Point charges of 5.00*µ*C5.00µC and –3.00*µ*C–3.00µC are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive?

To solve this problem, we need to find the position where a third charge can be placed so that the net force on it is zero. We will consider both scenarios: when the charges are of opposite signs and when both are positive.

### Given:
- Charge 1 (\(q_1\)) = \(+5.00 \, \mu\text{C}\)
- Charge 2 (\(q_2\)) = \(-3.00 \, \mu\text{C}\)
- Distance between \(q_1\) and \(q_2\) = \(0.250 \, \text{m}\)

Let's denote the third charge as \(q_3\), and assume it's placed at a distance \(x\) from \(q_1\). The distance from \(q_3\) to \(q_2\) would then be \(0.250 - x\).

### (a) When the charges have opposite signs

#### Finding the position \(x\):

The net force on \(q_3\) will be zero when the magnitudes of the forces exerted by \(q_1\) and \(q_2\) on \(q_3\) are equal. Using Coulomb's Law:

\[
F_1 = F_2
\]

\[
k \frac{|q_1 \cdot q_3|}{x^2} = k \frac{|q_2 \cdot q_3|}{(0.250 - x)^2}
\]

Since \(k\) and \(q_3\) cancel out, we have:

\[
\frac{q_1}{x^2} = \frac{|q_2|}{(0.250 - x)^2}
\]

Substituting the values:

\[
\frac{5.00 \, \mu\text{C}}{x^2} = \frac{3.00 \, \mu\text{C}}{(0.250 - x)^2}
\]

Now, solve for \(x\).

### (b) When both charges are positive

If both charges are positive, the third charge can be placed outside the segment joining the two charges because the forces will repel each other.

We'll also need to solve for \(x\) in this case.

Let's compute the results for both scenarios.### (a) When the Charges Have Opposite Signs

The possible solutions for the position \(x\) where the net force on the third charge would be zero are:
- \( x = 0.141 \, \text{m} \) from the \(+5.00 \, \mu\text{C}\) charge (between the two charges).
- \( x = 1.109 \, \text{m} \) from the \(+5.00 \, \mu\text{C}\) charge (beyond the \( -3.00 \, \mu\text{C} \) charge on the right side).

However, only the position **between the two charges** (\( x = 0.141 \, \text{m} \)) is physically meaningful in this context because the other solution would place the third charge beyond the first charge, which is not typical for such problems.

### (b) When Both Charges Are Positive

The possible solutions for the position \(x\) are:
- **Left side of the \(+5.00 \, \mu\text{C}\) charge:** \( x = -1.109 \, \text{m} \) (which places the third charge 1.109 meters to the left of the \(+5.00 \, \mu\text{C}\) charge).
- **Right side of the \(+3.00 \, \mu\text{C}\) charge:** \( x = 0.141 \, \text{m} \) (which is the same position as in part (a), but now on the right side and outside the segment).

In this case, the third charge must be placed either to the **left of the \(+5.00 \, \mu\text{C}\) charge** (1.109 m away) or to the **right of the \(+3.00 \, \mu\text{C}\) charge** (0.141 m away).

Would you like any further details or have additional questions?

Here are five related questions to explore this topic further:
1. What would happen if the distance between the two charges was doubled?
2. How does the magnitude of the third charge affect the location where the net force is zero?
3. What if one of the charges was much larger than the other, say 10 times larger?
4. How would the problem change if the charges were placed in a medium with a dielectric constant other than 1?
5. Can the third charge be at an infinite distance and still experience zero net force?

**Tip:** When solving problems with forces acting in opposite directions, consider the relative magnitudes of the forces and how distance affects each.

Explore how to find the equilibrium position of a third charge between two point charges of 5.00 µC and -3.00 µC separated by 0.250 meters. Learn about Coulomb's Law and the conditions under which the net force on the third charge is zero. Suitable for high school students studying electrostatics.

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