Ch 23: The Electric Field

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 23: The Electric Field

Problem 27b

Chapter 23, Problem 27b

Two 2.0-cm-diameter disks face each other, 1.0 mm apart. They are charged to ±10 nC. A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?

Verified step by step guidance

Determine the electric field between the two disks. Since the disks are parallel and closely spaced, the electric field can be approximated as uniform. Use the formula for the electric field between parallel plates:

E = \frac{σ}{ε}

₀, where

σ

is the surface charge density and

ε

₀ is the permittivity of free space. First, calculate

σ = \frac{Q}{A}

, where

Q

is the charge on one disk and

A

is the area of the disk.

Calculate the work done on the proton by the electric field as it moves from the negative disk to the positive disk. The work done is equal to the change in electric potential energy, which can be expressed as

W = q E d

, where

q

is the charge of the proton,

E

is the electric field, and

d

is the separation between the disks.

Relate the work done to the proton's initial kinetic energy. The proton must have just enough initial kinetic energy to overcome the electric potential energy difference. Use the work-energy principle:

K

_i = W, where

K

_i is the initial kinetic energy of the proton. The kinetic energy is given by

K

_i = 12mv², where

m

is the mass of the proton and

v

is its initial velocity.

Set the initial kinetic energy equal to the work done:

\frac{1}{2} m v

² = qEd. Solve for the initial velocity

v

v = \sqrt{\frac{2}{m} q E d}

Substitute the known values into the equation. Use the charge of the proton

q = 1.6 × 10

^-19 C, the mass of the proton

m = 1.67 × 10

^-27 kg, the electric field calculated in step 1, and the separation distance

d = 1.0 × 10

^-3 m. This will give the required launch speed.

Verified video answer for a similar problem:

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

Was this helpful?

Key Concepts

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

Electric Field

The electric field is a vector field surrounding charged particles that exerts a force on other charged objects. It is defined as the force per unit charge experienced by a positive test charge placed in the field. In this scenario, the electric field between the two charged disks will influence the motion of the proton as it moves from the negative to the positive disk.

Coulomb's Law

Coulomb's Law describes the force between two point charges. It states that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This law is essential for calculating the electric force acting on the proton as it travels through the electric field created by the charged disks.

Kinetic and Potential Energy

Kinetic energy is the energy of an object due to its motion, while potential energy is the stored energy based on its position in a field, such as an electric field. As the proton moves from the negative disk to the positive disk, it converts its initial kinetic energy into electric potential energy. To find the required launch speed, one must equate the kinetic energy of the proton to the potential energy it gains in the electric field.

Recommended video:

Guided course

06:35

Gravitational Potential Energy

Related Practice

Textbook Question

FIGURE EX23.25 shows a $1.5$ g ball hanging from a string inside a parallel-plate capacitor made with 12 cm × 12 cm electrodes. The electrodes are charged to±75 nC. What is the charge on the ball in nC?

views

Textbook Question

A proton is fired horizontally into a 1.0×10⁵ N/C vertical electric field. It rises 1.0 cm vertically after having traveled 5.0 cm horizontally. What was the proton's initial speed?

views

Textbook Question

Air 'breaks down' when the electric field strength reaches 3.0×10⁶ N/C, causing a spark. A parallel-plate capacitor is made from two 4.0 cm×4.0 cm electrodes. How many electrons must be transferred from one electrode to the other to create a spark between the electrodes?

views

Textbook Question

Electrostatic cleaners remove small dust particles and pollen grains from air by first ionizing them, then flowing the air between the plates of a parallel-plate capacitor, parallel to the plates, where electric forces deposit charged particles on one of the electrodes. A typical pollen grain has a mass of $5.0 \times 10^{- 10}$ g, the ionizer charges it with $750$ extra electrons, and a fan moves the air at $3.0$ m/s. Ignore air resistance and gravity. What minimum electric field strength is needed to deflect the grain by $3.0$ mm before it leaves the electrodes?

views

Textbook Question

You’ve hung two very large sheets of plastic facing each other with distance d between them, as shown in FIGURE EX23.19. By rubbing them with wool and silk, you’ve managed to give one sheet a uniform surface charge density $η_{1} = - η_{0}$ and the other a uniform surface charge density $eta sub 2 equals positive 3 eta sub 0$ . What are the electric field vectors at points 1, 2, and 3?

views

Textbook Question

INT The surface charge density on an infinite charged plane is −2.0×10⁻⁶ C/m². A proton is shot straight away from the plane at 2.0×10⁶ m/s. How far does the proton travel before reaching its turning point?

views