 In the coordinate system shown in the figure, particle 1 with charge q1 = 13.5 μC is located at coordinates (x1,y1) = (-3, 0) m and particle 2 with charge q2 = -8.5 μC is located at coordinates (x2,y2) = (1, 0) m.   Part (b)
Solve for the numerical value of x0. 

To solve for the numerical value of \( x_0 \), we need to understand the context of part (a) of the problem, which is typically about finding the point along the x-axis where the net electric field or potential is zero due to the charges \( q_1 \) and \( q_2 \).

### Problem Restatement:
- **Charge \( q_1 \)**: \( 13.5 \, \mu C \) located at \( (-3, 0) \) m.
- **Charge \( q_2 \)**: \( -8.5 \, \mu C \) located at \( (1, 0) \) m.

### Approach to Find \( x_0 \):
We want to find the point \( x_0 \) on the x-axis where the net electric field due to both charges is zero. The electric field due to a point charge \( q \) at a distance \( r \) is given by:

\[
E = \frac{k \cdot |q|}{r^2}
\]

Where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)). The direction of the electric field depends on the sign of the charge.

1. **For \( q_1 = 13.5 \, \mu C \) at \( x = -3 \)**:
   - Electric field \( E_1 \) at a point \( x_0 \) is:
   \[
   E_1 = \frac{k \cdot 13.5 \times 10^{-6}}{(x_0 + 3)^2}
   \]
   This field points away from \( q_1 \) since \( q_1 \) is positive.

2. **For \( q_2 = -8.5 \, \mu C \) at \( x = 1 \)**:
   - Electric field \( E_2 \) at a point \( x_0 \) is:
   \[
   E_2 = \frac{k \cdot 8.5 \times 10^{-6}}{(x_0 - 1)^2}
   \]
   This field points towards \( q_2 \) since \( q_2 \) is negative.

3. **Setting \( E_1 = E_2 \) for Net Electric Field to be Zero**:
   \[
   \frac{k \cdot 13.5 \times 10^{-6}}{(x_0 + 3)^2} = \frac{k \cdot 8.5 \times 10^{-6}}{(x_0 - 1)^2}
   \]

4. **Simplifying**:
   \[
   \frac{13.5}{(x_0 + 3)^2} = \frac{8.5}{(x_0 - 1)^2}
   \]
   \[
   13.5 \cdot (x_0 - 1)^2 = 8.5 \cdot (x_0 + 3)^2
   \]

5. **Solve for \( x_0 \)**:
   Expanding and solving this equation will yield the numerical value of \( x_0 \).

Let's perform the calculation.The solutions for \( x_0 \) are approximately:

1. \( x_0 = -0.77 \) meters
2. \( x_0 = 16.37 \) meters

### Interpretation:
- The first value \( x_0 = -0.77 \) is between the two charges, which is typically where the electric field would cancel out.
- The second value \( x_0 = 16.37 \) is far to the right, outside the range between the charges. This is less likely to be the point of interest unless specified.

Usually, the problem is asking for the point between the charges, so \( x_0 = -0.77 \) m would be the most relevant solution.

Do you want more details or have any other questions? Here are 5 related questions to consider:

1. How do the signs of the charges affect the direction of the electric fields?
2. Can we determine the electric potential at \( x_0 \)?
3. How would the point \( x_0 \) change if both charges were positive?
4. What would happen if the magnitudes of \( q_1 \) and \( q_2 \) were equal?
5. How do we find the point of zero electric potential between the charges?

**Tip:** To solve such problems more efficiently, always start by identifying the nature of the charges and their relative positions.

In the coordinate system shown in the figure, particle 1 with charge q1 = 13.5 μC is located at coordinates (x1,y1) = (-3, 0) m and particle 2 with charge q2 = -8.5 μC is located at coordinates (x2,y2) = (1, 0) m. Solve for the numerical value of x0.

Learn how to solve for the point x0 along the x-axis where the electric fields from two charges, q1 = 13.5 μC and q2 = -8.5 μC, cancel each other out. This solution involves electrostatic principles and algebra, suitable for students in Grades 11-12.

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