A small particle has charge -5.00 μC and mass 2.00 ×10^-4 kg . It moves from point A, where the

A small particle has charge -5.00 μC and mass 2.00 ×10^-4 kg...

A small particle has charge $-5.00 \mu \mathrm{C}$ and mass $2.00 \times$ $10^{-4} \mathrm{kg} .$ It moves from point $A,$ where the electric potential is $V_{A}=+200 \mathrm{V},$ to point $B,$ where the electric potential is $V_{B}=+800 \mathrm{V} .$ The electric force is the only force acting on the particle. The particle has speed 5.00 $\mathrm{m} / \mathrm{s}$ at point $A .$ What is its speed at point $B ?$ Is it moving faster or slower at $B$ than at $A$? Explain.

Key Concepts

Conservation of Mechanical Energy

This principle states that in the absence of non-conservative forces (such as friction), the total mechanical energy of a system, which is the sum of its kinetic and potential energies, remains constant. In problems involving charged particles in an electric field, any changes in electric potential energy manifest as changes in kinetic energy.

Electric Potential Energy

Electric potential energy is the energy a charged particle has due to its position in an electric field. It is calculated by multiplying the charge of the particle by the electric potential at a point. Changes in electric potential energy, as the particle moves between points with different potentials, result in a corresponding change in kinetic energy according to energy conservation.

Electric Potential Difference

The electric potential difference, or voltage, between two points in an electric field gives the work done per unit charge when moving a charge between these points. In energy conversion problems, this difference is crucial because it provides the driving force for the acceleration or deceleration of the particle.

Work-Energy Principle in Electric Fields

This concept connects the work done by the electric force on a charged particle with the change in its kinetic energy. When a particle moves in an electric field, the work done (or energy transferred) is equal to the product of the charge and the potential difference, leading to a gain or loss in kinetic energy depending on the sign of the charge and the potential gradient.

Transcript

00:01 So for this question, we're told that a small particle has a charge of 5 times 10 to the minus 6 coulomes.

00:19 And it has a mass of 2 .0 times 10 to the minus 4 coulomes or kilograms.

00:32 I'm going to fix this funny exponent.

00:36 So this should be a 6.

00:47 And it moves from a where the electric potential is 200 volts to b where it's 800 volts.

00:59 So va is 200, and vb is equal to 800 volts.

01:14 And the force is acting on the particle, and then this particle has a speed of, so it's 5 meters per second at a, and then our goal is to get what is the speed at b.

01:30 So i'll write that down.

01:32 Oh, this is confusing.

01:33 So this is little v at a is equal to 5 .00 meters per second.

01:42 And then our goal is to get little v at b equals question mark.

01:47 So if we have a charge moving within this potential, let's see, just going to check, is there any other information about the potential? so is it, is a potential changing linearly? well, i guess you don't even need that information because of this sort of path independence thing.

02:18 So for this problem, i think you want to more so use the principle.

02:23 That the change in potential energy, the magnitude of the change of the potential energy.

02:32 Actually, we shouldn't, we're probably not use magnitude since we care about if it's losing or gaining.

02:43 So the work done on a particle, i'll go back to that, the work done on a particle is equal.

02:53 I guess there's a couple ways to think about it.

02:55 So i'm going to think about it as there's, there's no, like i like to start off energy problems with the statement that w net external.

03:13 So the net external work done on a system can contribute to a change in kinetic energy or a change in potential energy.

03:22 So i'm going to say that the electric force is internal.

03:26 So i'll say that this thing is zero.

03:28 And so we get the equation that the change in kinetic energy is equal to minus the change in potential energy.

03:39 And so the, so let's go ahead and fill in either these sides.

03:43 So the changing kinetic energy, so it's starting at a, so it's one half m, v at a squared, or sorry, it ends actually.

03:53 So it should be changing kinetic energy is final minus initial...

Please give Ace some feedback