Question

Two positive point charges, each with charge q, separated by a distance d, repel each other with a force of magnitude 20 N. What is the magnitude of the force between two positive point charges of magnitude 2.58 q, separated by a distance 6.1 d in units of N? Enter a number with two digits behind the decimal point.

Answer 1

The magnitude of the force between two positive point charges of magnitude 2.58q, separated by a distance 6.1d, is approximately 2.54 units.

Step-by-step explanation:

The force between two point charges, q1 and q2, can be calculated using Coulomb's law:

F = k * (q1 * q2) / r^2

where F is the force, k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.

For the given question, we can calculate the magnitude of the force between the two positive point charges, q1 and q2, using the formula above:

F = k * (q1 * q2) / r^2

F = (9 * 10^9 Nm^2/C^2) * (q * q) / (d^2)

F = (9 * 10^9) * (q^2) / (d^2)

where q is the magnitude of the charge, d is the distance between the charges, and F is the force.

Given that the magnitude of the force is 20N, we can solve the equation for q in terms of F and d:

20 = (9 * 10^9) * (q^2) / (d^2)

Simplifying the equation,

q^2 = (20 * (d^2)) / (9 * 10^9)

Taking the square root of both sides,

q = √((20 * (d^2)) / (9 * 10^9))

Substituting the given values, q = √((20 * (6.1^2)) / (9 * 10^9))

q ≈ 2.54

Answer 2

F = 3.58 N

Step-by-step explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

`F = (k*q_(1)*q_(2))/(d^(2))` Formula (1)

`K=8.99*10^(9) (N*m^(2) )/(C^(2) )` : Coulomb constant

q1, q2: charge in Coulombs (C)

d: distance between charges (m)

We apply formula 1 for the two situations presented in the problem:

1)
`20 = k * (q*q)/(d^(2)) =  k * (q^(2))/(d^(2))` Equation (1)

2)
`F = k *(2,58q*2,58q)/((6.1d)^(2)) = (k*2.58^(2)*q^(2))/(6.1^(2)*d^(2))` Equation (2)

Solving the equation for d^2

`d^(2)=(k*q^(2))/(20)` Equation (3)

We replace d^2 from equation (3) in equation (2)

`F = (k*2.58^2*q^2)/(6.1^2*(k*q^2)/(20))` (We eliminate k and q^2)

`F = (2.58^2*20)/(6.1^2) = (6.6564*20)/(37.21) = 3.58 N`