/C\\E_{1y} = E_1\sin(\theta_1) = E_1\frac{12\times10^{-3}}{\sqrt{16^2 + (12\times 10^{-3})^2}} \sim 0[/tex]
The y-component of the electric field is negligible, because the point A is 12 mm away from the origin, whereas the charge is 16 m away form the origin. Since 16 m >> 12 mm, the y-component can be regarded as zero.
For the second particle,
[tex]E_2 = \frac{1}{4\pi\epsilon_0}\frac{6 \times 10^{-9}}{(9)^2 + (12\times 10^{-3})^2} = 0.669~N/C[/tex]
The x- and y-components of the E-field are
[tex]E_{2x} = E_2\cos(\theta_2) = E_2 \frac{9}{\sqrt{9^2 + (12\times 10^{-3})^2}} \sim 0.669~N/C\\E_{1y} = E_2\sin(\theta_2) = E_2\frac{9\times10^{-3}}{\sqrt{9^2 + (12\times 10^{-3})^2}} \sim 0[/tex]
Similarly, the y-component is negligible.
Finally, the total electric field can be written as
[tex]\vec{E} = \vec{E_1} + \vec{E_2}\\\vec{E} = (0.286 - 0.669)\^x\\\vec{E} = (-0.383~\^x)~ N/C[/tex]
The solution is ( 0.385 i ; 000 j) [N/C]
To get the electric field, due to a distribution of charges, it is necessary to calculate the electric field due to each one of the charges. The electric field is a vector therefore that condition must be taken into account in order to calculate the resulting field
- Calculating the electric field (E₁) in point A ( 0 mm ; 12 mm ) due to charge q₁ = 8 nC = 8×10⁻⁹ C
E₁ = K × q₁ )/ (d₁₂)² (1)
In that equation K = 8.98 × 10⁹ [N×m²/C²] , q₁ = 8×10⁻⁹ C and the distance is:
d₁₂ = √ (16)² + ( 0.012)² = √ 256 + 0.000144 = 16.0000044
d₁₂ = 16 m
It is necessary to point out that the value of the coordinate y for point A is extremely low when comparing with 16 m therefore we could anticipate the value of d₁₂ approximating to 16. ( 16 m >> 12 mm ) It is a moment to remark that the problem will be solved as point A where at the origin of the coordinates system. And for that situation, we will get only x components for the electric field.
Then by substitution in equation (1)
E₁ = 8.98 × 10⁹ [N×m²/C²] × 8×10⁻⁹ C / (16²) m²
E₁ = 0.2806 [N/C]
q₁ is a positive charge therefore any test (positive) charge will be rejected for the electric field due to q₁. Then E₁ will be in the direction of the negative x axis.
- Calculating the electric field (E₂) in point A ( 0 mm ; 12 mm ) due to charge q₂ = 6 nC = 6×10⁻⁹ C
Following the same procedure and considerations
E₂ = 8.98 × 10⁹ [N×m²/C²] × 6×10⁻⁹ C / (9²) m²
E₂ = 0.6651 [N/C]
E₂ direction is for the same reason explain when calculating E₁, in the direction of the positive x-axis
Then we neglecting the Ey components since the distance of 12 mm allow us to considered as point A was in the x-axis, as it was explained
Finally, E₁ and E₂ have both different directions we subtract them and the direction will be the direction of the bigger field
E₂ - E₁ = 0.6651 - 0.2806 = 0.3845 [N/C]
Et = 0.385 [N/C] in the direction of x-axis positive
