Electric Field due to Two Point Charges Two point charges are placed on the x ax

Question

Electric Field due to Two Point Charges

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.

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.

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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.

Figure 1Figure 2 of 2

Electric Field due to Two Point Charges

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.

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.

EOx,EOy =   N/C  

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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.

EOx,EOy =   N/C  

Figure 1Figure 2 of 2

6 nC O (0 m, 0 m) 42 (-9 m, 0 m) 8 nC 91 6 m, 0 m)

Explanation / Answer

q1 = 8 x 10-9 C

r1 = distance of q1 from origin = 16 m

q2 = 6 x 10-9 C

r2 = distance of q2 from origin = 9 m

E1 = electric field due to charge q1 = - k q1/r12 = - (9 x 109) (8 x 10-9 )/(16)2 = - 0.28125 N/C towards Left

E2 = electric field due to charge q2 = k q2/r22 = (9 x 109) (6 x 10-9 )/(9)2 = 0.67 N/C towards right

Net electric field is given as

E = E1 + E2 = - 0.28125 + 0.67 = 0.38875 N/C

b)

q1 = 8 x 10-9 C

r1 = distance of q1 from origin = 16 m

q2 = - 6 x 10-9 C

r2 = distance of q2 from origin = 9 m

E1 = electric field due to charge q1 = - k q1/r12 = - (9 x 109) (8 x 10-9 )/(16)2 = - 0.28125 N/C towards Left

E2 = electric field due to charge q2 = k q2/r22 = - (9 x 109) (6 x 10-9 )/(9)2 = - 0.67 N/C towards left

Net electric field is given as

E = E1 + E2 = - 0.28125 - 0.67 = - 0.95125 N/C

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