To determine the electric field created by the storm, we can use the relationship between electric field (E), potential difference (V), and distance (d). The formula that relates these quantities is: \[ E = \frac{V}{d} \] ### Step-by-Step Explanation 1. **Identify the Given Values**: - The potential difference \( V \) between the thundercloud and the ground is \( 10^7 \) volts (V). - The height \( d \) (which is the distance over which this potential difference occurs) is \( 4 \) meters (m). 2. **Substitute the Values into the Formula**: We can now substitute the values into the formula for the electric field: \[ E = \frac{10^7 \, \text{V}}{4 \, \text{m}} \] 3. **Perform the Calculation**: - First, calculate \( \frac{10^7}{4} \): \[ E = 2.5 \times 10^6 \, \text{N/C} \] 4. **Final Answer**: The electric field created by the storm is \( 2.5 \times 10^6 \, \text{N/C} \). ### Why Option B is Correct - The calculation shows that the electric field strength is \( 2.5 \times 10^6 \, \text{N/C} \), which corresponds to option B. ### Why the Other Options are Incorrect - **Option A: \( 2.0 \times 10^6 \, \text{N/C} \)**: - This value is lower than the calculated electric field. It does not account for the full potential difference over the given height. - **Option C: \( 1.0 \times 10^7 \, \text{N/C} \)**: - This value is incorrect because it suggests that the electric field is equal to the potential difference, which is not how electric fields are defined. The electric field is the potential difference divided by the distance, not equal to it. - **Option D: \( 4.0 \times 10^7 \, \text{N/C} \)**: - This value is significantly higher than the calculated electric field. It seems to misinterpret the relationship between potential difference and distance, suggesting an incorrect scaling. ### Common Pitfalls - **Misunderstanding the Formula**: Students sometimes confuse the electric field with potential difference. Remember, the electric field is the potential difference divided by the distance. - **Units**: Ensure that the units are consistent. Here, volts (V) and meters (m) are used correctly to yield newtons per coulomb (N/C). - **Calculation Errors**: Double-check arithmetic when dividing large numbers, as small mistakes can lead to incorrect answers. ### Revision Summary - The electric field \( E \) is calculated using \( E = \frac{V}{d} \). - For a potential difference of \( 10^7 \, \text{V} \) and a height of \( 4 \, \text{m} \), the electric field is \( 2.5 \times 10^6 \, \text{N/C} \). - Option B is correct; other options are incorrect due to miscalculations or misunderstandings of the relationship between potential difference and electric field. - Always check units and calculations to avoid common pitfalls.