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What is the strength of the electric field at the position indicated by the dot? (b=3.9 cm, a=1.37 cm, Q=14.5 nC, q=6.8 nC.)
What is the direction of the electric field at the position indicated by the dot? Give your answer as an angle measured ccw from the positive x-axis.

To solve the problem, we need to find the electric field at the point indicated by the black dot due to the three charges \(Q\), \(q\), and \(-q\).

### Step 1: Position of the charges
- The coordinates of the charges based on the given diagram:
  - \(Q\) is at the origin: \((0, 0)\)
  - \(-q\) is at \((a, 0)\)
  - \(q\) is at \((0, b)\)
  - The dot is at \((a, b)\)

### Step 2: Electric field contributions
The electric field due to a point charge \(q_i\) at a distance \(r_i\) from the point where we calculate the field is given by Coulomb's law:
\[
E = k_e \frac{|q_i|}{r_i^2}
\]
where \(k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\).

We need to calculate the electric field contributions from each charge and then find the net electric field at the point \((a, b)\).

1. **Electric field due to charge \(Q\) (at the origin):**
   - Distance from \(Q\) to the point \((a, b)\) is:
     \[
     r_Q = \sqrt{a^2 + b^2}
     \]
   - The electric field magnitude at \((a, b)\) is:
     \[
     E_Q = k_e \frac{Q}{r_Q^2} = k_e \frac{Q}{a^2 + b^2}
     \]
   - The direction of the field due to \(Q\) points away from \(Q\) because \(Q\) is positive. The angle of this field with the x-axis is:
     \[
     \theta_Q = \tan^{-1}\left(\frac{b}{a}\right)
     \]

2. **Electric field due to charge \(-q\) (at \((a, 0)\)):**
   - Distance from \(-q\) to the point \((a, b)\) is \(b\) (vertical distance).
   - The electric field magnitude is:
     \[
     E_{-q} = k_e \frac{q}{b^2}
     \]
   - The field due to \(-q\) points towards the charge (since it is negative), so the direction is downward, along the negative \(y\)-axis.

3. **Electric field due to charge \(q\) (at \((0, b)\)):**
   - Distance from \(q\) to the point \((a, b)\) is \(a\) (horizontal distance).
   - The electric field magnitude is:
     \[
     E_q = k_e \frac{q}{a^2}
     \]
   - The field due to \(q\) points away from \(q\), so the direction is to the right, along the positive \(x\)-axis.

### Step 3: Summing the electric fields
To find the total electric field, we sum the contributions from each charge as vectors.

1. **X-component of the total electric field**:
   \[
   E_x = E_q - E_Q \cos(\theta_Q)
   \]
   (since the \(E_Q\) field has a negative \(x\)-component).

2. **Y-component of the total electric field**:
   \[
   E_y = E_{-q} - E_Q \sin(\theta_Q)
   \]

### Step 4: Calculate values

I'll now calculate the total electric field and the angle measured counterclockwise from the positive \(x\)-axis.

Let me do these calculations.The strength of the electric field at the position indicated by the dot is approximately \(3.02 \times 10^5 \, \text{N/C}\).

The direction of the electric field is approximately \(-6.04^\circ\), measured counterclockwise from the positive \(x\)-axis (indicating a slight downward angle).

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How does the distance between the charges affect the strength of the electric field?
2. What would happen to the electric field if charge \(Q\) was negative instead of positive?
3. How does the direction of the electric field change when the charge at the origin is shifted?
4. What is the net electric field at a point midway between two charges of opposite signs?
5. How would the field change if all charges were doubled while keeping the distances constant?

**Tip**: In electrostatics, electric field vectors due to multiple charges can be combined using vector addition. It's important to consider both magnitude and direction for accurate results.

What is the strength of the electric field at the position indicated by the dot? (b=3.9 cm, a=1.37 cm, Q=14.5 nC, q=6.8 nC.) What is the direction of the electric field at the position indicated by the dot? Give your answer as an angle measured ccw from the positive x-axis.

Calculate the strength and direction of the electric field at a given point due to multiple charges. Includes step-by-step calculations using Coulomb's law, vector addition, and electric field components. Problem parameters: b = 3.9 cm, a = 1.37 cm, Q = 14.5 nC, q = 6.8 nC.

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