At a distance of 5 cm from the charge, the electric field strength is 2 * 10 ^ (5) V / m.

At a distance of 5 cm from the charge, the electric field strength is 2 * 10 ^ (5) V / m. find the magnitude of the charge creating this field.

r = 5 cm = 0.05 m.

E = 2 * 10 ^ 5 V / m.

k = 9 * 10 ^ 9 N * m2 / Cl2.

q -?

The strength of the electric field E is called the vector force characteristic of a given point of the field, which is determined by the ratio of the strength of the electrostatic field at a given point F to the value of the charge q1, which is placed in it: E = F / q.

We express the strength of the electrostatic field F by the Coulomb law: F = k * q1 * q / r ^ 2, k is the electric constant, r is the distance between the charges.

E = k * q1 * q / q1 * r ^ 2 = k * q / r ^ 2.

q = E * r ^ 2 / k.

q = 2 * 10 ^ 5 V / m * (0.05 m) 2/9 * 10 ^ 9 N * m2 / Cl2 = 5.5 * 10 ^ -8 Cl.

Answer: the electric field creates a charge of q = 5.5 * 10 ^ -8 C.