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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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