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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

What are the strength and direction of the electric field at the position indicated by the dot in FIGUREP23.37? Give your answer (a)in component form and (b)as a magnitude and angle measured cwor ccw(specify which) from the positive x-axis.

An infinite plane of charge with surface charge density 3.2μC/m2has a 20-cm-diameter circular hole cut out of it. What is the electric field strength directly over the center of the hole at a distance of 12cm?

Hint: Can you create this charge distribution as a superposition of charge distributions for which you know the electric field?

You have a summer intern position with a company that designs and builds nanomachines. An engineer with the company is designing a microscopic oscillator to help keep time, and you’ve been assigned to help him analyze the design. He wants to place a negative charge at the center of a very small, positively charged metal ring. His claim is that the negative charge will undergo simple harmonic motion at a frequency determined by the amount of charge on the ring.

a. Consider a negative charge near the center of a positively charged ring centered on the z-axis. Show that there is a restoring force on the charge if it moves along the z-axisbut stays close to the center of the ring. That is, show there’s a force that tries to keep the charge at z=0. b. Show that for small oscillations, with amplitude <<R, a particle of mass mwith charge-qundergoes simple harmonic motion with frequency f=12πqQ4πε0mR3,RandQare the radius and charge of the ring.

c. Evaluate the oscillation frequency for an electron at the center of a 2.0μmdiameter ring charged to 1.0×10-13C.

Two10cmdiameter charged rings face each other,20cmapart. Both rings are charged to+20nC. What is the electric field strength at (a) the midpoint between the two rings and (b) the center of the left ring?

The irregularly shaped area of charge in FIGURE Q23.7 has surface charge densityηi. Each dimension (x and y) of the area is reduced by a factor of 3.163.

a. What is the ratio ηf/ηi , where ηfis the final surface charge density?

b. An electron is very far from the area. What is the ratioFf/Fi of the electric force on the electron after the area is reduced to the force before the area was reduced?

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