Question

Помогите пожалуйста) Напряженность однородного электрического поля =5В/М . Точечный заряд25 нКл переместили на 10 см вдоль силовой линии поля.Какую работу при этом совершили силы поля ? Расчитайте напряжение между начальной и конечной точками,в которых находится заряд.

Answer 1

E=5 В/м                           работа поля: A=Eqd;
q=25 нКл=25 10⁻⁹Кл                             A=6*25 10⁻⁹*0,1=15 10⁻⁹ Дж;
d=10 см=0,1 м                напряжение: U=A/q;
__________________                          U=15 10⁻⁹/25 10⁻⁹=0,6 B
A-?                                         Ответ: А=15 10⁻⁹ Дж;
U-?                                                     U=0,6 B.

Answer 2

Calculation of Work Done by Electric Field

To calculate the work done by the electric field when moving a point charge, we can use the formula:

Work Done = Force x Distance

In this case, the force experienced by the point charge is given by the equation:

Force = Charge x Electric Field

The distance moved by the charge is given as 10 cm, which is equivalent to 0.1 meters.

Given that the electric field strength is 5 V/m and the charge is 25 nC (nanoCoulombs), we can calculate the work done by the electric field.

Calculation of Work Done

First, let's calculate the force experienced by the charge:

Force = Charge x Electric Field

Substituting the given values:

Force = (25 nC) x (5 V/m)

Converting the charge to Coulombs:

Force = (25 x 10^-9 C) x (5 V/m)

Force = 125 x 10^-9 N

Next, let's calculate the work done by the electric field:

Work Done = Force x Distance

Substituting the values:

Work Done = (125 x 10^-9 N) x (0.1 m)

Work Done = 12.5 x 10^-9 Nm

Work Done = 12.5 nJ (nanoJoules)

Therefore, the work done by the electric field when moving the point charge along the field lines is 12.5 nanoJoules.

Calculation of Voltage Difference

To calculate the voltage difference between the initial and final points where the charge is located, we can use the formula:

Voltage Difference = Electric Field x Distance

Given that the electric field strength is 5 V/m and the distance moved is 10 cm (0.1 m), we can calculate the voltage difference.

Voltage Difference = (5 V/m) x (0.1 m)

Voltage Difference = 0.5 V

Therefore, the voltage difference between the initial and final points where the charge is located is 0.5 Volts.

Please let me know if there is anything else I can help you with!