An electron is released from rest in a uniform electric fiel

An electron is released from rest in a uniform electric field. After travelling a distance of 0.001 m the electron has kinetic energy 1.6 times 10^-10J. Find the change in potential experienced by the electron. Find the magnitude of the electric field. Sketch the electron\'s displacement and indicate the direction of the electric field and the direction of increasing potential. A point charge q = -3 mu C is located at the origin. Calculate the electric field vector at the location r = -3i - 4j cm.

Solution

a)

Kinetic energy = potential difference * charge of the particle

So, KE = V*q

where q = charge of the electron = 1.6*10^-19 C

So, V = KE/q = 1.6*10^-19/1.6*10^-19 = 1 V <-----answer

b)

Potential difference, V = Electric field(E) * distance moved(d)

or, V = E*d

So, E = V/d = 1/0.001 = 1000 V/m

c)

If the charge is moving to the right, then Electric field is directed to the left and if you take the charge to be moving left then Electric field is towards right

d)

E = k*q/r^2

= 9*10^9*(3*10^-6)/(0.03^2 +0.04^2)

= 1.08*10^7 N/C

direction = atan(4/3)

= 53.1 deg  Counter clockwise from the +x axis