Question

на заряд q=-3*10^-7 Кл, который находится в точке С действует сила 0,035 H электрического поля создаваемый отрицательным электрическим зарядом -q. Определите какова напряженность поля в точке C.

Answer 1

Дано:

q = - 3·10⁻⁷ Кл

F = 0,035 Н

___________

E - ?

Напряженность поля по модулю:

E =  F / q = 0,035 / (3·10⁻⁷) ≈ 1,2·10⁵ В/м  или 120 кВ/м

Вектор напряженности поля Е направлен в ту же сторону, что и вектор силы F.

Answer 2

Calculation of Electric Field Intensity

To determine the electric field intensity at point C, we can use the formula:

Electric Field Intensity (E) = Force (F) / Charge (q)

Given: - Charge (q) = -3 * 10^-7 C - Force (F) = 0.035 N

Substituting the given values into the formula, we can calculate the electric field intensity:

E = F / q

E = 0.035 N / (-3 * 10^-7 C)

To calculate the value, we can use the following steps:

1. Convert the charge from Coulombs (C) to the elementary charge unit (e) using the formula: 1 C = 6.242 * 10^18 e

q = -3 * 10^-7 C * (6.242 * 10^18 e / 1 C)

2. Calculate the electric field intensity:

E = 0.035 N / q

Let's perform the calculations step by step.

Step 1: Conversion of Charge to Elementary Charge Unit

To convert the charge from Coulombs (C) to the elementary charge unit (e), we can use the conversion factor:

1 C = 6.242 * 10^18 e

Substituting the given charge value, we have:

q = -3 * 10^-7 C * (6.242 * 10^18 e / 1 C)

Calculating this, we find:

q ≈ -1.8726 * 10^12 e

Step 2: Calculation of Electric Field Intensity

Now, we can calculate the electric field intensity using the formula:

E = 0.035 N / q

Substituting the values, we have:

E ≈ 0.035 N / (-1.8726 * 10^12 e)

Calculating this, we find:

E ≈ -1.87 * 10^-14 N/e

Therefore, the electric field intensity at point C is approximately -1.87 * 10^-14 N/e.

Please note that the negative sign indicates the direction of the electric field.