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What are the forces (in N) on two charges of +0.800 C and −5.00 C, respectively, if they are separated by a distance of 5.00 m?

Answer :

The electrostatic force between two charges of +0.800 C and
−5.00 C separated by 5.00 m is calculated using Coulomb's Law, resulting in an attractive force of 1.438 * 10^9 Newtons.

The question requires using Coulomb's Law which states that the electrostatic force (F) between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is:

F = k * |q1 * q2| / r^2

Where:

  • k is Coulomb's constant (8.9875517873681764*10^9 N m^2/C^2),
  • q1 and q2 are the charges of the objects, and
  • r is the separation distance between the centers of the two charges.

Given that q1 = +0.800 C and q2 = -5.00 C, and they are separated by r = 5.00 m, we can calculate the force as follows:

F = (8.9875517873681764*10^9 N m^2/C^2) * |0.800 C * -5.00 C| / (5.00 m)^2

F = (8.9875517873681764*10^9) * (4) / 25

F = 8.9875517873681764 * 10^9 * 0.16 N

F = 1.438008285978028 * 10^9 N

The electrostatic force is therefore 1.438 * 10^9 Newtons, and since the charges are of opposite signs, the force will be attractive.

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