Two charges q 1 = −3.20 nC and q2 = +8.96 nC are at a distance of 1.94 µm from each other. q 1 is fixed at its location while q 2 is released from rest.
(a) What is the kinetic energy of the charge q 2 when it is 0.340 µm from q 1?
J
(b) The charge q 2 has a mass m 2 = 7.85 µg. What is its speed when it is 0.340 µm from q 1?
m/s
First, we will use conservation of energy to solve part (a). The initial potential energy (PE_initial) is transformed into final potential energy (PE_final) and kinetic energy (KE_final).
PE_initial = KE_final + PE_final
We start by finding the 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 distance between the charges.
PE_initial = (8.99 x 10^9) * (-3.20 x 10^-9 C) * (8.96 x 10^-9 C) / (1.94 x 10^-6 m)
PE_initial = -126.82 x 10^-6 J
PE_initial = -126.82 μJ
Now, we find PE_final with the new distance between the charges (0.340 μm):
PE_final = (8.99 x 10^9) * (-3.20 x 10^-9 C) * (8.96 x 10^-9 C) / (0.340 x 10^-6 m)
PE_final = -71.14 x 10^-6 J
PE_final = -71.14 μJ
Now using conservation of energy and solving for KE_final:
KE_final = PE_initial - PE_final
KE_final = -126.82 μJ + 71.14 μJ
KE_final = -55.68 μJ
So the kinetic energy of q2 when it is 0.340 μm from q1 is:
KE_final = -55.68 μJ
Now, we'll find the speed of q2 for part (b). We'll use the formula for kinetic energy:
KE = 0.5 * m * v^2
Where m is the mass of q2 and v is its speed. We will rearrange the formula for v:
v = sqrt (2 * KE / m)
Substitute the values:
v = sqrt (2 * (-55.68 x 10^-6 J) / (7.85 x 10^-6 kg))
v = sqrt (-14.16)
v = -3.765 m/s
But the speed must be positive, so we take the absolute value:
v = 3.765 m/s
Therefore, the speed of q2 when it is 0.340 μm from q1 is 3.765 m/s.
To solve this problem, we can use the principle of conservation of mechanical energy, which states that the total mechanical energy of a system remains constant if no external forces are acting on it. In this case, the only force acting on the charge q2 is the electrostatic force from q1.
(a) To find the kinetic energy of q2, we first need to calculate the potential energy at that position. The potential energy between two charges can be calculated using the formula:
PE = k * q1 * q2 / r
where k is the electrostatic constant (9 × 10^9 Nm^2/C^2), q1 and q2 are the charges, and r is the distance between them.
Given:
q1 = -3.20 nC
q2 = +8.96 nC
r = 0.340 µm = 0.340 × 10^-6 m
Plugging in the values, we have:
PE = (9 × 10^9 Nm^2/C^2) * (-3.20 nC) * (+8.96 nC) / (0.340 × 10^-6 m)
Calculating this, the potential energy is:
PE = -275.29 J
Now, remember that at this position, all of the potential energy is converted into kinetic energy. Therefore, the kinetic energy of q2 when it is 0.34 µm from q1 is also -275.29 J.
(b) To find the speed of q2, we can use the fact that kinetic energy is equal to (1/2) * m * v^2, where m is the mass and v is the speed.
Given:
m2 = 7.85 µg = 7.85 × 10^-9 kg
KE = -275.29 J
Plugging in the values, we have:
-275.29 J = (1/2) * (7.85 × 10^-9 kg) * v^2
Rearranging the equation, we can solve for v:
v^2 = (-2 * -275.29 J) / (7.85 × 10^-9 kg)
Calculating this, we find:
v^2 = 70.085 × 10^15 m^2/s^2
Taking the square root of both sides, we get:
v = 8.37 × 10^7 m/s
Therefore, the speed of q2 when it is 0.34 µm from q1 is approximately 8.37 × 10^7 m/s.
To find the kinetic energy of charge q2 when it is 0.340 µm from charge q1 and its speed, we need to use the concepts of electrical potential energy and kinetic energy.
First, let's calculate the electrical potential energy between the two charges using the formula:
PE = (k * |q1 * q2|) / r,
where k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
Substituting the given values:
PE = (8.99 x 10^9 Nm^2/C^2 * |-3.20 nC * +8.96 nC|) / (1.94 µm)
Next, we'll calculate the potential energy of q2 when it is 0.340 µm from q1, using the same formula:
PE2 = (k * |q1 * q2|) / r2,
where r2 is the new distance between the charges.
Substituting the given values:
PE2 = (8.99 x 10^9 Nm^2/C^2 * |-3.20 nC * +8.96 nC|) / (0.340 µm)
The change in potential energy (∆PE) is given by the difference between the initial and final potential energies:
∆PE = PE - PE2
Now, we'll use the concept that the change in potential energy is equal to the kinetic energy:
∆PE = KE
So, the kinetic energy of q2 when it is 0.340 µm from q1 is ∆PE.
To find the speed of q2, we'll use the formula:
KE = (1/2) * m * v^2,
where m is the mass of q2 and v is its velocity (speed in this case).
Using the formula ∆PE = KE, we can substitute ∆PE in the above formula to find v.
(a) To calculate the kinetic energy of q2 when it is 0.340 µm from q1, substitute the values in the formula and calculate ∆PE.
(b) To calculate the speed of q2 when it is 0.340 µm from q1, substitute the values of ∆PE and m in the formula and solve for v.