An electron is very far from the area. What is the ratio Ff/Fi of the electric force on the electron after the area is reduced to the force before the area was reduced?

Respuesta :

The given question is incomplete. The complete question is as follows.

The irregularly shaped area of charge in the figure has surface charge density ηi. Each dimension (x and y) of the area is reduced by a factor of 3.68.

What is the ratio ηf/ηi where ηf is the final surface charge density? I found this value to be 13.5 and it is correct

An electron is very far from the area. What is the ratioFf/Fi of the electric force on the electronafter the area is reduced to the force before the area wasreduced? I found this to also be 13.5, but it is wrong.

Explanation:

Let us consider that expression for area will be as follows.

             Area ([tex]\Delta A[/tex]) = [tex]\Delta x \times \Delta y[/tex]

As,       [tex]\frac{\eta_{f}}{\eta_{i}} = (\frac{1}{3.68})^{2}[/tex]

                        = 13.5

Also, [tex]\frac{F_{f}}{F_{i}}[/tex] = 1

As the electron is very far from the area. Hence, it can be considered as a point charge. Whereas charge is also constant as the force is not changing.

In this exercise we have to have knowledge about force to calculate the reduced area in this way, we can say that this corresponds to:

[tex]A=13.5[/tex]

Knowing that the reduced area can be calculated by the following formula, we find that:

[tex]A=\Delta X * \Delta Y\\ n_f/n_i=13.5[/tex]

As the power happen very far from the field. Hence, it maybe thought-out as a point charge. Whereas charge is in addition to perpetual as extrasensory perception happen not changeful.

See more about force at brainly.com/question/26115859

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