В однородном электрическом поле напряженностью 1 кВ/м переместили заряд -25 нКл в направлении силовой линии на 2 см. Найти работу поля, изменение потенциальной энергии заряда и напряжение между начальной и конечной точками перемещения.
Внимание!! Математические вычисления могут быть не точны!
Solution:The work done by an electric field can be calculated by multiplying the magnitude of the electric field by the distance moved in the direction of the field. In this case, the magnitude of the electric field is given as 1 kV/m and the distance moved is 2 cm. Therefore, the work done by the electric field is 2 × 10-6 J.
To calculate the change in potential energy, we can use the formula ΔU = qΔV, where q is the charge and ΔV is the change in voltage. Here, the charge is given as -25 nC and the change in voltage can be calculated using the formula ΔV = Ed, where E is the electric field and d is the distance moved. Substituting the values, we get ΔV = 1 kV/m × 0.02 m = 0.02 kV. Therefore, the change in potential energy is ΔU = (-25 nC) × (0.02 kV) = -0.5 nJ.
Finally, the voltage between the initial and final points can be calculated by dividing the change in potential energy by the charge. In this case, the voltage is (-0.5 nJ) / (-25 nC) = 0.02 kV.
Note: The negative signs in the calculations indicate that the electric field and potential energy are acting in opposite directions to the motion of the charge.
Внимание! Используйте решение от нейросети аккуратно, т.к они могут быть не точными. Обязательно проверьте ответы самостоятельно! Особенно это касается математических вычислений
