An object with charge qo = 1.5μC and mass 0.7 g starts from rest at the +6V equipotential line. Calculate its change in potential energy and speed when it reaches the +2V line.

The acquired kinetic energy equals the potential energy drop, which equals the charge times the voltage decrease. That will be in Joules.

Then use Kinetic Energy = (1/2) M V^2 to solve for V

To calculate the change in potential energy and speed of the object, we can use the formulas:

Change in potential energy (ΔPE) = qΔV

Where q is the charge of the object, and ΔV is the change in voltage.

Speed (v) can be calculated using the conservation of energy:

Potential energy (PE) + Kinetic energy (KE) = Total mechanical energy (E)

Initially, the object is at the +6V equipotential line. At this position, we can say that the initial potential energy is 0 since there is no change in potential when starting from a potential line.

Let's calculate the change in potential energy and then the speed when it reaches the +2V line.

Given:
qo = 1.5 μC (microCoulombs)
m = 0.7 g (grams)
ΔV = -4V (potential difference from +6V to +2V)

First, let's convert the charge to Coulombs and the mass to kilograms for consistency:
q = qo * 10^(-6) = 1.5 * 10^(-6) C
m = 0.7 * 10^(-3) kg

Now we can calculate ΔPE:
ΔPE = q * ΔV
= (1.5 * 10^(-6) C) * (-4V)
= -6 * 10^(-6) J

The change in potential energy is -6 * 10^(-6) Joules.

Next, let's calculate the speed of the object.

Since the initial potential energy is 0, we have:
PE + KE = E
0 + KE = E
Therefore, the total mechanical energy, E, is equal to the kinetic energy, KE.

We can write the equation for kinetic energy as:
KE = (1/2)mv^2

Rearranging the equation, we can find v:
v = √(2KE / m)

Substituting the known values:
v = √(2 * |-6 * 10^(-6)| J / 0.7 * 10^(-3) kg)

Taking the absolute value of ΔPE since kinetic energy cannot be negative:
v = √(12 * 10^(-6) J / 0.7 * 10^(-3) kg)
v = √(17.14 m^2/s^2)
v ≈ 4.14 m/s

The speed of the object when it reaches the +2V line is approximately 4.14 m/s.

To calculate the change in potential energy and speed of the object, we'll need to use the equations related to electric potential energy and kinetic energy.

1. First, let's calculate the change in potential energy (ΔPE):
- The formula for electric potential energy is PE = q * ΔV, where PE is the potential energy, q is the charge, and ΔV is the change in voltage.
- In this case, the charge q = 1.5 μC, and the change in voltage ΔV = +6 V - +2 V = 4 V.
- Substitute these values into the formula: ΔPE = q * ΔV = (1.5 μC) * (4 V).

Note: We need to convert the charge from microcoulombs to coulombs, and the final result will be in joules.

2. Next, let's calculate the speed (v) of the object at the +2V line:
- The formula for kinetic energy is KE = (1/2) * m * v^2, where KE is the kinetic energy, m is the mass, and v is the velocity of the object.
- In this case, the mass m = 0.7 g, which needs to be converted to kilograms (kg).
- Since the object starts from rest, the initial kinetic energy is 0.
- The final kinetic energy is equal to the change in potential energy (ΔPE) calculated in the previous step.
- Substitute the values into the formula: (1/2) * m * v^2 = ΔPE.

Now let's perform the calculations:

1. Converting charge from microcoulombs to coulombs:
1.5 μC = 1.5 x 10^(-6) C

ΔPE = (1.5 x 10^(-6) C) * (4 V) = 6 x 10^(-6) J (joules)

2. Converting mass from grams to kilograms:
0.7 g = 0.7 x 10^(-3) kg

(1/2) * (0.7 x 10^(-3) kg) * v^2 = 6 x 10^(-6) J

Rearranging the equation:
v^2 = (2 * 6 x 10^(-6) J) / (0.7 x 10^(-3) kg)

Calculating the square root of both sides:
v = √((2 * 6 x 10^(-6) J) / (0.7 x 10^(-3) kg))

v ≈ 0.032 m/s (rounded to three decimal places)

Therefore, the change in potential energy is 6 x 10^(-6) J (joules), and the speed of the object when it reaches the +2V line is approximately 0.032 m/s.