A charge Q = +4.5 uC 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 uC 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 (change in VAB = VB-VA). (b) Calculate the speed of the small object when it reaches point B

To calculate the potential difference between point B and point A, we need to use the equation:

ΔV = VB - VA

(a) Calculate the potential difference between point B and point A (change in VAB = VB-VA):

Step 1: Calculate the electric potential at point A (VA) due to the charge Q at the origin.

The formula for the electric potential due to a point charge is given by:

V = kQ / r

where V is the electric potential, k is the Coulomb's constant (9 x 10^9 N m^2/C^2), Q is the charge, and r is the distance.

Given that the charge Q is +4.5 uC (microCoulombs) and located at the origin, the distance between point A and the origin is 2.0 cm. Converting the charge to Coulombs gives Q = 4.5 x 10^-6 C.

Substituting these values, we have:

VA = kQ / r
= (9 x 10^9 N m^2/C^2) x (4.5 x 10^-6 C) / (2.0 x 10^-2 m)
= 2.025 x 10^3 V

So, VA = 2.025 x 10^3 V.

Step 2: Calculate the electric potential at point B (VB) also due to the charge Q at the origin.

Since point B is further along the x-axis from the origin, the distance between point B and the origin is 10.0 cm. Converting this to meters, we have r = 10.0 x 10^-2 m.

Using the formula for electric potential, we have:

VB = kQ / r
= (9 x 10^9 N m^2/C^2) x (4.5 x 10^-6 C) / (10.0 x 10^-2 m)
= 4.5 x 10^2 V

So, VB = 4.5 x 10^2 V.

Step 3: Calculate the potential difference:

ΔV = VB - VA
= (4.5 x 10^2 V) - (2.025 x 10^3 V)
= -1.575 x 10^3 V

Therefore, the potential difference between point B and point A is -1.575 x 10^3 V.

(b) Calculating the speed of the small object at point B:

The potential difference between two points in an electric field can be related to the change in kinetic energy of a charged object. The equation is given by:

ΔK = qΔV

where ΔK is the change in kinetic energy, q is the charge of the object, and ΔV is the potential difference.

Since the small object is released from rest, it will have an initial kinetic energy of zero.

We know that the charge q is +8.0 uC (microCoulombs) and the potential difference ΔV is -1.575 x 10^3 V. Converting the charge to Coulombs gives q = 8.0 x 10^-6 C.

Substituting these values, we have:

ΔK = qΔV
= (8.0 x 10^-6 C) x (-1.575 x 10^3 V)
= -1.26 x 10^-2 J

The change in kinetic energy is negative because the potential difference is negative.

The change in kinetic energy is also equal to the final kinetic energy (Kf) minus the initial kinetic energy (Ki):

ΔK = Kf - Ki

Since the object starts from rest, the initial kinetic energy (Ki) is zero.

Therefore:

-1.26 x 10^-2 J = Kf - 0
Kf = -1.26 x 10^-2 J

The speed (v) is related to kinetic energy (K) by:

K = (1/2)mv²

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

Rearranging the equation, we get:

v = √(2K / m)

Substituting the values, we have:

v = √(2 x (-1.26 x 10^-2 J) / (2.0 x 10^-3 kg)
= √(-6.3) m/s (since the kinetic energy is negative)

As the square root of a negative number is imaginary, we conclude that the speed of the small object when it reaches point B is imaginary.