A small particle has charge −5.00-5.00 μμC and mass 2.00×10−42.00\$$times10^{-4} kg. It $$moves from point AA, where the electric potential is VA=+200V_A = +200 V, to point BB, where the electric potential is VB=+800V_B = +800 V. The electric force is the only force acting on the particle. The particle has speed 5.005.00 m/s at point AA. What is its speed at point BB? Is it moving faster or slower at BB than at AA? Explain.

Ch 23: Electric Potential

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 23: Electric Potential

Problem 13

Chapter 23, Problem 13

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

Verified step by step guidance

First, understand that the problem involves the conservation of energy, specifically the conservation of mechanical energy, which includes kinetic and electric potential energy.

Calculate the initial kinetic energy at point A using the formula:

K_{A} = \frac{1}{2} m v^{2}

, where

m

is the mass and

v

is the speed at point A.

Calculate the initial electric potential energy at point A using the formula:

U_{A} = q V_{A}

, where

q

is the charge and

V_{A}

is the electric potential at point A.

Apply the conservation of energy principle: the total mechanical energy at point A is equal to the total mechanical energy at point B. Therefore,

K_{A} + U_{A} = K_{B} + U_{B}

, where

K_{B}

and

U_{B}

are the kinetic and potential energies at point B.

Solve for the speed at point B using the rearranged equation:

K_{B} = K_{A} + U_{A} - U_{B}

, and then use

K_{B} = \frac{1}{2} m v^{2}

to find the speed

v

at point B. Compare the speeds at points A and B to determine if the particle is moving faster or slower at point B.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Potential Energy

Electric potential energy is the energy a charged particle possesses due to its position in an electric field. It is calculated using the formula U = qV, where q is the charge and V is the electric potential. As the particle moves from point A to point B, the change in electric potential energy will affect its kinetic energy and thus its speed.

Conservation of Energy

The conservation of energy principle states that the total energy of an isolated system remains constant. In this scenario, the particle's initial kinetic energy and electric potential energy at point A will equal its final kinetic energy and electric potential energy at point B, allowing us to solve for the particle's speed at point B.

Kinetic Energy

Kinetic energy is the energy of motion, given by the formula KE = 0.5mv^2, where m is mass and v is velocity. As the particle moves from point A to point B, changes in electric potential energy will result in changes in kinetic energy, affecting the particle's speed. By comparing kinetic energies at both points, we can determine if the particle is moving faster or slower.

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