Find the electric field at the origin, point O. Give the x and y components of t

Question

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 point to the right is positive and a y component that upward is positive. Now assume that change q_2 = -6 nC. What is the next electric field at the origine, point? 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 components that point upward is positive.

Explanation / Answer

Charges q1 = 8 nC = 8x10 -9 C

            q2 = 6 nC = 6 x10 -9 C

Electric field at origin due to charge q1 is E1 = Kq1/x 2

Where K = Coulomb's constant = 8.99 x10 9 Nm 2/ C 2

           x = Distance between origin and q1

              = 16 m

Substitutue values you get E1 = ( 8.99 x10 9 )( 8x10 -9 )/16 2

                                              = 0.2809 N/C

It is along negative x direction

So, E1 = 0.2809 (-i) N/C

Where - i = unit vector along negative x axis

Electric field at origin due to charge q2 is E2 = Kq2/x ' 2

Where K = Coulomb's constant = 8.99 x10 9 Nm 2/ C 2

           x ' = Distance between origin and q2

              = 9 m

Substitutue values you get E2 = ( 8.99 x10 9 )( 6x10 -9 )/9 2

                                              = 0.6659 N/C

It is along positive x direction.

So, E2 = 0.6659 (i) N/C

Where i = unit vector along positive x axis

So, net electric field at origin E = E1+E2

                                              = -0.2809 i +0.6659 i

                                              = 0.385 i

Resultant is along positive x axis.

(b).

Electric field at origin due to charge q1 is E1 = Kq1/x 2

Where K = Coulomb's constant = 8.99 x10 9 Nm 2/ C 2

           x = Distance between origin and q1

              = 16 m

Substitutue values you get E1 = ( 8.99 x10 9 )( 8x10 -9 )/16 2

                                              = 0.2809 N/C

It is along negative x direction

So, E1 = 0.2809 (-i) N/C

Where - i = unit vector along negative x axis

Electric field at origin due to charge q2 is E2 = Kq2/x ' 2

Where K = Coulomb's constant = 8.99 x10 9 Nm 2/ C 2

           x ' = Distance between origin and q2

              = 9 m

Substitutue values you get E2 = ( 8.99 x10 9 )( 6x10 -9 )/9 2

                                              = 0.6659 N/C

It is also along negative x direction.

So, E2 = 0.6659 (-i) N/C

Where - i = unit vector along negative x axis

So, net electric field at origin E = E1+E2

                                              = -0.2809 i -0.6659 i

                                              = -0.9468 i N/C

resultant is along negative x axis

Resultant is along positive x axis.

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