Three bees are oriented as shown in the figure. B1 has +17 µC of charge, B2 has −5 µC of charge, and B3 has +26 µC of charge.
B1 to b2 5 cm
B2 to b3 10 com

Bee number 3 is stationed at the observation location for this problem, and we want to find the net electric field at B3. We'll do this in a few steps.
(a) What are the x and y components of the electric field

at the observation location (Bee number 3) due to B1? Remember that the components can be positive or negative depending on their directions along the x or y axis.
E1x =
E1y =

(b) What are the x and y components of the electric field

at the observation location (Bee number 3) due to B2? Again, remember that the components can be positive or negative depending on their directions along the x or y axis.
E2x = N/C
E2y =

(c) What are the magnitude and direction of the net electric field at the observation location where B3 is resting?

magnitude

N/C

direction

° counterclockwise from the +x-axis

(d) What is the magnitude of the force on B3 due to this net electric field?

The x and y components of the electric field due to B₁ are equal to 6.5 × 10⁶ N/C and -4.5 × 10⁶ N/C respectively.
The x and y components of the electric field due to B₂ are equal to -4.5 × 10⁶ N/C and 0 N/C respectively.
The magnitude of the net electric field at B₃ is equal to 6.04 × 10⁶ N/C.
The direction of the net electric field at the observation location is equal to 49.76°.
The magnitude of the force on B₃ due to this net electric field is equal to 157.04 Newton.

Given the following data:

Charge of B₁ = +17 µC to C = 17 × 10⁻⁶ C.

Charge of B₂ = -5 µC to C = -5 × 10⁻⁶ C.

Charge of B₃ = +26 µC to C = 26 × 10⁻⁶ C.

Radius of B₁ to B₂ = 5 cm to m = 0.05 meter.

Radius of B₂ to B₃ = 10 cm to m = 0.1 meter.

How to determine the x and y components of the electric field?

First of all, we would calculate the electric field due to B₁ to B₃ and the electric field due to B₂ to B₃ respectively.

The electric field due to B₁ to B₃ is given by:

E₁ = kq₁/r₁²

E₁ = (9 × 10⁹ × 17 × 10⁻⁶)/(0.05² + 0.1²)

E₁ = 153 × 10³/0.0125

Electric field, E₁ = 12.24 × 10⁶ N/C.

The electric field due to B₂ to B₃ is given by:

E₂ = kq₂/r₂²

E₂ = (9 × 10⁹ × 5 × 10⁻⁶)/(0.1²)

E₂ = 45 × 10³/0.01

Electric field, E₂ = 4.5 × 10⁶ N/C.

Also, the magnitude of the angle formed is given by tan trigonometry:

Tanθ = 5/10

Tanθ = 0.5

θ = tan⁻¹(0.5)

θ = 26.56°.

Next, we would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁x = E₁cosθ - E₂

E₁x = 12.24 × 10⁶ × cos(26.56) - 4.5 × 10⁶

E₁x = 10.95 × 10⁶ - 4.5 × 10⁶

E₁x = 6.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁y = -E₁sinθ

E₁y = -12.24 × 10⁶ × sin(26.56)

E₁y = -12.24 × 10⁶ × 0.4471

E₁y = -5.5 × 10⁶ N/C.

Also, the magnitude of the electric field is given by:

Exy = √(E₁x² + E₁y²)

Exy = √(6.5 × 10⁶)² + (-5.5 × 10⁶)²)

Exy = √4.225 × 10¹³ + 3.025 × 10¹³)

Exy = √(7.25 × 10¹³)

Exy = 8.5 × 10⁶ N/C.

Part B.

We would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₂:

E₂x = -E₂sinθ

E₂x = -4.5 × 10⁶ × sin(90)

E₂x = -4.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₂y = E₂cosθ

E₂y = -4.5 × 10⁶ × cos(90)

E₂y = 0 N/C.

How to calculate the net electric field at the observation location?

The magnitude of the net electric field at B₃ is given by:

Eₙ = √(E₁x + E₂x)² + E₁y²)

Eₙ = √(6.5 × 10⁶ + (-4.5 × 10⁶))² + (-5.5 × 10⁶)²)

Eₙ = √(2.5 × 10⁶)² + (3.025 × 10¹³)

Eₙ = √(6.25 × 10¹²) + (3.025 × 10¹³)

Eₙ = √(3.65 × 10¹³)

Eₙ = 6.04 × 10⁶ N/C.

For the direction, we have:

Tanθ = x/y

Tanθ = 6.5/-5.5

Tanθ = -1.1818

θ = tan⁻¹(-1.1818)

θ = 49.76°.

Part C.

The magnitude of the force on B₃ due to this net electric field is given by:

F = B₃ × Eₙ

F = 26 × 10⁻⁶ × 6.04 × 10⁶

F = 157.04 Newton.

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