Question

Найдите напряженность и потенциал в точке, лежащей посередине между двумя зарядами по 5•10 -8 Кл, расположенными на расстоянии 1 м в вакууме.

Answer 1

E = q / 4peR^2, E = U/ R дальше подставляешь

 

Answer 2

Calculation of Electric Field Intensity and Electric Potential

To find the electric field intensity and electric potential at a point located midway between two charges of 5 x 10^-8 C, separated by a distance of 1 m in a vacuum, we can use the formulas for electric field intensity and electric potential due to a point charge.

The electric field intensity (E) at a point due to a point charge (q) is given by Coulomb's law:

E = k * (q / r^2)

where: - E is the electric field intensity - k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2) - q is the charge - r is the distance from the charge to the point

The electric potential (V) at a point due to a point charge (q) is given by the formula:

V = k * (q / r)

where: - V is the electric potential - k is the electrostatic constant - q is the charge - r is the distance from the charge to the point

Let's calculate the electric field intensity and electric potential at the point using the given values.

Calculation:

The charges are located at a distance of 1 m from each other, and the point of interest is located midway between them. Therefore, the distance from each charge to the point is 0.5 m.

Using the formula for electric field intensity:

E = k * (q / r^2)

Substituting the values: - k = 8.99 x 10^9 Nm^2/C^2 - q = 5 x 10^-8 C - r = 0.5 m

We can calculate the electric field intensity (E) at the point.

E = (8.99 x 10^9 Nm^2/C^2) * (5 x 10^-8 C / (0.5 m)^2)

Calculating the value of E:

E = 8.99 x 10^9 Nm^2/C^2 * (5 x 10^-8 C / (0.5 m)^2)

E = 8.99 x 10^9 Nm^2/C^2 * (5 x 10^-8 C / 0.25 m^2)

E = 8.99 x 10^9 Nm^2/C^2 * 2 x 10^-7 C / m^2

E = 1.798 x 10^3 N/C

Therefore, the electric field intensity at the point is 1.798 x 10^3 N/C.

Using the formula for electric potential:

V = k * (q / r)

Substituting the values: - k = 8.99 x 10^9 Nm^2/C^2 - q = 5 x 10^-8 C - r = 0.5 m

We can calculate the electric potential (V) at the point.

V = (8.99 x 10^9 Nm^2/C^2) * (5 x 10^-8 C / 0.5 m)

Calculating the value of V:

V = 8.99 x 10^9 Nm^2/C^2 * (5 x 10^-8 C / 0.5 m)

V = 8.99 x 10^9 Nm^2/C^2 * 1 x 10^-7 C / m

V = 8.99 x 10^2 Nm/C

Therefore, the electric potential at the point is 8.99 x 10^2 Nm/C.

Answer:

The electric field intensity at the point is 1.798 x 10^3 N/C, and the electric potential at the point is 8.99 x 10^2 Nm/C.