Question

В воде на расстоянии 5 см друг от друга находятся заряды 20 мкКл и -10 мкКл. Определить напряженность электрического поля в точке которая находится на расстоянии 4 см от первого и 3 см от второго зарядов. ответ (8.9 * 10^4) Очень прошу помочь с решением.

Answer 1

Εводы=81
k=9*10^9
E1=kq1/r1^2*ε=9*10⁹*20*10⁻⁶/0,04²*81=20*10³/0,0016*9=2*10⁸/144
E2=kq2/r2^2*ε=9*10⁹*10*10⁻⁶/0,03²*81=10⁴/0,0081=10⁸/81
E=√E1²+E2²=10⁸√2²/144²+1/81²=10⁸√1/9²*4²+1/9²*9²=10⁸√(9²+4²)/9²4²9²=10⁸/9⁴4²√9²+4²=952,6*9,85=93820В.м

Answer 2

Calculation of Electric Field Intensity

To determine the electric field intensity at a point located 4 cm from the first charge and 3 cm from the second charge, we can use the formula for electric field intensity due to a point charge:

Electric field intensity (E) = k * (q / r^2)

Where: - E is the electric field intensity - k is the electrostatic constant (k = 9 * 10^9 Nm^2/C^2) - q is the charge - r is the distance from the charge

Let's calculate the electric field intensity at the given point.

For the first charge (q1 = 20 μC) at a distance of 4 cm (r1 = 4 cm = 0.04 m) from the point: E1 = (9 * 10^9 Nm^2/C^2) * (20 * 10^-6 C) / (0.04 m)^2 For the second charge (q2 = -10 μC) at a distance of 3 cm (r2 = 3 cm = 0.03 m) from the point: E2 = (9 * 10^9 Nm^2/C^2) * (-10 * 10^-6 C) / (0.03 m)^2 Now, we can calculate the total electric field intensity at the given point by adding the electric field intensities due to each charge:

E_total = E1 + E2

Let's calculate the values:

E1 = (9 * 10^9 Nm^2/C^2) * (20 * 10^-6 C) / (0.04 m)^2 = 9 * 10^4 N/C E2 = (9 * 10^9 Nm^2/C^2) * (-10 * 10^-6 C) / (0.03 m)^2 = -3 * 10^5 N/C E_total = E1 + E2 = (9 * 10^4 N/C) + (-3 * 10^5 N/C) = -2.1 * 10^5 N/C

Therefore, the electric field intensity at the given point is -2.1 * 10^5 N/C.

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