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Chapter 23: 39 - Excercises And Problems (page 655)

Charges -qand +2qin FIGUREP23.39are located at x=±a. Determine the electric field at points 1-4. Write each field in component form.

Short Answer

Expert verified

The Electric field component is ,E=14πε0q55a2(3i^2j^).

Step by step solution

01

Step: 1 Electric field:

The Electric field charge is

E=14πε0qr2

The following diagram depicts the electromagnetic lines at node one due to these two charges, which can be used to compute the electric field.

02

Step: 2 Finding distance:

Charge from node one as

r12=a2+(2a)2r12=a2+4a2r12=5a2

The distance from charge -qis

r22=a2+(2a)2r22=a2+4a2r22=5a2

03

Step: 3 Finding angle:

The value of sine angle is

sinθ=ar2sinθ=aa5sinθ=15

The value of cosine angle is

cosθ=2ar2cosθ=2aa5cosθ=25

04

Step; 4 Magnitude at node one:

The magnitude at electric field is at node one is

E1=14πε02qr12

E1=14πε02q5a2

The components as

E1x=E1sinθ

Substitutingsinθ

E1x=14πε02q5a215E1x=14πε02q55a2

05

Step: 5 Equating:

The component yas

E1y=14πε02q5a225E1y=14πε04q55a2

E1y=E1cosθ

Substitutingcosθ

E1y=14πε02q5a225E1y=14πε04q55a2

06

Step: 6 Finding component:

The xcomponent at field as

E2x=E2sinθE2x=14πε0q5a215E2x=14πε0q55a2

The ycomponent at field as

E2y=E2cosθE2y=14πε0q5a225E2y=14πε02q55a2

The negative sign denotes field in downward side.

07

Step: 7 Finding nodes:

The field at node two +2qis

Ex,2q=14πε02qa2

The field at node two -qis

Ex,q=14πε0q(3a)2Ex,q=14πε0q9a2

Because these two variables are solely on the xaxis, their ycomponents are zero.

08

Step: 8 Resultant field at node two:

The resultant field at node two is

E=Ex,2q+Ex,qi^

Substituting

E=14πε02qa2+14πε0q9a2i^E=14πε0qa2179i^E=1736πε0qa2i^

09

Step: 9 Resultant field at node three:

The resultant charges is

E=Ex,2q+Ex,qi^

Substituting

E=14πε02q9a214πε0qa2i^E=14πε0qa2291i^E=736πε0qa2i^

The Diagram as

10

Step: 10 Finding charge:

The distance from charge 2qis

r12=a2+(2a)2r12=a2+4a2r12=5a2

The distance from charge -qis

r22=a2+(2a)2r22=a2+4a2r22=5a2

11

Step: 11 Finding angle:

The sine angle as

sinθ=ar2sinθ=aa5sinθ=15

The cosine angle as

cosθ=2ar2cosθ=2aa5cosθ=25

12

Step: 12 Finding component:

Te field at component is

Ex=E1x+E2x

Substituting

Ex=14πε02q55a2+14πε0q55a2Ex=14πε03q55a2

Similarlly,

Ey=14πε04q55a2+14πε02q55a2Ey=14πε02q55a2

13

Step: 13 Finding electric field: 

The net field in component as

E=Exi^+Eyj^

Substituting the values

E=14πε03q55a2i^+14πε02q55a2j^E=14πε0q55a2(3i^2j^).

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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 Eat 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.

A sphere of radius Rand surface charge density ηis positioned with its center distance 2R from an infinite plane with surface charge density η. At what distance from the plane, along a line toward the center of the sphere, is the electric field zero?

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FIGUREP23.41 is a cross section of two infinite lines of charge that extend out of the page. Both have linear charge density λ. Find an expression for the electric field strength E at height y above the midpoint between the lines.

One type of ink-jet printer, called an electrostatic ink-jet printer, forms the letters by using deflecting electrodes to steer charged ink drops up and down vertically as the ink jet sweeps horizontally across the page. The ink jet forms30μm diameter drops of ink, charges them by spraying 800,000 electrons on the surface, and shoots them toward the page at a speed of 20m/s. Along the way, the drops pass through two horizontal, parallel electrodes that are 6.0mmlong,4.0mm wide, and spaced 1.0mm apart. The distance from the center of the electrodes to the paper is 2.0cm. To form the tallest letters, which have a height of 6.0mm, the drops need to be deflected upward (or downward) by 3.0mm. What electric field strength is needed between the electrodes to achieve this deflection? Ink, which consists of dye particles suspended in alcohol, has a density of 800kg/m3 .

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