Question

Между двумя параллельными пластинами, вертикально расположенными, подвешен шарик массой 0,1 г. Пластины заряжены и при напряженности 45000 В/м шарик отклонился от вертикали на 10градусов. Найти заряд шарика

Answer 1

Дано: (жирным показаны векторные величины)

m = 10^-4 кг

E = 4.5 * 10^4 В/м что равносильно Н/Кл(в поисковике есть информация по переводу величин)

a = 10*

------------------

Найти: q

В первую очередь стоит нарисовать рисунок изобразив действующие на шарик силы — сила создаваемая подвешенным шариком Fшар направленная вниз; сила воздействия электрического поля на заряд Fполя, направленная горизонтально, перпендикулярно к силе Fшар и их результирующая сила Fрез, схематически, немного отклонённая от Fшар.

При этом Fшар = m*g, Fполя = E*q. Схематически можем изобразить прямоугольный треугольник с катетами m*g и E*q.

Используем соотношение тангенса для составления уравнения:

$tg(\alpha) =\frac{Eq}{mg}$

Выразим заряд:

$q = \frac{tg(\alpha)*m*g}{E} )$

Соответствие величин:

[Кл] = (кг * м/c^2) / (Н/Кл) = Н / H/Кл = [Кл]

Подставим и рассчитаем:

q = (tg(10) * 10^-4 кг * 10 м/с^2) / 4.5 * 10^4 H/Кл = 3.92 * 10^-9 Кл

Ответ: 3.92 нКл

Answer 2

Calculation of the Charge on the Ball

To find the charge on the ball, we can use the concept of electric field and the force experienced by a charged particle in an electric field.

The force experienced by a charged particle in an electric field is given by the equation:

F = qE

Where: - F is the force experienced by the charged particle - q is the charge on the particle - E is the electric field strength

In this case, the ball is suspended between two parallel plates with a vertical orientation. The ball is deflected from the vertical by an angle of 10 degrees. The electric field strength between the plates is given as 45000 V/m.

The force experienced by the ball can be decomposed into two components: the vertical component and the horizontal component. The vertical component of the force is balanced by the gravitational force acting on the ball, while the horizontal component of the force causes the deflection.

Since the ball is in equilibrium, the vertical component of the force is equal to the weight of the ball:

F_vertical = mg

Where: - F_vertical is the vertical component of the force - m is the mass of the ball - g is the acceleration due to gravity

The horizontal component of the force can be calculated using trigonometry:

F_horizontal = F * sin(θ)

Where: - F_horizontal is the horizontal component of the force - F is the total force experienced by the ball - θ is the angle of deflection

Since the force experienced by the ball is the product of the charge on the ball and the electric field strength, we can rewrite the equation as:

F = qE

Combining the equations for the horizontal and vertical components of the force, we have:

qE = F_horizontal = F * sin(θ)

Substituting the equation for the vertical component of the force, we get:

qE = (mg) * sin(θ)

Now we can solve for the charge on the ball (q):

q = (mg) * sin(θ) / E

Given that the mass of the ball (m) is 0.1 g (or 0.0001 kg), the angle of deflection (θ) is 10 degrees, and the electric field strength (E) is 45000 V/m, we can substitute these values into the equation to find the charge on the ball.

Let's calculate it:

q = (0.0001 kg * 9.8 m/s^2) * sin(10 degrees) / 45000 V/m

Using a calculator, we find that the charge on the ball is approximately 2.18 x 10^-9 Coulombs.

Please note that this calculation assumes that the electric field is uniform between the plates and that the ball is a point charge.