A clear plastic pen 12 cmlong is rubbed all over with wool, and acquires a negative charge of -2nC. You want to figure out the electric field a distance of 18 mmfrom the pen, near the middle of the pen. (a) You decide to model the pen as a rod consisting of a series of five segments, each of which you will consider to be approximately point-like. What is the length of each segment in meters? (b) What is the amount of charge Qon each of the five segments? (c) In general, if the rod has a length Land total charge Q, and you divide the rod into Nsegments, what is the amount of chargeQon each piece? (d) If the length of each segment is dL, write a symbolic expression for the number of pieces Nin terms of the length of the rod Land the length of one piece dL.(e) Now write a symbolic expression for the amount of charge on each piece in terms of the length of the rod and the length of a small piece.

Short Answer

Expert verified

(a) The length of each segment is 0.024 m .

(b) The amount of charge in each of the five segments is -4×10-10C.

(c) The amount of charge Qon each piece is QN.

(d) The symbolic expression for the number of piecesN is LdL.

(e) The symbolic expression for the amount of charge on each piece is Q=QLdL.

Step by step solution

01

Identification of the given data

The given data is listed below as:

  • The length of the plastic pen is, l=12cm.
  • The plastic pen has acquired change of Q=-2nC.
  • The electric field is at a distance of 18 mm from the pen.
02

Significance of the electric field

The electric field is type of a region that is beneficial for the electrically charged particle to exert a particular force on another particle.

In this problem, the concept of the electric field gives the length of each segment, and amount of charge on each piece.

03

(a) Determination of the length of each segment

The equation of the length of each segment can be expressed as:

l1=l5

Here, lis the length of the plastic pen andl1 is the length of each segment.

Substitute 12 cm forl in the above equation.

l1=12cm51m100cm=0.12m5=0.024m

Thus, the length of each segment is 0.024m.

04

(b) Determination of the amount of charge of each of the five segments

The equation of the amount of charge in each of the five segment is expressed as:

Q1=Q5

Here,Q is the charge of the plastic pen andQ1 is the amount of charge on each segment

Substitute -2nCfor Qin the above equation.

Q1=-2nC5=-2nC51×10-9C1nC=-2×10-9C5=-4×10-10C

Thus, the amount of charge in each of the five segments is -4×10-10C.

05

(c) Determination of the amount of charge on each piece

The algebraic expression for the charge per unit length that is divided into segments can be expressed as:

L=LN …(i)

Here, Lis the expression for the charge per unit length, Lis the total length of the rod and Nis the number of segments.

The expression for the definition for the charge per unit length is expressed as:

Q=QLL

Here, Qis the amount of the charge, Qis the total charge, Lis the expression for the charge per unit length and Lis the total length of the rod.

SubstituteLN for Lin the above equation.

Q=QL×LN=QN …(ii)

Thus, the amount of charge Qon each piece is QN.

06

(d) Determination of the symbolic expression for the number of pieces N

From the equation (i), the equation of the number of pieces can be expressed as:

N=LL

Here, Lis the expression for the charge per unit length, Lis the total length of the rod andN is the number of segments.

Substitute LNfor Lin the above equation.

N=LdL …(iii)

Thus, the symbolic expression for the number of pieces Nis LdL.

07

(e) Determination of the symbolic expression of the charge on each piece

From the equation (ii), the equation of the amount of charge on each piece can be expressed as:

Q=QN

Here,Q is the amount of the charge, Qis the total charge andN is the number of pieces of the rod.

Substitute the value ofN forLdLas obtained from the equation (iii) in the above equation.

Q=QLdL=QLdL

Thus, the symbolic expression for the amount of charge on each piece is Q=QLdL.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that the radius of a disk is R=20, and the total charge distributed uniformly all over the disk isrole="math" localid="1656058758873" Q=6×10-6C. Use the exact result to calculate the electric fieldfrom the center of the disk, and alsofrom the center of the disk. Does the field decrease significantly?

Two rings of radius5Cm are24 apart and concentric with a common horizontal x axis. The ring on the left carries a uniformly distributed charge of+31nC , and the ring on the right carries a uniformly distributed charge of-31nC. (a) What are the magnitude and direction of the electric field on the x axis, halfway between the two rings? (b) If a charge of-9nC were placed midway between the rings, what would be the force exerted on this charge by the rings?

A thin plastic spherical shell of radius 5 cmhas a uniformly distributed charge of -25nCon its outer surface. A concentric thin plastic spherical shell of radius 8 cmhas a uniformly distributed charge of+64nC on its outer surface. Find the magnitude and direction of the electric field at distances of, 3 cm, 7 cm and 10 cmfrom the center. See Figure 15.63.

If the magnitude of the electric field in air exceeds roughly 3 × 106 N/C, the air brake down and a spark form. For a two-disk capacitor of radius 47 cm with a gap of 1 mm, what is the maximum charge (plus and minus) that can be placed on the disks without a spark forming (which would permit charge to flow from one disk to the other)?

If the magnitude of the electric field in air exceeds roughly3×106N3, the air break down and a spark forms. For a two-disk capacitor of radius 51 cm with a gap of 2 mm, if the electric field inside is just high enough that a spark occurs, what is the strength of the fringe field just outside the center of the capacitor?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free