charge Q = +4.5 ƒÝC is located at the origin and is fixed so that it cannot move. A small object with mass m = 2.0 g and charge q = +8.0 ƒÝC is released from rest at point A on the x-axis at x = 2.0 cm. The small object moves without friction along the x-axis under the influence of the charge Q only (gravity can be neglected) and eventually reaches point B on the x-axis at x = 10.0 cm. (a) Calculate the potential difference between point B and point A (ƒ´VAB = VB ƒ{ VA). (b) Calculate the speed of the small object when it reaches point B

To solve this problem, we need to consider both the gravitational potential energy and the electrostatic potential energy.

(a) Calculating the potential difference:

The potential difference between two points is given by the formula:

ΔV = VB - VA

Since point A is at the origin (x = 0) and point B is at x = 10.0 cm, we need to calculate the potential at each point.

At point A, the potential is due to the charge Q. The electrostatic potential energy is given by:

PE = k * (|Q1| * |Q2|) / r

where k is the Coulomb constant (9.0 x 10^9 Nm^2/C^2), |Q1| and |Q2| represent the magnitudes of the charges, and r is the distance between the charges.

In this case, |Q1| = 4.5 μC and |Q2| = 8.0 μC, and the distance between them is 2.0 cm = 0.02 m.

PE_A = k * (|Q| * |q|) / r = (9.0 x 10^9 Nm^2/C^2) * ((4.5 x 10^-6 C) * (8.0 x 10^-6 C)) / 0.02 m

Now, the potential at point A is given by the formula:

VA = PE_A / q

where q is the charge of the small object.

VA = PE_A / q = [((9.0 x 10^9 Nm^2/C^2) * ((4.5 x 10^-6 C) * (8.0 x 10^-6 C)) / 0.02 m)] / (8.0 x 10^-6 C)

Similarly, we can calculate the potential at point B:

PE_B = k * (|Q| * |q|) / r = (9.0 x 10^9 Nm^2/C^2) * ((4.5 x 10^-6 C) * (8.0 x 10^-6 C)) / 0.1 m

VB = PE_B / q = [((9.0 x 10^9 Nm^2/C^2) * ((4.5 x 10^-6 C) * (8.0 x 10^-6 C)) / 0.1 m)] / (8.0 x 10^-6 C)

Finally, we can calculate the potential difference:

ΔV = VB - VA

(b) Calculating the speed at point B:

The change in potential energy of the small object as it moves from A to B is equal to the change in kinetic energy:

ΔPE = ΔKE

The change in kinetic energy can be written as:

ΔKE = (1/2) * m * v^2

where m is the mass of the small object and v is its speed.

Using the principle of conservation of energy, we can equate the change in potential energy to the change in kinetic energy:

ΔPE = (1/2) * m * v^2

Substituting the potential difference, ΔV, as the change in potential energy:

q * ΔV = (1/2) * m * v^2

Now we can rearrange the equation to solve for v:

v = √[(2 * q * ΔV) / m]

Substituting the previously calculated values for ΔV, q, and m will give us the speed of the small object at point B.