Question

Помогите пожалуйста Найти напряженность электрического поля в точке, удаленной на 20 см от точечного заряда 40 Кл.

Answer 1

Найдите напряженность электрического поля

 

E = 1/ԑ * kq/r^2 = 1/2 *9*10^9 *2.5*10^-8 / 0.05^2 = 45 000 В/м =45 кВ/м

 

ОТВЕТ 45 000 В/м =45 кВ/м

Answer 2

Calculating Electric Field Intensity

To calculate the electric field intensity (\(E\)) at a point located at a distance of 20 cm from a point charge of 40 kC, we can use the formula:

\[E = \frac{{kq}}{{r^2}}\]

Where: - \(k\) is the Coulomb's constant (\(8.9875 \times 10^9 \, Nm^2/C^2\)) - \(q\) is the charge (40 kC or \(40 \times 10^3 C\)) - \(r\) is the distance from the point charge (20 cm or \(0.20 m\))

Substituting the given values into the formula:

\[E = \frac{{(8.9875 \times 10^9 \, Nm^2/C^2) \times (40 \times 10^3 C)}}{{(0.20 m)^2}}\]

\[E = \frac{{(8.9875 \times 10^9) \times (40 \times 10^3)}}{{0.04}}\]

\[E = \frac{{(8.9875 \times 40 \times 10^9 \times 10^3)}}{{0.04}}\]

\[E = \frac{{(8.9875 \times 40 \times 10^{12})}}{{0.04}}\]

\[E = \frac{{(3.595 \times 10^{14})}}{{0.04}}\]

\[E = 8.9875 \times 10^{15} N/C\]

Therefore, the electric field intensity at a point 20 cm away from a point charge of 40 kC is 8.9875 x 10^15 N/C.