two large parallel metal plates are 1.50 cm apart and have charges of equal magnitude but opposite signs on…

two large parallel metal plates are 1.50 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. take the potential of the positive plate (located at the origin) to be zero, with the other plate located at x = 1.50 cm. if the potential difference between the plates is 9.32 v, find the electric field between the plates. enter a positive number if the field points along positive x and a negative number if the field points along negative x.

two large parallel metal plates are 1.50 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. take the potential of the positive plate (located at the origin) to be zero, with the other plate located at x = 1.50 cm. if the potential difference between the plates is 9.32 v, find the electric field between the plates. enter a positive number if the field points along positive x and a negative number if the field points along negative x.

Answer

Explanation:

Step1: Recall electric - field formula

The formula for the electric field between two parallel plates is $E=\frac{\Delta V}{d}$, where $\Delta V$ is the potential difference between the plates and $d$ is the distance between them.

Step2: Convert distance to SI units

The distance $d = 1.50\ cm=0.0150\ m$, and $\Delta V = 9.32\ V$.

Step3: Calculate the electric - field magnitude

Substitute the values into the formula: $E=\frac{\Delta V}{d}=\frac{9.32\ V}{0.0150\ m}\approx621\ V/m$. Since the field points from the positive to the negative plate and we are asked for a positive number when the field points along negative $x$ (which means the positive plate is at the larger $x$ - value), the value of the electric field is positive.

Answer:

$621$