Located 10 centimeters apart from each other are two charged particles. Both have equal mass. The second object has a charge that is twice as large as the charge on the first object. They experience a repulsive force of 95N due to the electrical interactions. What is the magnitude of the smaller charge?

The mass has nothing to do with this problem since acceleration is not asked for

F = k (2q)q/.1^2

To solve this problem, we can use Coulomb's Law, which states that the magnitude of the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Let's break down the information given:

Distance between the particles, r = 10 cm = 0.1 m (converted to meters)
Force between the particles, F = 95 N
Charge on the first object, q1 = ?
Charge on the second object, q2 = 2q1 (twice as large as q1)

Using Coulomb's Law, the equation for the force is:

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

where k is the electrostatic constant.

Now, let's rearrange the equation to solve for q1:

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

To find q1, we need to know the value of k, which is approximately 8.99 x 10^9 N m^2/C^2.

Plugging in the given values:

q1 = (95 N * (0.1 m)^2) / (8.99 x 10^9 N m^2/C^2 * 2q1)

Simplifying the equation:

q1 = (0.01 N m^2) / (17.98 x 10^9 N m^2/C^2)

Dividing top and bottom of the fraction by 10^9, we get:

q1 = 0.01 / 17.98 C

Calculating the value on the right side of the equation:

q1 ≈ 0.000556 C

Therefore, the magnitude of the smaller charge (q1) is approximately 0.000556 Coulombs.