Question

Определить напряженность E электрического поля, создаваемого точечным зарядом q=100 нКл на расстоянии r = 5 см от него. Диэлектрическая проницаемость равна 3. Ответ дайте в кВ/м.

Answer 1

Ответ:

E = 120 кВ/м

Объяснение:

Дано:

$q = 100 \cdot 10^{-9}$ Кл

$r =$ 0,05 м

$\epsilon = 3$

$k = 9 \cdot 10^{9}$ H · м² / Кл²

Найти:

$E \ - \ ?$

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$E = \dfrac{kq}{\epsilon r^{2}}$

$E =$ $(9 \cdot 10^{9}$ H · м² / Кл² * $100 \cdot 10^{-9}$ Кл) / (3 * 0,0025 м²) = 120 000 В/м =

= 120 кВ/м

Ответ: E = 120 кВ/м.

Answer 2

Calculation of Electric Field Intensity

To determine the electric field intensity (E) created by a point charge (q) of 100 nC at a distance (r) of 5 cm, we can use the formula:

E = (k * q) / (r^2)

Where: - E is the electric field intensity - k is the electrostatic constant (k = 1 / (4πε₀)) - q is the charge - r is the distance from the charge

Given that the charge (q) is 100 nC and the distance (r) is 5 cm, we need to convert these values to SI units before calculating the electric field intensity.

Converting the charge to SI units: - 1 nC = 1 × 10^(-9) C - Therefore, 100 nC = 100 × 10^(-9) C = 1 × 10^(-7) C

Converting the distance to SI units: - 1 cm = 0.01 m - Therefore, 5 cm = 5 × 0.01 m = 0.05 m

Now we can substitute the values into the formula and calculate the electric field intensity.

Using the given value of the dielectric constant (ε = 3), we can calculate the electric field intensity in kV/m.

Calculation:

E = (k * q) / (r^2)

Where: - k = 1 / (4πε₀) - ε₀ = 8.854 × 10^(-12) F/m (permittivity of free space)

Substituting the values: - k = 1 / (4π * 8.854 × 10^(-12) F/m) - q = 1 × 10^(-7) C - r = 0.05 m

Calculating k: - k ≈ 8.99 × 10^9 Nm^2/C^2

Calculating the electric field intensity (E): - E = (8.99 × 10^9 Nm^2/C^2 * 1 × 10^(-7) C) / (0.05 m)^2

Simplifying the equation: - E = (8.99 × 10^9 Nm^2/C^2 * 1 × 10^(-7) C) / 0.0025 m^2 - E = 3.596 × 10^14 N/C

Converting the electric field intensity to kV/m: - 1 N/C = 0.001 kV/m - Therefore, 3.596 × 10^14 N/C = 3.596 × 10^11 kV/m

Answer:

The electric field intensity created by the point charge of 100 nC at a distance of 5 cm is approximately 3.596 × 10^11 kV/m.

Please note that the calculation assumes a vacuum or free space as the medium. If the medium has a different dielectric constant, the electric field intensity may be affected.