Question

Заряд 15 мкКл перемістився в електричному полі із точки з потенціалом 40 В у точку з потенціалом 30 В. Яку роботу виконало поле?

Answer 1

Ответ:

0.00015 (Дж) или 150 мкДж

Объяснение:

Дано:

q = 15 мкКл = 15·10⁻⁶ Кл

φ₁ = 40 В

φ₂ = 30 В

Найти:

А - работу поля

Решение:

А = q · (φ₁ - φ₂) = 15·10⁻⁶ · (40 - 30) = 0.00015 (Дж) = 150 мкДж.

Answer 2

Electric Field and Work Done

When a charge moves in an electric field from a point with a higher potential to a point with a lower potential, the electric field does work on the charge. In this case, a charge of 15 µC has moved from a point with a potential of 40 V to a point with a potential of 30 V.

Work Done by the Electric Field: The work done by the electric field can be calculated using the formula: \[ W = q \cdot \Delta V \] Where: - \( W \) = work done by the electric field - \( q \) = charge - \( \Delta V \) = change in potential

Substituting the given values: \[ W = 15 \times 10^{-6} \, C \times (40 - 30) \, V \] \[ W = 15 \times 10^{-6} \, C \times 10 \, V \] \[ W = 0.15 \times 10^{-4} \, C \cdot V \] \[ W = 1.5 \times 10^{-5} \, J \]

Result: The electric field has done 1.5 × 10^-5 J of work on the charge as it moved from the point with a potential of 40 V to the point with a potential of 30 V.