Answer:
Ex= -23.8 N/C Ey = 74.3 N/C
Explanation:
As the electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.
So, we can first the field due to q1, as follows:
Due to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:
E₁ = k*(4.55 nC) / r₁²
If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:
E₁ₓ = 0 E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)
For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.
It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:
r₂² = 1²m² + (0.6)²m² = 1.36 m²
The magnitude of the electric field due to q2 can be found as follows:
E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)
Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.
In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:
E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ
the cosine of θ, is just, by definition, the opposite of x/r₂:
⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855
By the same token, sin θ can be obtained as follows:
sin θ = - (0.6 m / 1.17 m) = -0.513
⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)
⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)
The total x and y components due to both charges are just the sum of the components of Ex and Ey:
Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C
From (1) and (4), we can get Ey:
Ey = E₁y + E₂y = 88.6 N/C + (-14.3 N/C) =74.3 N/C
