Question

Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the potential difference between the initial and final points

Answer 1

The potential difference between the initial and final point is 3.0 V.

The given parameters:

• distance between the plates, d = 0.2 m
• voltage across the plates, V = 12 V
• position of the electron from negative plate, x₁ = 0.1 m
• position of the electron from the positive plate, x₂ = 0.0 5m

The potential difference between the initial and final point is calculated as follows;

$E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}$

where:

$d_2$ is the distance of the electron between the positive and negative plate

$0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m$

$V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V$

Thus, the potential difference between the initial and final point is 3.0 V.

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