Two identical +9.0 µC point charges are initially spaced 5.2 cm from each other. If they are released at the same instant from rest, how fast will they be moving when they are very far away from each other? Assume they have identical masses of 1.0 mg.
Potential energy of the charges is
PE=k•q^2/r=9 •10^9•(9•10^-6)^2/0.052=14 Joules.
According to the Law of conservation of energy
PE=2KE
KE=mv^2/2.
2KE= mv^2 =PE
v=sqroot(PE/m)=sqroot(14/10^-3)=118.3 m/s.
To determine the final velocity of the two charges when they are very far away from each other, we can use the principle of conservation of energy. Initially, the charges are at rest, so their initial kinetic energy is zero.
The final kinetic energy of the charges, when they are very far away from each other, will also be zero since they will come to a stop due to the mutual electrostatic repulsion.
The electrostatic potential energy, U, between two point charges can be given by the equation:
U = k * (q1 * q2) / r
Where:
- k is the Coulomb's constant, approximately equal to 9 × 10^9 N m^2/C^2
- q1 and q2 are the charges
- r is the distance between the charges
Since the charges are identical (+9.0 µC), we can substitute the values into the equation:
U = (9 × 10^9 N m^2/C^2) * ((9.0 × 10^-6 C)^2) / (0.052 m)
Simplifying the equation:
U = (9 × 10^9 N m^2/C^2) * (81 × 10^-12 C^2) / (0.052 m)
U = (9 × 81 × 10^-3) * (10^9 / 0.052) N m
U = (729 × 10^-3) * (10^9 / 0.052) N m
U = (729 × 10^6 / 0.052) N m
U ≈ 140,192,307 N m
Since the final kinetic energy is zero, the total mechanical energy of the system is equal to the potential energy. Therefore, the mechanical energy is also 140,192,307 N m.
The mechanical energy, E, can be expressed as:
E = (1/2) * (m1 * v1^2) + (1/2) * (m2 * v2^2)
Where:
- m1 and m2 are the masses of the charges (1.0 mg each)
- v1 and v2 are their respective final velocities
Substituting the values into the equation:
140,192,307 N m = (1/2) * (1.0 × 10^-6 kg) * (v1^2) + (1/2) * (1.0 × 10^-6 kg) * (v2^2)
Since the masses and charges are identical, v1 = v2 = v:
140,192,307 N m = (1/2) * (1.0 × 10^-6 kg) * (v^2) + (1/2) * (1.0 × 10^-6 kg) * (v^2)
140,192,307 N m = (1.0 × 10^-6 kg) * v^2
Simplifying the equation:
v^2 = (140,192,307 N m) / (1.0 × 10^-6 kg)
v^2 ≈ 140,192,307,000 m^2/s^2
Taking the square root of both sides:
v ≈ √(140,192,307,000 m^2/s^2)
v ≈ 374,121 m/s
Therefore, when the charges are very far away from each other, they will have a final velocity of approximately 374,121 m/s in opposite directions.
To find the final speed of the charges when they are very far away from each other, we can make use of the principle of conservation of energy.
The charges will initially possess potential energy due to their electrical potential energy. As they move apart, this potential energy is converted into kinetic energy. When they are very far away from each other, the potential energy will be zero and all the initial potential energy will be converted into kinetic energy.
First, let's calculate the initial potential energy of the system using the formula:
Potential Energy (PE) = (k * q1 * q2) / r
Where k is the electrostatic constant (k = 8.99 x 10^9 N * m^2 / C^2), q1 and q2 are the charges, and r is the separation distance between the charges.
Given that the charges are identical and have a magnitude of +9.0 µC = 9.0 x 10^-6 C, and the separation distance is 5.2 cm = 0.052 m, we can calculate the initial potential energy:
PE = (8.99 x 10^9 N * m^2 /C^2) * (9.0 x 10^-6 C)^2 / 0.052 m
= 1.1598 J
Now, since the potential energy is converted into kinetic energy, we can equate the initial potential energy to the final kinetic energy:
PE = 1/2 * m * v^2
Where m is the mass of each charge (1.0 mg = 1.0 x 10^-6 kg) and v is the final velocity of the charges.
Substituting the values:
1.1598 J = 1/2 * (1.0 x 10^-6 kg) * v^2
v^2 = (2 * 1.1598 J) / (1.0 x 10^-6 kg)
v^2 = 2.3196 x 10^6 m^2/s^2
Finally, taking the square root of both sides to solve for v:
v = sqrt(2.3196 x 10^6 m^2/s^2)
v ≈ 1522.6 m/s
Therefore, when the charges are very far apart, they will be moving with a final velocity of approximately 1522.6 m/s.