Electric Field at a Rock During a Volcano Eruption
Question:
Often times during volcano eruptions the clouds have large static charge buildups that cause lightning to happen. If a rock is ejected by the volcano and it flies to a height of 2345 m, and the clouds have a charge concentration of +40.0 C at a height of 3000 m and -40.0 C at a height of 1000 m, what is the electric field at the ejected rock?
Answer:
The rock and both clouds lie on a vertical line (assume the z-axis). We apply Coulomb’s Law to find the electric field at the location of the rock.
Electric field due to a point charge:
\( E = \frac{k |Q|}{r^2} \), where \( k = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)
\( E = \frac{k |Q|}{r^2} \), where \( k = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)
Step 1: Determine distances
- Distance from rock to positive cloud: \( r_+ = 3000 – 2345 = 655 \, \text{m} \)
- Distance from rock to negative cloud: \( r_- = 2345 – 1000 = 1345 \, \text{m} \)
Step 2: Direction of Electric Fields
- Positive charge: field points away from the charge → downward
- Negative charge: field points towards the charge → also downward
Electric Field from Positive Cloud:
\( E_+ = \frac{8.99 \times 10^9 \times 40.0}{655^2} \)\( = \frac{359.6 \times 10^9}{429025} \approx \)
8.38 × 105 N/C
Electric Field from Negative Cloud:
\( E_- = \frac{8.99 \times 10^9 \times 40.0}{1345^2} \)\( = \frac{359.6 \times 10^9}{1,809,025} \approx \)
1.99 × 105 N/C
Net Electric Field at Rock:
Since both point downward, we add them:\( E_{\text{net}} = E_+ + E_- = 8.38 \times 10^5 + 1.99 \times 10^5 \)
= 1.04 × 106 N/C downward
Final Answer:
Magnitude: 1.04 × 106 N/C
Direction: Downward