A `-4.0 mu C` charge is located 0.30 m to the left of a + `+ 6.0 mu C` charge. What is the magnitude and direction of the electrostatic force on the positive charge ?

Options

A. `2.4 N` to the right
B. `2.4 N` to the left
C. `4.8 N` to the right
D. `4.8 N` to the left

Answer

To solve the problem, we will use Coulomb's Law, which states that the electrostatic force \( F \) between two point charges is given by the formula:

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

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

### Step 1: Identify the charges and distance
We have:
- Charge \( q_1 = +6.0 \, \mu C = +6.0 \times 10^{-6} \, C \)
- Charge \( q_2 = -4.0 \, \mu C = -4.0 \times 10^{-6} \, C \)
- Distance \( r = 0.30 \, m \)

### Step 2: Substitute values into Coulomb's Law
Using the values in Coulomb's Law:

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

Substituting the values:

\[ F = 9 \times 10^9 \frac{|(6.0 \times 10^{-6}) \cdot (-4.0 \times 10^{-6})|}{(0.30)^2} \]

### Step 3: Calculate the product of the charges
Calculate \( |q_1 \cdot q_2| \):

\[ |q_1 \cdot q_2| = |(6.0 \times 10^{-6}) \cdot (-4.0 \times 10^{-6})| = 24.0 \times 10^{-12} \, C^2 \]

### Step 4: Calculate \( r^2 \)
Calculate \( r^2 \):

\[ r^2 = (0.30)^2 = 0.09 \, m^2 \]

### Step 5: Substitute into the equation
Now substitute back into the equation for \( F \):

\[ F = 9 \times 10^9 \frac{24.0 \times 10^{-12}}{0.09} \]

### Step 6: Calculate the force
Now perform the calculation:

\[ F = 9 \times 10^9 \cdot \frac{24.0 \times 10^{-12}}{0.09} = 9 \times 10^9 \cdot 2.6667 \times 10^{-10} = 2.4 \, N \]

### Step 7: Determine the direction of the force
Since the positive charge (+6.0 µC) is attracted to the negative charge (-4.0 µC), the direction of the force on the positive charge will be towards the left (towards the negative charge).

### Final Answer
The magnitude of the electrostatic force on the positive charge is \( 2.4 \, N \) and the direction is towards the left.

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