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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

Three charges are placed at the corners of the triangle in FIGURE Q23.15. The ++charge has twice the quantity of charge of the two - charges; the net charge is zero. Is the triangle in equilibrium? If so, explain why. If not, draw the equilibrium orientation.

An electric field can induce an electric dipole in a neutral atom or molecule by pushing the positive and negative charges in opposite directions. The dipole moment of an induced dipole is directly proportional to the electric field. That is, p=αE, where αis called the polarizability of the molecule. A bigger field stretches the molecule farther and causes a larger dipole moment.

a. What are the units of α?

b. An ion with charge qis distancerfrom a molecule with polarizability α. Find an expression for the force Fionondipole.

An electret is similar to a magnet, but rather than being permanently magnetized, it has a permanent electric dipole moment. Suppose a small electret with electric dipole moment 1.0×10-7Cm is 25cmfrom a small ball charged to +25nC, with the ball on the axis of the electric dipole. What is the magnitude of the electric force on the ball?

Your physics assignment is to figure out a way to use electricity to launch a small 6.0-cm-long plastic drink stirrer. You decide that you’ll charge the little plastic rod by rubbing it with fur, then hold it near a long, charged wire, as shown in FIGURE P23.56. When you let go, the electric force of the wire on the plastic rod will shoot it away. Suppose you can uniformly charge the plastic stirrer to 10nCand that the linear charge density of the long wire is 1.0×10-7C/m. What is the net electric force on the plastic stirrer if the end closest to the wire is 2.0cmaway?

Hint: The stirrer cannot be modeled as a point charge; an integration is required.

11. II A small glass bead charged to +6.0nCis in the plane that bisects a thin, uniformly charged, 10-cm-long glass rod and is 4.0cmfrom the rod's center. The bead is repelled from the rod with a force of 840μN. What is the total charge on the rod?

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