Question

Електричне поле, переміщуючись ділянкою кола заряд 60 Кл, виконує таку саму роботу, яку виконує сила тяжіння під час падіння масою 200г з висоти 360 м. Чому дорівнює напруга на ділянці? Срочно надо оченьь!!!!

Answer 1

За перевод не отвечаю.
Используем закон сохранения энергии:
m*g*h=q*U
U=m*g*h/q=0,2*10*360/60=12 B

Answer 2

Electric Field and Work Done

To answer your question, let's first understand the concept of electric field and work done.

Electric Field: An electric field is a region around a charged object or collection of charged objects where an electric force can be exerted on other charged objects. It is a vector quantity, meaning it has both magnitude and direction. The electric field is created by the presence of electric charges.

Work Done: In physics, work is defined as the transfer of energy that occurs when a force is applied to an object and it moves in the direction of the force. The work done on an object is equal to the force applied multiplied by the distance over which the force is applied.

Comparison of Electric Field and Gravitational Force

Now, let's compare the work done by the electric field while moving through a section of a circle with a charge of 60 C to the work done by the gravitational force during the fall of a mass of 200 g from a height of 360 m.

The work done by the electric field is given by the formula:

Work = Force x Distance

In this case, the force is the electric force exerted on the charge and the distance is the length of the section of the circle.

Similarly, the work done by the gravitational force is given by the formula:

Work = Force x Distance

In this case, the force is the gravitational force exerted on the mass and the distance is the height from which the mass falls.

To find the voltage on the section of the circle, we need to equate the work done by the electric field to the work done by the gravitational force.

Calculation of Voltage on the Section of the Circle

Let's calculate the voltage on the section of the circle.

Given: - Charge on the section of the circle = 60 C - Mass of the object = 200 g = 0.2 kg - Height from which the object falls = 360 m

To find the voltage, we need to equate the work done by the electric field to the work done by the gravitational force.

Work done by the electric field = Work done by the gravitational force

Electric Force x Distance = Gravitational Force x Distance

The distance cancels out on both sides of the equation, so we are left with:

Electric Force = Gravitational Force

The electric force is given by:

Electric Force = Electric Field x Charge

The gravitational force is given by:

Gravitational Force = Mass x Acceleration due to gravity

Substituting these values into the equation, we get:

Electric Field x Charge = Mass x Acceleration due to gravity

Simplifying the equation, we can solve for the electric field:

Electric Field = (Mass x Acceleration due to gravity) / Charge

Plugging in the given values, we have:

Electric Field = (0.2 kg x 9.8 m/s^2) / 60 C

Calculating this, we find:

Electric Field = 0.0327 N/C

Therefore, the voltage on the section of the circle is 0.0327 N/C.

Please note that the above calculation assumes a uniform electric field and neglects any other forces or factors that may be present.