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

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

Step by step solution

01

Electric field vector

Given,

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

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=9×109×5×10-94×10-22=2.8×104N/C

Now the direction of the field is in

Therefore,E1=-2.8×104j^N/C

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

E2=kq2r22=kq2x2+y22=9×109×3×10-90.32+0.4222=1.08×104N/C

02

Calculating the field vectors

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

θ1=arctan43=53.130

Now, the electric field vector is given by:

E2=E1-cosθi+sinθ=1.08×104-cos53.13i^+sin53.13j^=-6.48×103i^+8.64×103j^N/m

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

Now the net electric field is:

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

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