Answer:
1950.65 m
Explanation:
Parameters given:
First charge, Q1 = 43.2 C
Second charge, Q2 = -38.7 C
Magnitude of force between them, F = 3.95 * 10^6 N
The magnitude of the electric force between two electric charges at a distance r from one another is given as:
|F| = |k * Q1 * Q2|/r²
3.95 * 10^6 = |8.99 * 10^9 * 43.2 * -38.7| / r²
Making r² the subject of the formula and solving:
r² = 3805025.189
r = 1950.65 m
Answer:
1950.65m
Explanation:
Coulomb's law states that the force F of attraction between two charges q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance d between them. Mathematically;
|F| = k |q1||q2| /r²
Given |q1| = |43.2| C = 43.2C
|q2| = |-38.7|C = 38.7C
F = 3.95 x 106 N
d = ?
k is the coloumbs constant = 8.99 ×109 Nm2 / C2
Substituting this values in the formula to get distance d we have;
3.95x10^6= 8.99×10^9×43.2×38.7/d²
3.95x10^6d² = 15,029,841,600,000
d² =15,029,841,600,000/3,950,000
d² = 3,805,023.189
d = √3,805,023.189
d = 1950.65m
Average separation between the charges is 1950.65m
