A space probe approaches a planet, taking measurements as it goes. If it detects a potential difference of $6000 \mathrm { MV }$ between the altitudes of $253,000 \mathrm {~km}$ and $276,000 \mathrm {~km}$ above the planet's surface, what is the approximate electric field strength produced by the planet at $264,500 \mathrm {~km}$ above the surface? Assume the electric field strength 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 points. Here, $\Delta V = 6000 \times 10^6 \, \text{V}$ 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 6000MV6000 \mathrm { MV }6000MV

A Space Probe Approaches a Planet, Taking Measurements as It $6000 \mathrm { MV }$