Answer to Question #126486 in Physics for Tshego

Question

What are the magnitude and direct of the acceleration of an electron at a point where the electric field has magnitude 7200N/C

Answer 1

If the electric field has magnitude $E = 7200N/C$ , then the force acting on the electron at that point will be:

$F = eE$

where $e = -1.6\times 10^{-19}C$ is the electron charge.

According to the Newton's second law, the acceleration of the electron will be:

$a = \dfrac{F}{m} = \dfrac{eE}{m}$

where $m = 9.1\times 10^{-31} kg$ is the mass of electron.

Substituting numerical values, obtain:

$a = \dfrac{eE}{m} = \dfrac{-1.6\times10^{-19}\cdot 7200}{9.1\times10^{-31}} \approx -1.27\times 10^{15}m/s^2$

Sign minus reflects the fact, that the acceleration and electric field directed to the opposite sides.

Answer. -1.27*10^15 m/s^2, opposite to the electric field.