The electric field is formed by a point charge of 24 nC. Determine at what distance from

| September 7, 2021 | education

The electric field is formed by a point charge of 24 nC. Determine at what distance from it is the point at which the field strength is 15 kN / C. With what force at this point does the field act on the test charge of 1.8 nC?

q1 = 24 nC = 24 * 10 ^ -9 C
E = 15 kN / C = 15 * 10 ^ 3 N / C
q2 = 1.8 nC = 1.8 * 10 ^ -9 C
r-?
F-?
The electric field strength E is determined by the formula E = (1/4 * P * e) * q1 / r ^ 2, where P = 3.14 is constant pi, e = 8.85 * 10 ^ -12 Cl ^ 2 / Nm ^ 2 is the electrical constant, q1 is the amount of charge, r is the distance to the charge. Let us express the distance r = square root of the expression (q1 / 4 * P * e * E). Substituting the data, we get r = square root of 24 * 10 ^ -9 C / 4 * 3.14 * 8.85 * 10 ^ -12 C ^ 2 / Nm ^ 2 * 15 * 10 ^ 3 N / C = 0.12 m.
According to Coulomb’s law F = (1/4 * P * e) * q1 * q2 / r ^ 2, substitute the data F = (1/4 * 3.14 * 8.85 * 10 ^ -12 Cl ^ 2 / Nm ^ 2) * 24 * 10 ^ -9 C * 1.8 * 10 ^ -9 C / 0.12 m ^ 2 = 26.7 * 10 ^ -6 N.
Answer: distance r = 0.12 m, F = 26.7 * 10 ^ -6 N



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