Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00nC is on
ID: 1586446 • Letter: P
Question
Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00nC is on the x -axis at x = 3.00cm. Point P is on the y-axis at y = 4.00cm. A. Calculate the electric fields E 1 and E 2 at point P due to the charges q1 and q2. 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^. 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^, j^.
Explanation / Answer
1) We have given that one Point charge q1= -5.00*10^-9 C is at the origin
Other Point charge q2 = +3.00*10^-9 C is at x= 3*10^-2 m.
The point P is on the y-axis at y= 4*10^-2 m.
Electric field due to q1 at P=E1 =9*10^9*5*10^-9 /(4*10^-2)^2
E1 =85.5 /(16*10^-4)
E1=2.8125*10^4 N/C
As the charge q1 is negative E1 is in negative y direction
In vector notation , E1 =[ 2.8125*10^4 N/C ](-j) ……….Ans.
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The Point charge q2= +3.00nC is on the x-axis at x = 3.00cm.
The distance of q2 from point P = r!=5 cm = 5*10^-2 m
Vector r! = -0.03i^ +0.04j^
The Electric field due to q2at P=E2 =9*10^9*3*10^-9 /(5*10^-2)^2
E2 = 1.08*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*10^4 cosO
Component of E2 =1.08*10^4 *[0.6] =6.48*10^3 N/C
Component of E2 in negative x axis direction=6.48*10^3N/C(-i)
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Component of E2 in +y axis direction =1.08*10^4 sinO
Component of E2 in +y axis direction =1.08*10^4*[0.8]
Component of E2 in +y axis direction =8.64*10^3N/C( j)
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In vector form ,E2 =6.48*10^3N/C(-i)+8.64*10^3N/C( j) ……..Ans.
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b) The resultant field at P, expressed in unit vector form,
E = E1+E2
E= 2.8125*10^4 N/C(-j)+6.48*10^3N/C(-i)+8.64*10^3N/C( j)
E= 28.125*10^3 N/C(-j)+6.48*10^3N/C(-i)+8.64*10^3N/C( j)
The resultant field at P= - 6.48*10^3N/C(i)-19.485*10^3N/C(j)
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Resultant field of magnitude 20.5342*10^3 N/C make angle of 71.6 degree anticlockwise with negative x - axis
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