Part A Determine the magnitude of the electric field E at the origin 0 in the fi

Question

Part A

Determine the magnitude of the electric field E at the origin 0 in the figure(Figure 1) due to the two charges at A and B.

Express your answer in terms of the variables Q, l, k, and appropriate constants.

Part B

Determine the direction of the electric field E at the origin 0 in the figure due to the two charges at A and B.

Part C

Repeat A, but let the charge at B be reversed in sign.

Express your answer in terms of the variables Q, l, k, and appropriate constants.

Part D

Repeat B, but let the charge at B be reversed in sign.

Part A

Determine the magnitude of the electric field E at the origin 0 in the figure(Figure 1) due to the two charges at A and B.

Express your answer in terms of the variables Q, l, k, and appropriate constants.

Part B

Determine the direction of the electric field E at the origin 0 in the figure due to the two charges at A and B.

Part C

Repeat A, but let the charge at B be reversed in sign.

Express your answer in terms of the variables Q, l, k, and appropriate constants.

Part D

Repeat B, but let the charge at B be reversed in sign.

Explanation / Answer

EAx = 0

EAy = -k*Q/l^2

EBx = -k*Q*cos30/l^2

EBy = -k*Q*sin30/l^2 = -k*Q*/2l^2

Ex = EAx + EBx = -k*Q*sqrt3/2l^2

Ey = EAy + EBy = -sqrt3k*Q/2l^2

E =sqrt(Ex^2+Ey^2)

E = sqrt3*k*Q/l^2

++++++++++++++

part B)

direction = tan^-1(Ey/Ex) = 210 degrees counter clockwise with +x axis

_____________

part C)

EAx = 0

EAy = -k*Q/l^2

EBx = k*Q*cos30/l^2

EBy = k*Q*sin30/l^2 = k*Q/2l^2

Ex = EAx + EBx = k*Q*sqrt3/2l^2

Ey = EAy + EBy = -k*Q/2l^2

E =sqrt(Ex^2+Ey^2)

E = k*Q/l^2

(d)

direction = 30 with + x axis

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