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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=14πε0QL2(21)[i^+j^].

Step by step solution

01

Step: 1 Electric field:

The Electric fiels at point charge is

E=14πε0qr2r^

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=14πε0QL2

Substituting and solving,

E1=14πε0QL2(i^)E1=14πε0QL2i^

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=14πε04QBP2

Substituting BP=2L

E2=14πε04Q(2L)2E2=14πε04Q2L2

04

Step; 4 Equating:

The horizontal component is

E2x=E2cos45

Substituting,

E2x=14πε04Q2L2cos45E2x=14πε04Q22L2

The vertical component is

E2y=E2sin45

Substituting,

E2y=14πε04Q2L2sin45E2y=14πε04Q22L2

05

Step: 5 Expression field:

The electric charge at corner is

E3=E3(j^)

The magnitude at charge is

E3=14πε0QL2

Substituting

E3=14πε0QL2(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=14πε0QL2+14πε04Q22L2i^+14πε0QL2+14πε04Q22L2j^E=14πε0QL2(1+2)i^+14πε0QL2(1+2)j^E=14πε0QL2(21)(i^+j^)

The net electric filed isP=14πε0QL2(21)[i^+j^].

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