Question

during a thunderstorm, a meteorologist measures earths electric field near the surface, and finds that equipotential planes 1300 v are spaced a distance of 7.50 cm apart from each other and are parallel to earths surface. assuming the electric field is uniform, if the equipotential planes increase voltage with altitude, what is the electric field at this location? enter a positive number if the field points up and negative if the field points down.

Explanation:

Step1: Recall electric - field formula

The relationship between electric field $E$, potential difference $\Delta V$, and distance $d$ in a uniform electric - field is $E=\frac{\Delta V}{d}$.

Step2: Convert distance to SI units

The distance $d = 7.50\ cm=0.075\ m$. The potential difference $\Delta V = 1300\ V$.

Step3: Calculate the electric field

Substitute $\Delta V = 1300\ V$ and $d = 0.075\ m$ into the formula $E=\frac{\Delta V}{d}$, we get $E=\frac{1300}{0.075}\ V/m\approx1.73\times 10^{4}\ V/m$.

Answer:

$1.73\times 10^{4}$