Question

An electric field of strength E = 4700 N/C is directed along the +x-axis as shown.

An electric field of strength E = 4700 N/C is directed along the +x-axis as shown. What is the magnitude of the potential difference between point B and point A if the distance between these two points is 10 cm? An electron is initially at rest at point B. How much kinetic energy (in electron-Volts, or eV) will the electron gain by moving to the higher potential at point A? What is the kinetic energy of the electron at point A measured in Joules?

Answer 1

a)

Relation between electric field and potential difference is given as,

$E = \left | \frac{dV}{dx} \right |$

So the potential difference between the points A and B is,

$dV = E \times dx$

$dV = 4700 \times 0.1$

b) Change in energy is given by,

$dw= q \times dV$

$dw= e \times 470V = 470 eV$

c) Kinetic energy in Joules

$dw=1.602\times 10^{-19}\times 470 = 7.53\times 10^{-17}J$