Question

2. Напряженность электрического поля заряда, помещенного в керосин (Екеросина=2), в некотрой точке равна Ек, как изменится наряженность поля в этой точке если заряд поместить в воду (Еводы=81)?

Answer 1

Е=kq/r^2×e
e- диэлектрическая проницаемость
Стоит в знаменателе
В керосине
Е=kq/r^2×2
В воде
Е=kq/r^2×81

Уменьшится в 40,5 раза

Answer 2

Electric Field Intensity in Different Media

When a charge is placed in a medium, the electric field intensity at a certain point is determined by the properties of that medium. The electric field intensity in kerosene (E_kerosene) is given as 2, and in water (E_water) it is given as 81.

Effect of Changing the Medium on Electric Field Intensity:

When a charge is placed in a different medium, the electric field intensity at a certain point changes. The relationship between the electric field intensity in different media can be described using the concept of relative permittivity (also known as dielectric constant).

The electric field intensity in a medium is related to the electric field intensity in vacuum (E_vacuum) by the formula:

E_medium = E_vacuum / ε

Where: - E_medium is the electric field intensity in the medium - E_vacuum is the electric field intensity in vacuum - ε is the relative permittivity of the medium

Applying the Concept to the Given Scenario

Given: - E_kerosene = 2 - E_water = 81

We can calculate the relative permittivity of water (ε_water) compared to vacuum using the formula:

ε_water = E_vacuum / E_water

Substituting the given values: ε_water = 2 / 81

Calculating: ε_water ≈ 0.0247

Now, if the charge is placed in water (E_water = 81), the electric field intensity at a certain point in water will change. The electric field intensity in water will be reduced compared to that in kerosene due to the higher relative permittivity of water.

Therefore, the electric field intensity in the given point will decrease when the charge is placed in water compared to when it was in kerosene.

This change in electric field intensity is a result of the difference in relative permittivity between the two media.

In summary, the electric field intensity in the given point will decrease when the charge is placed in water (E_water = 81) compared to when it was in kerosene (E_kerosene = 2).