How much work is required to carry a `6muC` charge from the negative terminal to the positive terminal of a 9 V battery?

Options

A. `54xx10^(-3)J`
B. `54xx10^(-6)J`
C. `54xx10^(-9)J`
D. `54xx10^(-12)J`

Answer

To find the work required to carry a charge from the negative terminal to the positive terminal of a battery, we can use the formula for work done in moving a charge in an electric field:

### Step-by-Step Solution:

1. **Identify the given values:**
- Charge (Q) = 6 µC = \(6 \times 10^{-6}\) C
- Potential difference (V) = 9 V

2. **Understand the relationship between work, charge, and potential difference:**
The work done (W) in moving a charge in an electric field is given by the formula:
\[ W = Q \times V \]
where:
- \(W\) is the work done,
- \(Q\) is the charge,
- \(V\) is the potential difference.

3. **Substitute the values into the formula:**
\[ W = (6 \times 10^{-6} \, \text{C}) \times (9 \, \text{V}) \]

4. **Calculate the work done:**
\[ W = 54 \times 10^{-6} \, \text{J} \]
or
\[ W = 54 \, \mu\text{J} \]

5. **Final answer:**
The work required to carry a 6 µC charge from the negative terminal to the positive terminal of a 9 V battery is \(54 \, \mu\text{J}\).