A space probe approaches a planet,taking measurements as it goes.If it detects a potential difference of 6000 MV between the altitudes of 253,000 km and 276,000 km above the planet's surface,what is the approximate electric field strength produced by the planet at 264,500 km above the surface? Assume the $\vec { E }$ field is approximately constant at these altitudes.

Question

Quizplus · Accepted Answer

The electric field strength $ E $ can be calculated using the formula $ E = \frac{\Delta V}{\Delta d} $, where $ \Delta V $ is the potential difference and $ \Delta d $ is the distance between the two altitudes. Here, $ \Delta V = 6000 \times 10^6 $ volts and $ \Delta d = 276,000 \, \text{km} - 253,000 \, \text{km} = 23,000 \, \text{km} = 23,000,000 \, \text{m} $. Thus, $ E = \frac{6000 \times 10^6}{23,000,000} \approx 261 \, \text{N/C} $.

A Space Probe Approaches a Planet,taking Measurements as It Goes E⃗\vec { E }E

A Space Probe Approaches a Planet,taking Measurements as It Goes $\vec { E }$