Ejercicio Simulation 2 · Física II URV 2024

Tema: Electrostatics: Coulomb, electric field, Gauss's law and potential · Del examen Exámenes de Física II: Teoría y Laboratorio (Mayo 2024) (2024)

Enunciado

Simulation 2. Electric field and potential of a -1 μC charge

Place a charge q=−1 μCq = -1\,\mu\text{C} at the origin (0,0)(0, 0) and repeat the measurements of point 1 at (100 m,100 m)(100\text{ m}, 100\text{ m}) and (100 m,−100 m)(100\text{ m}, -100\text{ m}). Compare the outcomes with the positive charge case: what differences are observed?

1.

Measurements at position (100 m,100 m)(100\text{ m}, 100\text{ m}):

V

N/C

N/C

N/C

°

2.

Measurements at position (100 m,−100 m)(100\text{ m}, -100\text{ m}):

V

N/C

N/C

N/C

°

3.

Compare the outcomes with the positive charge case: what differences are observed?

Enunciado: Ejercicio Simulation 2 · Física II URV 2024

Solución

Según los criterios oficiales de corrección

1. Potential V at (100 m, 100 m)

Solución: -63.6 V

1. Field magnitude E at (100 m, 100 m)

Solución: 0.45 N/C

1. Ex at (100 m, 100 m)

Solución: -0.32 N/C

1. Ey at (100 m, 100 m)

Solución: -0.32 N/C

1. Angle alpha at (100 m, 100 m)

Solución: -135°

2. Potential V at (100 m, -100 m)

Solución: -63.64 V

2. Field magnitude E at (100 m, -100 m)

Solución: 0.44 N/C

2. Ex at (100 m, -100 m)

Solución: -0.32 N/C

2. Ey at (100 m, -100 m)

Solución: 0.32 N/C

2. Angle alpha at (100 m, -100 m)

Solución: 134° / 135°

3. Compare the outcomes with the positive charge case: what differences are observed?

Solución: The magnitude of the electric field E and the absolute value of the electric potential |V| remain unchanged because the distance and magnitude of the source charge are identical. However, the potential becomes strictly negative (V < 0), and the direction of the electric field vector reverses by 180°, pointing radially inward toward the negative source charge rather than radially outward.

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