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A plastic rod with linear charge density λis bent into the quarter circle shown in FIGURE. We want to find the electric field at the origin.

a. Write expressions for the x- and y-components of the electric field at the origin due to a small piece of charge at angle θ.

b. Write, but do not evaluate, definite integrals for the x- and y-components of the net electric field at the origin.

c. Evaluate the integrals and write Enetin component form

Short Answer

Expert verified

(a) Expression for components of field,dEx=kλRanddEx=kλRsinθΔθ

(b) Electrical field at the origin, Ex=kλR0π/2cosθΔθand Ey=kλR0π/2sinθΔθ

(c)Ein component form,E=14πε02QπR2(i^+j^)

Step by step solution

01

Find Expressions for component of field (part a)

The force experienced by such a unit positive charge put at a spot is the electric field intensity at that point. The intensity of an electric field is a vector quantity.

E=kqR2

where Rdenotes distance, kdenotes Coulomb's constant, and qdenotes charge

dE=kΔqR2

and

x=Rθ,Δx=RΔθ

(From figure)

q=λx,Δq=λ(Δx)

(charge in terms of density)

Δx=RΔθand Δq=λ(RΔθ)

Fill in the blanks in the equation with these values.,

dE=kλΔθR

As a result, the electric field's xcomponent will be,

dEx=dEcosθ=kλΔθRcosθ

dEx=kλRcosθΔθ

As a result, the electric field's ycomponent will be,

dEy=kλRsinθΔθ

02

Find Electrical field at the origin (part b)

xcomponent equation,

dEx=kλRcosθΔθ

Integrate this equation,

Ex=kλR0π/2cosθΔθ (1)

ycomponent equation,

dEy=kλRsinθΔθ

Integration of this equation,

Ey=kλR0π/2sinθΔθ (2)

Linear charge density of the rod,

λ=2QπR

substitute this value in equation (1)and (2)

Ex=k2QπR20π/2cosθΔθ

Ey=k2QπR20π/2sinθΔθ

Calculate the integral using both equations,

0π/2cosθdθ=[sinθ]0π/2=+1

0π/2sinθdθ=[-cosθ]0π/2=+1

03

Find electric field in component form

the electric field can be written as:

Enet=Exi^+Eyj^

Expand:

Exi^=14πε02QπR20π/2cosθΔθi^

Eyj^=Exi^=14πε02QπR20π/2sinθΔθj^

the sum of equations will be,

localid="1650218257527" Enet=14πε02QπR2(i^+j^)

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Most popular questions from this chapter

The combustion of fossil fuels produces micron-sized particles of soot, one of the major components of air pollution. The terminal speeds of these particles are extremely small, so they remain suspended in air for very long periods of time. Furthermore, very small particles almost always acquire small amounts of charge from cosmic rays and various atmospheric effects, so their motion is influenced not only by gravity but also by the earth's weak electric field. Consider a small spherical particle of radius r, density ρ, and charge q. A small sphere moving with speed v experiences a drag force Fdrag=6πηrv, where η is the viscosity of the air. (This differs from the drag force you learned in Chapter 6 because there we considered macroscopic rather than microscopic objects.)

a. A particle falling at its terminal speed vtermis in equilibrium with no net force. Write Newton's first law for this particle falling in the presence of a downward electric field of strength E, then solve to find an expression for vterm.

b. Soot is primarily carbon, and carbon in the form of graphite has a density of 2200kg/m3. In the absence of an electric field, what is the terminal speed in mm/s of a 1.0-μm-diameter graphite particle? The viscosity of air at 20°C is 1.8×10-5kg/ms.

c. The earth's electric field is typically (150 N/C , downward). In this field, what is the terminal speed in mm/s of a 1.0 μm-diameter graphite particle that has acquired 250 extra electrons?

What are the strength and direction of the electric field at the position indicated by the dot in FIGURE EX23.2? Specify the direction as an angle above or below horizontal.

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aAt what distance along the z-axis is the electric field strength a maximum?

bWhat is the electric field strength at this point?

Two2.0cmdiameter insulating spheres have a6.0cmspace between them. One sphere is charged to+10nC, the other to-15nC. What is the electric field strength at the midpoint between the two spheres?

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