Question

What is the magnitude of the acceleration experienced by an electron in an electric field of 1200 N/C? How does the direction of the acceleration depend on the direction of the field at that point?

Answer 1

Explanation:

It is given that,

Electric field experienced by an electron, $E=1200\ N/C$

Charge on electron, $q=-1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ C$

The electric force is balanced by the force acting on electron as,

ma = qE

$a=\dfrac{qE}{m}$

$a=\dfrac{-1.6\times 10^{-19}\times 1200}{9.1\times 10^{-31}}$

$a=-2.1\times 10^{14}\ m/s^2$

So, the acceleration of the electron is $-2.1\times 10^{14}\ m/s^2$. As the electron is a negatively charged particle, the field lines is from negative to positive charge.