A 50 µC charged sphere of mass 0.050kg is stationary and 2.0 m away from a -30µC charged sphere. The 50µC charged sphere released and allowed to approach the other. When it is 1.2 m away, what is the speed of this sphere?
To solve this problem, we can use the principle of conservation of mechanical energy. We assume that there is no external force acting on the charged sphere system, hence the total mechanical energy is conserved.
The mechanical energy of the system is given by:
E = K + U
Where:
E is the total mechanical energy,
K is the kinetic energy of the charged sphere, and
U is the electrical potential energy.
Since the charged sphere starts from rest, its initial kinetic energy is zero. Thus, the initial mechanical energy is equal to the initial electrical potential energy:
Ei = Ui
When the charged sphere is 1.2 m away from the other sphere, we can calculate the final mechanical energy (Ef) and electrical potential energy (Uf) using the given information.
Ef = Kf + Uf
The electrical potential energy can be calculated using the formula:
U = k * (|q1 * q2|) / r
Where:
k is the electrostatic constant (9 x 10^9 Nm^2/C^2),
q1 and q2 are the charges of the spheres (50µC and -30µC respectively),
and r is the distance between the spheres (1.2 m).
Substituting the values:
Uf = (k * |50µC * -30µC|) / 1.2 m
Next, we equate the initial electrical potential energy (Ui) to the final mechanical energy (Ef) to calculate the final kinetic energy:
Ui = Ef
Since the initial kinetic energy is zero, we have:
Ui = Kf + Uf
Substituting the values, solve for Kf:
50µC * V = (k * |50µC * -30µC|) / 1.2 m
Solve for V (speed) by rearranging the equation:
V = ((k * |50µC * -30µC|) / (1.2 m)) / 50µC
V = (k * |50µC * -30µC|) / (1.2 m * 50µC)
Now, substitute the given values:
V = (9 x 10^9 Nm^2/C^2 * |50µC * -30µC|) / (1.2 m * 50µC)
After performing the calculations, you will find the speed of the sphere when it is 1.2 m away from the other sphere.