Question

A uniform electric field of magnitude 375 n/c pointing in the positive x - direction acts on an electron, which is initially at rest. after the electron has moved 3.20 cm, what is (a) the work done by the field on the electron, (b) the change in potential energy associated with the electron, and (c) the velocity of the electron?

Answer 1

(a) The force exerted by the electric field on the electron is given by the product between the electron charge q and the intensity of the electric field E:
$F=qE=(1.6 \cdot 10^{-19}C)(375 N/C)=6\cdot 10^{-17}N$
Under the action of this force, the electron moves by:
$\Delta x = 3.20 cm=0.032 m$
And the work done by the electric field on the electron is equal to the product between the magnitude of the force and the displacement of the electron. The sign has to be taken as positive, because the direction of the force is the same as the displacement of the electron, so:
$W=F \Delta x= (6\cdot 10^{-17}N)(0.032 m)=1.9 \cdot 10^{-18}J$

(b) The electron is initially at rest and it starts to move under the action of the electric field. This means that as it moves, it acquires kinetic energy and it loses potential energy. The change in potential energy is the opposite of the work done by the electric field:
$\Delta U = U_f - U_i = -1.9 \cdot 10^{-18} J$
Where Uf and Ui are the final and initial potential energy of the electron.

(c) For the conservation of energy, the sum of the kinetic energy and potential energy of the electron at the beginning of the motion and at the end must be equal:
$U_i + K_i = U_f + K_f$ (1)
where Ki and Kf are the initial and final kinetic energies.
The electron is initially at rest, so Ki =0, and we can rewrite (1) as 
$U_i - U_f = - \Delta U = K_f = \frac{1}{2}m_e v_f^2$
and by using the mass of the electron me, we can find the value of the final velocity of the electron:
$v_f= \sqrt{ -\frac{2 \Delta U}{m_e} }= \sqrt{- \frac{2(-1.9 \cdot 10^{-18} J)}{9.1 \cdot 10^{-31} kg} } =2.04 \cdot 10^6 m/s$