the concept of vector electric field, the potential electric as a scalar we can find the total electric field, the electric potential and the force on a charge at a given point, the answers are
b) E = (1.61 i ^ + 5.92 j ^) 10¹⁴ N/C
E = 6.135 10¹⁴ N /C, θ = 15.2º
c) V = - 1.3 109 V
d) F = 1.23 10⁹ N, θ = 15.2º
Given parameters
The charges q₁ = + 3 10⁻⁶ C and q₂ = -5 10⁻⁶C and q₃ = +2 10⁻⁶ C The distance between charges 1 and 2 x = 5 10⁻⁶ m The distance between load 1 and the test point y = 10 10⁻⁶ m
to find
a) system scheme
b) the electric field at the test point
c) The electric potential at the test point
d) The force on q₃ at the test point
a) In the adjoint we can see a diagram of the charges and the distances in the exercise there are also the electric field vectors, the force vectors have the same direction as the electric field vectors.
b) The electric field is a vector magnitude that for point charges is given by
E = k [tex]\frac{q_i}{r^2}[/tex]
Where E is the electric field, k is the Coulomb constant (k=8.99 10⁹N m²/C²), q the charge and r the distance between the charge and the positive test charge
The electric field at test point P is the vector sum of the fields created by each
E_{total} = E₁ + E₂
E₁ = k q₁ / y²
E₂ = k q₂ / r²
bold indicates vectors
let's find the distances
distance from q₁ to the test point
r = y = 10 10⁻⁶ m
distance from q₂ to test point, we use Pythagoras' theerema
r² = x² + y²
r² = (5² + 10² )10⁻¹²
r² = 125 10⁻¹² m²
r = 11.18 10⁻⁶ m
we look for every electric field
E₁ = 9 10⁹ [tex]\frac{3 \ 10-6 }{(10 \ 10-6)^2 }[/tex]
E₁ = 27 10¹³ N / C
E₂ = 9 10⁹ 5 10⁻⁶ / 125 10⁻¹²
E₂ = 3.6 10¹⁴ N / C
One of the easiest way to shorten the total field is by using the sum of the components of each field
E_{total} = Eₓ i ^ + E_y j ^
Eₓ = E₂ₓ
E_y = E₁ + E_{2y}
use trigonometry to find the direction of the E₂ field
tan θ = y / x
θ = tan⁻¹ [tex]\frac{y}{x}[/tex]
θ = tan⁻¹ (10/5)
θ = 63.4º
cos 63.4 = [tex]\frac{E_{2x}}{E_2}[/tex]
sin 63.4 = [tex]\frac{E_{2y}}{E_2}[/tex]
E₂ₓ = E2 cos 63.4
E_{2y} = E2 sin 63.4
E₂ₓ = 3.6 10¹⁴ cos 63.4 = 1.61 10¹⁴ N / C
E_{2y} = 3.6 10¹⁴ sin 63.4 = 3.22 10¹⁴ N / C
The total field is the sum of the component of the electric field
Eₓ = E₂ₓ = 1.61 10¹⁴ N / C
E_y = E₁ + E_{2y} = 2.7 10¹⁴ + 3.22 10¹⁴
E_y = 5.92 10¹⁴ N / C
E = (1.61 i ^ + 5.92 j ^) 10¹⁴ N / C
c) the electric potential is a scalar quantity given by
V = k ∑ [tex]\frac{q_i}{r}[/tex]
for this case
V = k ( [tex]\frac{q_1}{y} + \frac{q_2}{r}[/tex])
V = 9 10⁹ ( [tex]\frac{3}{10} - \frac{5}{11.18 }[/tex] )
V = 9 10⁹ (0.3 - 0.447)
V = - 1.3 10⁹ V
d) The electric force is given by the relation
F = q₀ E
where F is the force, qo the test charge and E the total electric field
Let's find the modulus and direction of the electric field using the Pythagorean theorem
E = [tex]\sqrt{E_x^2 +E_y^2 }[/tex]
E = [tex]\sqrt{1.61^2 + 5.92^2 }[/tex] 10¹⁴
E = 6.135 10¹⁴ N / c
the direction is found with trigonometry
tan θ = [tex]\frac{E_y}{E_x}[/tex]
θ = tan⁻¹ [tex]\frac{E_y}{E_x}[/tex]
θ = tan⁻¹ [tex]\frac{1.61}{5.92}[/tex]
θ = 15.2º
the force is
F = 2 10⁻⁶ 6.135 10¹⁴
F = 1.23 10⁹ N
this force has the same direction of the electric field θ = 15,2º measured counterclockwise from the x axis
In conclusion with the concept of vector electric field, the potential electric as a scalar we can find the total electro field, the electric potential and the force on a charge at a given point, the answers are
b) E = (1.61 i ^ + 5.92 j ^) 10¹⁴
E = 6.135 10¹⁴ N /C, θ = 15.2º
c) V = - 1.3 10⁹ V
d) F = 1.23 10⁹ N, θ = 15.2º
learn more about electric field here:
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