A particle (m = 2.0 $\mu$g,q = -5.0 $\mu$C)has a speed of 30 m/s at point A and moves (with only electric forces acting on it)to point B where its speed is 80 m/s.Determine the electric potential difference V_A - V_B. A)-2.2 kV B)+1.1 kV C)-1.1 kV D)+2.2 kV E)+1.3 kV

Question

Quizplus · Accepted Answer

We can start by using the conservation of energy principle. The change in kinetic energy of the particle is equal to the work done by the electric force:

ΔK = W

ΔK = Kf - Ki = (1/2)mvf^2 - (1/2)mvi^2

W = -ΔU = -qΔV

where ΔU is the change in potential energy, ΔV is the change in electric potential, and q is the charge of the particle. Since the charge q is negative, the work done by the electric force is also negative.

Combining these equations, we get:

(1/2)mvf^2 - (1/2)mvi^2 = qΔV

Substituting the given values, we get:

(1/2)(2.0 kg)(80 m/s)^2 - (1/2)(2.0 kg)(30 m/s)^2 = (-5.0 μC) ΔV

Solving for ΔV, we get:

ΔV = (1/5.0 μC) [(1/2)(2.0 kg)(80 m/s)^2 - (1/2)(2.0 kg)(30 m/s)^2]

ΔV = 1.1 kV

The change in electric potential between points A and B is therefore +1.1 kV (i.e., the electric potential increases as the particle moves from A to B). The correct answer is choice C.

A Particle (M = 2 μ\muμ G,q = -50 μ\muμ C)has a Speed of 30 M/s at Point a and C)has

A Particle (M = 2 $\mu$ G,q = -50 $\mu$ C)has a Speed of 30 M/s at Point a and C)has