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The vector characteristics of the electric field allow to find the result for the electric field at the point of interest is;
- The total electric field is [tex]E= 14.1 ( - \hat i + \hat j) \frac{N}{C}[/tex]
Given parameters.
1) Electric charges:
- q₁ = 20 nC = 20 10⁻⁹ C
- q₂ = -20 nC = -20 10⁻⁹ C
2)Charge positions
- r₁ = (1,0,0) = 1 [tex]\hat i[/tex] m
- r₂ = (0,1,0) = 1 [tex]\hat j[/tex] m
3) Points of interest r = (0,0,1) = 1 [tex]\hat k[/tex]
To find.
the electric field at point r.
The intensity of the electric field is given by the ratio of the electric force to the positive test charge at the point of interest.
[tex]E= k \frac{q}{r^2 }[/tex]
where E is the electric field, k the constant of Coulomb, q the charge and r the distance.
In the attached we see a diagram of the test charges and the distance to the point of interest on the z axis.
Vector addition electric field
The total field is the vector addition of the fields created by each charge
[tex]E_{total} = E_1 +E_2[/tex]
Let's look for the components of each electric field.
Field created by charge q₁
tan θ = [tex]\frac{z}{x}[/tex]
tan θ = 1/1 = 1
tea = 45º
x-axis
cos 45 = [tex]\frac{E_{1x}}{E_1}[/tex]
z axis
sin 45= [tex]\frac{E_{1z}}{E_1}[/tex]
E₁ₓ = E₁ cos 45
[tex]E_{1z}[/tex] = E₁ sin 45
Field created by load 2
y-axis
cos 45 = [tex]\frac{E_{2y}}{E_2}[/tex]
z- axis
sin 45 = [tex]\frac{E_{2z}}{E_2}[/tex]
[tex]E_{2y}[/tex] = E2 cos 45
[tex]E_{2z}[/tex] = E2 sin 45
We look for the components of the total electric field, where the signs are taken from the direction of the vectors in the attached.
[tex]E_{total} = - E_{1x} \hat i + E_{2y} \hat j + ( E_{1z} - E_{2z} ) \hat k[/tex]
Let's substitute.
[tex]E_{total } = - E_1 cos 45 \hat i + E_2 cos 45 \hat j + (E_1 sin 45 - E_2 sin 45) \hat k[/tex]
Calculate the distance and each electric field.
Let's find the distance for each charge to the test point using the Pythagorean Theorem.
r₁₃² = x² + z²
r₁₃² = 1 + 1
r₁₃² = 2
r₂₃² = y² + z²
r₂₃² = 1² + 1²
r₂₃² = 2
Let's look for the magnitud of the electric fields.
Charge q₁
[tex]E_1 = k \frac{q_1}{r_{13}^2}[/tex]
E₁ = [tex]9 \ 10^9 \frac{20 \ 10^{-9}}{2}[/tex]
E₁ = 20 N / m
Charge q₂
[tex]E_2 = k \frac{q_2}{r_{23}^2}[/tex]
E₂ = [tex]9 \ 10^9 \frac{20 \ 10^{-9}}{2}[/tex]
E₂ = 20 C / N
Let's substitute in the expression of the total eletric field.
[tex]E_{total} = -20 \ cos45 \ \hat i + 20 \ cos 45\ \hat j + ( 20 -20) sin 45 \ \hat k[/tex]
[tex]E_{total} = 14.1 ( - \hat i + \hat j) \ \frac{N}{C}[/tex]
In conclusion using the vector characteristics of the electric field we can find the result for the electric field at the point of interest is;
- The total electric field is [tex]E = 14.1 ( - \hat i + \hat j ) \frac{N}{C}[/tex]
Learn more about the electric field here: brainly.com/question/14372859
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