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Chapter 23: Q. 38- Excercises And Problems (page 655)

FIGUREP23.38shows three charges at the corners of a square. Write the electric field at point Pin component form.

Short Answer

Expert verified

The Electric filed at point isP=140QL2(21)[i^+j^].

Step by step solution

01

Step: 1 Electric field:

The Electric fiels at point charge is

E=140qr2r^

Electric field lines due to provided energies on a position Pand associated resolution along the x,yaxes are depicted in the diagram below.

02

Step: 2 Solving:

The electric field at point Ais,

E1=E1(i^)

The electric field magnitude of charge is

E1=140QL2

Substituting and solving,

E1=140QL2(i^)E1=140QL2i^

03

Step: 3 Solvingfield at B:

The electric filed at corner Bis,

E2=E2xi^+E2yj^

The distance BPfrom the diagram is

BP=BC2+CP2

Substituting L=BC=CP

BP=L2+L2BP=2L

The magnitude at charge Bis

E2=1404QBP2

Substituting BP=2L

E2=1404Q(2L)2E2=1404Q2L2

04

Step; 4 Equating:

The horizontal component is

E2x=E2cos45

Substituting,

E2x=1404Q2L2cos45E2x=1404Q22L2

The vertical component is

E2y=E2sin45

Substituting,

E2y=1404Q2L2sin45E2y=1404Q22L2

05

Step: 5 Expression field:

The electric charge at corner is

E3=E3(j^)

The magnitude at charge is

E3=140QL2

Substituting

E3=140QL2(j^)

06

Step: 6 Net electric field:

The resultant horizontal direction is

Ex=E1+E2x

The net electric field is

E=Exi^+Eyj^

solving,

E=140QL2+1404Q22L2i^+140QL2+1404Q22L2j^E=140QL2(1+2)i^+140QL2(1+2)j^E=140QL2(21)(i^+j^)

The net electric filed isP=140QL2(21)[i^+j^].

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Most popular questions from this chapter

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=E, where is 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 ?

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

An infinite plane of charge with surface charge density 3.2C/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?

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

a. What is the ratio f/i , where fis 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?

A 2.0-mm-diameter glass sphere has a charge of +1.0nC. What speed does an electron need to orbit the sphere 1.0mm above the surface?

A parallel-plate capacitor is formed from two 6.0-cm-diameter electrodes spaced 2.0mmapart. The electric field strength inside the capacitor is 1.0106N/C. What is the charge (in nC ) on each electrode?

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