Ch 25: The Electric Potential

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 25: The Electric Potential

Problem 50b

Chapter 25, Problem 50b

The electric potential in a region of space is given by V=V₀[(x²+2y²)/(0.10 m)²], where V₀ is a constant. A proton released from rest at (x, y)=(20 cm, 0 cm) reaches the origin with a speed of 7.5×10⁵ m/s. At what value of y on the y-axis should a He⁺ ion (charge +e, mass 4 u) be released from rest to reach the origin with the same speed?

Verified step by step guidance

Understand the problem: The electric potential V is given as a function of x and y. A proton and a He+ ion are released from rest at different points and reach the origin with the same speed. The goal is to find the y-coordinate for the He+ ion's release point.

Step 1: Write the expression for the change in electric potential energy (ΔU) for a charge q moving in an electric potential V. The change in potential energy is ΔU = qΔV, where ΔV is the change in electric potential between the initial and final positions.

Step 2: Use the work-energy principle. The change in potential energy is converted into kinetic energy. For a particle of mass m and charge q, the kinetic energy at the origin is given by K = (1/2)m*v². Equate ΔU = K to find the relationship between the potential difference and the speed of the particle.

Step 3: For the proton, substitute its charge (q = +e), mass (m = m_p), and speed (v = 7.5×10^5 m/s) into the energy equation. Use the given potential function V to calculate the potential difference ΔV between the initial position (x = 20 cm, y = 0 cm) and the origin (x = 0, y = 0).

Step 4: For the He+ ion, repeat the process. Substitute its charge (q = +e), mass (m = 4u, where u is the atomic mass unit), and the same speed (v = 7.5×10^5 m/s). Use the potential function V to calculate the y-coordinate on the y-axis where the He+ ion should be released. Solve for y by equating the potential difference ΔV to the kinetic energy of the He+ ion.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

11m

Was this helpful?

Key Concepts

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

Electric Potential (V)

Electric potential, denoted as V, is the amount of electric potential energy per unit charge at a point in an electric field. It is a scalar quantity that indicates the work done to move a charge from a reference point to a specific point in the field without any acceleration. In this problem, the electric potential is given as a function of position, which influences the motion of charged particles like protons and He+ ions.

Kinetic Energy and Conservation of Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated as KE = 0.5 * m * v², where m is mass and v is velocity. The principle of conservation of energy states that the total energy in a closed system remains constant. In this context, the potential energy lost by the charged particle as it moves through the electric field is converted into kinetic energy, allowing us to relate the initial potential energy to the final kinetic energy at the origin.

Motion of Charged Particles in Electric Fields

Charged particles, such as protons and He+ ions, experience a force when placed in an electric field, described by F = qE, where q is the charge and E is the electric field strength. This force causes the particles to accelerate according to Newton's second law, F = ma. The trajectory and final speed of the particles depend on their initial conditions and the characteristics of the electric field, which is defined by the electric potential in this problem.

Recommended video:

Guided course

06:28

Electric Field due to a Point Charge

Related Practice

Textbook Question

A 0.25 pg dust particle with 50 excess electrons is sitting at rest on top of a 5.0-cm-diameter metal sphere. Closing a switch charges the sphere almost instantaneously. To what potential must the sphere be charged to launch the dust particle to a height of 5.0 m? Ignore air resistance.

views

Textbook Question

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.2 cm apart with a 25 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential. With what speed does an electron exit the electron gun if its entry speed is close to zero?

views

Textbook Question

Three electrons form an equilateral triangle 1.0 nm on each side. A proton is at the center of the triangle. What is the potential energy of this group of charges?

views

Textbook Question

A room with 3.0-m-high ceilings has a metal plate on the floor with V=0 V and a separate metal plate on the ceiling. A 1.0 g glass ball charged to +4.9 nC is shot straight up at 5.0 m/s. How high does the ball go if the ceiling voltage is +3.0×10⁶ V?

Textbook Question

What is the escape speed of an electron launched from the surface of a 1.0-cm-diameter glass sphere that has been charged to 10 nC?

Textbook Question

A 2.0-mm-diameter glass bead is positively charged. The potential difference between a point 2.0 mm from the bead and a point 4.0 mm from the bead is 500 V. What is the charge on the bead?

views