a point charge q1=+9μc is held fixed at the origin. A second point charge with a charge of-5μC and mass of 3.2 x 10^-4 kg is placed on the + x-axis, 0.90 m from the origin. a.) What is the potential energy of the pair of charges? (consider the potential to be zero at infinite.) b.) if the second point charge is released from rest, what is it's speed 0.24 m from the origin along the x-axis?

To solve this problem, we need to use two concepts: potential energy and conservation of mechanical energy.

a) Calculation of Potential Energy:
The potential energy of a system of two charges is given by the formula:

U = (k * q1 * q2) / r

Where:
- U is the potential energy
- k is Coulomb's constant (k ≈ 8.99 x 10^9 Nm^2/C^2)
- q1 and q2 are the charges of the point charges
- r is the distance between the charges

In this case, q1 = +9μC and q2 = -5μC, and the distance between them is 0.90 m. We need to convert the charges to Coulombs:

q1 = +9μC = 9 x 10^-6 C
q2 = -5μC = -5 x 10^-6 C

Plugging these values into the potential energy formula:

U = (8.99 x 10^9 Nm^2/C^2) * (9 x 10^-6 C) * (-5 x 10^-6 C) / (0.90 m)

Now calculate the potential energy.

b) Calculation of Speed:
To find the speed of the second point charge when it is 0.24 m from the origin, we can use the principle of conservation of mechanical energy. Initially, the system has potential energy U (obtained in part a), and when the charge moves to a distance of 0.24 m, its potential energy will be zero.

The initial mechanical energy (Ei) of the system is the sum of the potential energy (U) and the kinetic energy (Ki) of the second charge:

Ei = U + Ki

When the charge moves to 0.24 m, its potential energy (U) becomes zero, so the total mechanical energy (Ef) remains constant:

Ei = Ef = Ki'

At 0.24 m, the only energy the charge possesses is kinetic energy. Therefore, the final kinetic energy (Ki') can be calculated.

Ki' = Ef = 1/2 * m * v^2

Where:
- Ki' is the final kinetic energy
- m is the mass of the second charge
- v is the speed of the second charge

Now we can substitute the values into the equation and solve for v.

a.) To calculate the potential energy of the pair of charges, we can use the formula:

Potential Energy = (k * |q1 * q2|) / r

Where:
- k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2),
- q1 is the first charge (+9μC),
- q2 is the second charge (-5μC),
- r is the distance between the charges (0.90 m).

Let's substitute the values into the formula:

Potential Energy = (8.99 x 10^9 Nm^2/C^2 * |(+9μC) * (-5μC)|) / 0.90 m

Note: The absolute value is taken because potential energy is a scalar quantity and does not consider the direction of charges.

Calculating the multiplication of the charges:

Potential Energy = (8.99 x 10^9 Nm^2/C^2 * 45 x 10^-12 C^2) / 0.90 m
Potential Energy = 4.5 x 10^-3 Nm

Therefore, the potential energy of the pair of charges is 4.5 x 10^-3 Nm.

b.) To determine the speed of the second point charge when it is 0.24 m from the origin along the x-axis, we can use the principle of conservation of mechanical energy, which states that the initial mechanical energy (potential energy) is equal to the final mechanical energy (kinetic energy).

Initial mechanical energy = Potential energy = 4.5 x 10^-3 Nm

Final mechanical energy = Kinetic energy

The kinetic energy of an object can be calculated using the formula:

Kinetic Energy = (1/2) * m * v^2

Where:
- m is the mass of the second point charge (3.2 x 10^-4 kg),
- v is the speed of the second point charge.

Let's rearrange the formula to solve for v:

v = √[(2 * Kinetic Energy) / m]

v = √[(2 * Final mechanical energy) / m]

Since the final mechanical energy is equal to the potential energy:

v = √[(2 * 4.5 x 10^-3 Nm) / (3.2 x 10^-4 kg)]

Calculating the expression inside the square root:

v = √[28.125]

v ≈ 5.30 m/s

Therefore, the speed of the second point charge when it is 0.24 m from the origin along the x-axis is approximately 5.30 m/s.

a. PE=kqq/r

b. the speed will depend on the KE.
KE=OriginalPE-finalPE
=kqq(1/.9 -1/.66)
(q=(9micro, -.5micro)

then 1/2 mv^2=above, solve for v.