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Let's break down the problem:

### Given:
- Two charges:
  - \(q_1 = -2.5 \, \mu C\)
  - \(q_2 = 6 \, \mu C\)
- Separation between the charges = \(50 \, \text{cm} = 0.5 \, \text{m}\)

### Part (a): At what spot along a line through the charges is the net electric field zero?

We need to find the point along the line where the electric fields from both charges cancel each other out. 

#### Step 1: Electric Field due to a point charge

The electric field due to a point charge \(q\) at a distance \(r\) is given by:
\[
E = \frac{k|q|}{r^2}
\]
where \(k\) is Coulomb’s constant \((k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2)\).

#### Step 2: Setting up the equation

Let the point where the electric field is zero be a distance \(x\) from the 6 \(\mu C\) charge. The total distance between the two charges is 0.5 m, so the distance from the \( -2.5 \, \mu C\) charge would be \((0.5 - x)\).

For the net electric field to be zero, the magnitudes of the electric fields due to the two charges at this point must be equal:
\[
\frac{k|q_1|}{(0.5 - x)^2} = \frac{k|q_2|}{x^2}
\]

Since \(k\) cancels out, we are left with:
\[
\frac{|q_1|}{(0.5 - x)^2} = \frac{|q_2|}{x^2}
\]
Substitute the values of \(q_1\) and \(q_2\):
\[
\frac{2.5}{(0.5 - x)^2} = \frac{6}{x^2}
\]

#### Step 3: Solving for \(x\)

Cross-multiply to solve for \(x\):
\[
2.5x^2 = 6(0.5 - x)^2
\]

Expand both sides:
\[
2.5x^2 = 6(0.25 - x + x^2)
\]
\[
2.5x^2 = 6(0.25) - 6x + 6x^2
\]
\[
2.5x^2 = 1.5 - 6x + 6x^2
\]

Move all terms to one side:
\[
0 = 6x^2 - 2.5x^2 - 6x + 1.5
\]
\[
0 = 3.5x^2 - 6x + 1.5
\]

This is a quadratic equation:
\[
3.5x^2 - 6x + 1.5 = 0
\]

We can solve this using the quadratic formula:
\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3.5)(1.5)}}{2(3.5)}
\]
\[
x = \frac{6 \pm \sqrt{36 - 21}}{7}
\]
\[
x = \frac{6 \pm \sqrt{15}}{7}
\]
\[
x \approx \frac{6 \pm 3.87}{7}
\]

So:
\[
x_1 \approx \frac{6 + 3.87}{7} \approx 1.41 \, \text{m} \quad (\text{Not possible, since } x > 0.5 \, \text{m})
\]
\[
x_2 \approx \frac{6 - 3.87}{7} \approx 0.30 \, \text{m}
\]

Therefore, the point where the electric field is zero is approximately 0.30 m from the 6 \(\mu C\) charge.

### Part (b): Force on a +14 \(\mu C\) charge at this point

Once we have the point where the electric field is zero, placing any charge there would result in zero net electric field acting on it. Since electric field and force are related by:
\[
F = qE
\]
If the net electric field is zero, then the force on the charge will also be zero.

Thus, the force on a +14 \(\mu C\) charge placed at this spot would be **zero**.

---

Would you like any more details or clarification?

### 5 Related Questions:
1. How do we calculate the electric field for multiple charges in different geometries?
2. What happens to the electric field at points far away from the charges?
3. How does the direction of electric field vary for positive and negative charges?
4. Can the electric field be zero between two like charges?
5. How do we find the electric potential at a point between two charges?

### Tip:
When solving for points where electric fields cancel, always remember that the distances must be measured from the appropriate charge, and the ratio of magnitudes of the charges plays a critical role in determining the point.

Two charges, -2.5 μC and 6 μC, are fixed in place and separated by 50 cm. (a) At what spot along a line through the charges is the net electric field zero? Locate this point relative to the positive charge. (b) What would be the force on a charge of +14 μC placed at this spot?

Solve a physics problem involving two charges of -2.5 μC and 6 μC separated by 50 cm. Learn how to find the point where the electric field is zero using Coulomb's law and quadratic equations, and calculate the force on a +14 μC charge at this point.

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