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Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00 nC is on the x-axis at x = 3.00 cm. Point P is on the y-axis at y = 4.00 cm. (a) Calculate the electric fieldsandat point P due to the charges q1 and q2. Express your results in terms of unit vectors (see Example 21.6). (b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form.

Short Answer

Expert verified

Answer:

(a) The electric field E1 and E2 at the point P due to the point q1 and q2 is ,

E1=-2.8ร—104j^N/C,-6.48ร—103i^+8.64ร—103j^N/m

(b) The resultant field at P isrole="math" localid="1658483325561" -6.48ร—103i^+1.94ร—104j^N/m

Step by step solution

01

Electric field vector

Given,

q1=-5.0nCq2=3.0nCx=3.0ร—10-2my=4.0ร—10-2m

The electric field due to the charge is,

E1=kq1r12

Now, the electric field due to the charge at origin is given by:

E1=kq1r12=9ร—109ร—5ร—10-94ร—10-22=2.8ร—104โ€ŠN/C

Now the direction of the field is in

Therefore,E1=-2.8ร—104j^N/C

Now, the electric field due to the second charge is given by:

E2=kq2r22=kq2x2+y22=9ร—109ร—3ร—10-90.32+0.4222=1.08ร—104โ€ŠN/C

02

Calculating the field vectors

Therefore, the angle and the electric field and the horizontal is:

ฮธ1=arctan43=53.130

Now, the electric field vector is given by:

E2=E1-cosฮธi+sinฮธ=1.08ร—104-cos53.13i^+sin53.13j^=-6.48ร—103i^+8.64ร—103j^N/m

Therefore, the resultant field at P is-6.48ร—103i^+8.64ร—103j^N/m

Now the net electric field is:

E=E1+E2=-2.8ร—104j^-6.48ร—103i^+8.64ร—103j^=-6.48ร—103i^-1.94ร—104j^N/m

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Most popular questions from this chapter

Ordinary household electric lines in North America usually operate at 120 V . Why is this a desirable voltage, rather than a value considerably larger or smaller? On the other hand, automobiles usually have 12 V electrical systems. Why is this a desirable voltage?

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current in the wire? (b) What is the magnitude of thedrift velocity of the

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Fig. E25.30.

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