Two identical small charged spheres, each having a mass of 5.0x102 kg and having the

magnitudes of -3µC and -4µC, hang in equilibrium. The angle θ is
15.0°. Find distance that separates the charge.( Important constants: Charge 8.99x109
, acceleration due gravity 9.8m/s2).

1 answer

We can use Coulomb's law to find the force between the two charges:

F = k * q1 * q2 / r^2

where k is Coulomb's constant (8.99x10^9), q1 and q2 are the magnitudes of the charges (-3µC and -4µC), and r is the distance between them.

Since the spheres are in equilibrium, the net force on each sphere must be zero. We can draw a free body diagram for one of the spheres:

- There is a downward gravitational force of mg (where m is the mass of the sphere and g is the acceleration due to gravity)
- There is an upward electrostatic force from the other sphere

The angle θ is given in the diagram, so we can split the electrostatic force into horizontal and vertical components:

F_horizontal = F * sin(θ)
F_vertical = F * cos(θ)

Since the sphere is in equilibrium, the vertical forces must balance:

F_vertical = mg

We can solve for F using this equation, and then use the horizontal component to find the distance between the charges:

F = mg / cos(θ)

r = √(k * q1 * q2 / F)

Plugging in the given values:

F = (5.0x10^2 kg) * (9.8 m/s^2) / cos(15°) ≈ 2486 N

r = √(8.99x10^9 N*m^2/C^2 * (-3x10^-6 C) * (-4x10^-6 C) / 2486 N) ≈ 0.0186 m

Therefore, the distance between the charges is approximately 0.0186 meters.