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FIGURE shows a thin rod of length Lwith total charge Q. Find an expression for the electric fieldE→ at point P. Give your answer in component form.

Short Answer

Expert verified

Electrical field in component form,E=Q4πϵon×1n2+L2i-Q4πϵoLn1-nn2-L2j^

Step by step solution

01

Electrical charge

A fundamental feature of matter held by some constituent particles that affects how the particles react to an external magnetic field Positive and negative electric charge exists in separate natural units and cannot be manufactured or destroyed.

02

Evaluate the field

At point, take a small amount of energy due to the axis charge electrical field.

Ey=14πϵo×QLdy×yn2+y23/2

Applying Integration to the equation,

Ey=14πϵo×QL∫0Lydyn2+y23/2

Substituting the limits;

Ey=Q4πϵoL-1n2+y2

Ey=Q4πϵoL-1n2+L2+1n

Ey=-Q4πϵo2n1-n2L2+L2j^

03

Explain by diagram

Applying the equation,

En=Ecosθ

cosθ=nn2+y2

En=14πϵo×Q2×dy×nn2+y23/2

Applying Integration,

En=14πϵo×QLn∫0Ldyn2+y23/2

Substituting the limits value,

En=Q4πϵo×1nn2+L2i^

E=E→n+E→y

Applying the vectors value,we get

E=Q4πϵon×1n2+L2i-Q4πϵoLn1-nn2-L2j^

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Most popular questions from this chapter

Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in FIGURE. 23.47

a. Find an expression for the electric field E→at the center of the semicircle.

Hint: A small piece of arc length Δsspans a small angle Δθ=Δs/R, where Ris the radius.

b. Evaluate the field strength if localid="1651169583117" L=10cmand localid="1651169587457" Q=30nC.

What are the strength and direction of the electric field at the position indicated by the dot in FIGURE EX23.1? Specify the direction as an angle above or below horizontal

FIGURE is a cross section of two infinite lines of charge that extend out of the page. The linear charge densities are ±λ. Find an expression for the electric field strength Eat height yabove the midpoint between the lines.

An electron traveling parallel to a uniform electric field increases its speed from 2.0×107m/sto4.0×107m/s over a distance of 1.2cm. What is the electric field strength?

A problem of practical interest is to make a beam of electrons turn a 90°corner. This can be done with the parallel-plate capacitor shown in FIGURE. An electron with kinetic energy 3.0×10-17Jenters through a small hole in the bottom plate of the capacitor.

a. Should the bottom plate be charged positive or negative relative to the top plate if you want the electron to turn to the right? Explain.

b. What strength electric field is needed if the electron is to emerge from an exit hole 1.0cmaway from the entrance hole, traveling at right angles to its original direction?

Hint: The difficulty of this problem depends on how you choose your coordinate system.

c. What minimum separation dminmust the capacitor plates have?

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