An electron is shot horizontally with a speed of 350 m/s into a uniform electric

Question

An electron is shot horizontally with a speed of 350 m/s into a uniform electric field of 6.20 N/C directed downward, as shown, 2.00 mu s after it has entered the field, how far in the vertical direction will the electron have moved? (Indicate whether it moves upward or downward.) Two protons are placed 10.0 cm away from each other, as shown. Find the electric field in unit-vector notation at point * 40.0 cm to the right of the origin. One of the protons from the previous part of the problem is released from rest. How fast is this proton moving when it has traveled a distance of 30.0 cm away from its starting position?

Explanation / Answer

given,

distance between protons = 10 cm

distance between point and origin = 40 cm

electric field = kq/r^2

electric field = 9 * 10^9 * 1.6 * 10^-19 / (0.4^2 + 0.05^2)

electric field = 8.861 * 10^-9 N/C

angle between two electric field = tan^-1(5/40)

angle between two electric field = 7.125 degree

x component = 2 *  8.861 * 10^-9 * cos(7.125)

x component = 1.758 * 10^-8 N/C

y component = 0 N/C

electric field in unit vector notation = 1.758 * 10^-8i + 0j

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