previous 5of5 return to assignme Exercise 21.44 Part A Pont charge qi=-500nClSat

Question

previous 5of5 return to assignme Exercise 21.44 Part A Pont charge qi=-500nClSatthe orgnand pant charge 300mCE onthe z-ans at z- 3.00cm Point P is on the y-axis at y 4.00em Calculate the electric fields Ea and Ea at point P due to the charges gh and h Express your results in terms of unit vectors (see example 21.6 in the textbook) Express your answer in terms of the unit vectors i, j. Enter your answers separated by a comma h, N/C Submit My Answers Give Up Incorrect, Try Again; 7 attempts remaining Part B Use the results of part (a) to obtain the resultant field at P. expressed in unit vector form Express your answer in terms of the unit vectors i N/C Submit My Answers Give Up Incorrect: Try Again; 9 attempts remaining Provide Feedback C

Explanation / Answer

1)

q1= -5 x 10^-9 C is at the origin

q2 = 3 x 10^-9 C is at x = 3 x 10^-2 m.

Point P is on the y-axis at y = 4 x 10^-2 m.

Electric field due to q1at P = E1 = (9 x 10^9 x 5 x 10^-9) / (4 x 10^-2)^2

E1 = 45 / (16 x 10^-4)

E1 = 2.8125 x 10^4 N/C

As charge q1 is negative E1 is in negative y direction

In vector notation , E1 = [ 2.8125 x 10^4 N/C ](-j)

Point charge q2= +3 nC is on the x-axis at x = 3 cm.

Distance of q2 from point P = r! = 5 cm = 5 x 10^-2 m

Vector r! = -0.03i^ +0.04j^

Electric field due to q2 at P = E2 = (9 x 10^9 x 3 x 10^-9) / (5 x 10^-2)^2

E2 = 1.08 x 10^4 N/C in a direction from q2 to P making angle O with negative x axis

tan(O) = 4/3

Angle = O = 53 degree,

cos53 = 0.6, and sin53 = 0.8

Component of E2 in negative x axis direction =1.08 x 10^4 cosO

Component of E2 =1.08 x 10^4*[0.6] = 6.48 x 10^3 N/C

Component of E2 in negative x axis direction = 6.48 x 10^3 N/C(-i)

Component of E2 in +y axis direction = 1.08 x 10^4 sinO

Component of E2 in +y axis direction = 1.08 x 10^4*[0.8]

Component of E2 in +y axis direction = 8.64 x 10^3 N/C( j)

In vector form,E2 = 6.48 x 10^3 N/C(-i) + 8.64 x 10^3 N/C( j)

b)

The resultant field at P, expressed in unit vector form,

E = E1 + E2

E = 2.8125 x 10^4 N/C(-j) + 6.48 x 10^3 N/C(-i) + 8.64 x 10^3 N/C( j)

E = 28.125 x 10^3 N/C(-j) + 6.48 x 10^3 N/C(-i) + 8.64 x 10^3 N/C( j)

The resultant field at P = - 6.48 x 10^3N/C(i) - 19.485 x 10^3 N/C(j)

magnitude = sqrt[( 6.48 x 10^3)^2 + (19.485 x 10^3)^2] = 20.5342 x 10^3 N/C

angle = tan^-1(-19.485 / -6.48) = 71.6 degree.

Resultant field of magnitude 20.5342 x 10^3 N/C makes an angle of 71.6 degree anticlockwise with negative x - axis.

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