Two point charges are placed on the x axis.(Figure 1) The first charge, q 1 = 8.

Question

Two point charges are placed on the x axis.(Figure 1) The first charge, q1 = 8.00 nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis.

Fig1

Fig2

Part A

Find the electric field at the origin, point O.

Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

Part B

Now, assume that charge q2 is negative; q2=6nC. (Figure 2) What is the net electric field at the origin, point O?

Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

Explanation / Answer

Fig 1

Given charges q1 = 8*10^-9 C at 16 m on x-axis from origin and

   q2 = 6*10^-9 C at 9 m on -ve x-axis from origin
we know that

here
the direction of the electric fields of charge q2 is along +ve x-axis and
the direction of the electric fields of charge q1 is along -ve x-axis and
  
   the electric field at origin is

       E1 = kq1/r1^1
       = 9*10^9*8*10^-9/16^2 N/C

       = 0.281 N/C

         
E1 is negative along -ve direction

and

       E2 = kq2/r2^1
       = 9*10^9*6*10^-9/9^2 N/C

       = 0.667 N/C
      
E2 is positive along +ve direction      

Ex = 0.667-0.281 N/C = 0.386 N/C, so E at origin is E = 0.386 N/C

fig 2

Given charges q1 = 8*10^-9 C at 16 m on x-axis from origin and

   q2 = - 6*10^-9 C at 9 m on -ve x-axis from origin
we know that

here
the direction of the electric fields of charge q2 is along -ve x-axis and
the direction of the electric fields of charge q1 is along -ve x-axis and
  
   the electric field at origin is

       E1 = kq1/r1^1
       = 9*10^9*8*10^-9/16^2 N/C

       = 0.281 N/C

         
E1 is negative along -ve direction

and

       E2 = kq2/r2^1
       = 9*10^9*(-6*10^-9)/9^2 N/C

       = -0.667 N/C

electric field is E = E1+E2 = -0.281-0.667 N/C = -0.948 N/C

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