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b) Draw electric field lines. Note the 2 charges are not the same magnitude. Note also there are certain points where the direction of the field can be determined easily.

c) If an electron is placed at the point indicated by the open circle on the x axis, what is the magnitude of the force on the electron? An electron's charge is -1.6 x 10^-19 C.

Solutions: a) I found that the Ex=0 N/C because of symmetry making the net field completely in the y direction.

For the y direction I determined that the r would be 2.83 m by using pythagorean's theorem. But i was getting confused on how to add the components of this direction. Would it be something like this: Using E = kq/r^2

E1 = (9x10^9)(2x10^-6)/2.83^2 = 2247.5 N/C E1y=E1sin(49.97 degrees)=1588.39 N/C

E2=(9x10^9)(1x10^-6)/2.83^2=1123.75 N/C E2y=E2sin(49.97 degrees)=-794.19 N/C

Ey=794.2 N/C

b) I wasn't really sure how to go about this other than knowing that electric field lines point towards negative charges, and the 2 charges will repel each other.

c) I wasn't sure how to go about this.