Question

a uniform electric field has equipotential planes parallel to the xy - plane as shown below. if the planes are separated by 2.96 cm, find the electric field. enter a positive number if the field points up and negative if the field points down.
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-100
-120
-140 v
v/m

Explanation:

Step1: Identify the potential difference

The potential difference $\Delta V$ between the two equipotential - planes is $\Delta V=|- 100-(-140)|V = 40V$. The distance between the planes $d = 2.96cm=0.0296m$.

Step2: Use the formula for the electric field

The formula for the magnitude of a uniform electric field is $E=\frac{\Delta V}{d}$.
Substitute $\Delta V = 40V$ and $d = 0.0296m$ into the formula: $E=\frac{40}{0.0296}\frac{V}{m}\approx1351.35\frac{V}{m}$. Since the field points down (negative $z$ - direction as the lower - potential is below the higher - potential), the value is negative.

Answer:

$- 1351.35$