A test charge of  +2μC is placed halfway between a charge of +6μC and another of  +4μC
separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the  +6μC charge)?

To solve this problem, we need to use Coulomb's Law, which gives the force between two point charges. The formula for Coulomb's Law is:

\[
F = k \frac{|q_1 q_2|}{r^2}
\]

where:
- \( F \) is the magnitude of the force between the charges,
- \( k \) is Coulomb's constant \((8.99 \times 10^9 \, \text{N m}^2/\text{C}^2)\),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and
- \( r \) is the distance between the charges.

### Given:
- Test charge (\(q_t\)) = \(+2 \, \mu\text{C}\)
- Charge 1 (\(q_1\)) = \(+6 \, \mu\text{C}\)
- Charge 2 (\(q_2\)) = \(+4 \, \mu\text{C}\)
- Distance between \(q_1\) and \(q_2\) = 10 cm = 0.1 m

Since the test charge is placed halfway between the two charges, the distance from the test charge to each of the other charges is \( r = 0.05 \, \text{m} \).

### (a) Magnitude of the Force on the Test Charge

Let's calculate the force on the test charge due to each of the other charges.

1. **Force due to \(q_1\) (6μC):**
   \[
   F_1 = k \frac{|q_1 q_t|}{r^2} = 8.99 \times 10^9 \, \frac{|6 \times 10^{-6} \times 2 \times 10^{-6}|}{(0.05)^2} \, \text{N}
   \]

2. **Force due to \(q_2\) (4μC):**
   \[
   F_2 = k \frac{|q_2 q_t|}{r^2} = 8.99 \times 10^9 \, \frac{|4 \times 10^{-6} \times 2 \times 10^{-6}|}{(0.05)^2} \, \text{N}
   \]

Now, let's compute these values.

### (b) Direction of the Force

Since both \(q_1\) and \(q_2\) are positive, the forces on the test charge due to each will be repulsive, meaning they will push the test charge away from themselves.

- The force due to \(q_1\) will push the test charge away from \(q_1\), which means toward \(q_2\).
- The force due to \(q_2\) will push the test charge away from \(q_2\), which means toward \(q_1\).

Since the charges are unequal, the forces will not cancel out. The net force will be in the direction of the weaker force.

Let's calculate the magnitudes to determine the net force and its direction.### (a) Magnitude of the Force on the Test Charge

- The force on the test charge due to the \(+6 \, \mu\text{C}\) charge (\(q_1\)) is \( F_1 = 43.15 \, \text{N} \).
- The force on the test charge due to the \(+4 \, \mu\text{C}\) charge (\(q_2\)) is \( F_2 = 28.77 \, \text{N} \).

### (b) Direction of the Force

Since \( F_1 > F_2 \), the net force on the test charge will be in the direction of \( F_1 \), which is away from the \(+6 \, \mu\text{C}\) charge and toward the \(+4 \, \mu\text{C}\) charge.

### Final Answer:
- **Magnitude of the Net Force:** The net force on the test charge is the difference between \( F_1 \) and \( F_2 \):

\[
F_{\text{net}} = F_1 - F_2 = 43.15 \, \text{N} - 28.77 \, \text{N} = 14.38 \, \text{N}
\]

- **Direction of the Force:** The force is directed **toward the \( +4 \, \mu\text{C} \) charge**.

Would you like further details or have any other questions?

Here are five related questions to expand on this topic:
1. How would the net force change if the test charge was placed closer to the \(+6 \, \mu\text{C}\) charge?
2. What would happen if the \(+4 \, \mu\text{C}\) charge was replaced with a \( -4 \, \mu\text{C} \) charge?
3. How does the force on the test charge change if the charges are moved farther apart?
4. What is the electric field at the position of the test charge due to the two charges?
5. How would the situation change if the test charge was negative instead of positive?

**Tip:** Remember that the direction of the force in Coulomb's law depends on the sign of the charges involved—like charges repel, unlike charges attract.

Learn how to calculate the magnitude and direction of the force on a test charge placed between two point charges (+6μC and +4μC) separated by 10 cm. Understand the application of Coulomb's Law and solve for the net force using detailed step-by-step calculations. Suitable for undergraduate students studying electricity and magnetism.

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