Four electric charges are arranged as shown. Each charge is placed a distance d

Question

Four electric charges are arranged as shown. Each charge is placed a distance d = 3.0 cm from the origin, and the values of thechargesareq1 =1.5C,q2 =3.0C,q3 =4.5C,and

q4 = 6.0 C.

(a) Determine the magnitude of the electric field E at the origin.

(b) Determine the direction of the electric field E at the origin. Express your answer as an angle , measured counterclockwise from the +x direction.

Now an electron of mass m = 9.11×10-31 kg and charge q = 1.60 ×10-19 C is placed at the origin and released from rest.
(c) Calculate the magnitude the initial acceleration of the electron.

(d) Determine the direction of the initial acceleration of the electron.

Explanation / Answer

All charges are +ve, note q4 =4q1 q3 =3 q1 q2 =2q1

Field in the +ve x Ex = (3-1)x1.5x10-6 x9x109/(3 x10-2 )2 =2 x 1.5x107 N/C

Field in the +ve y direction=(4-2)x1.5x10-6 X9X109 /(3x10-2 )=2x1.5x107 N/C

Both the Ex & Ey fields are equal= 3X 107 N/C will be at an angle 450 to +ve X-axis

Electric field E=(Ex2 + Ey2 )=1.414x3x107 N/C

magnitude of Force on electron =1.414x3x107 x1.6x10-19 N

Acceleration= 1.414x3x107 x1.6x10-19 N/9.11x10-31 m/s2 =.745x 1019 m/s2

direction is opposite to the direction of the field ;450 anti clock wise from - ve x -axix

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