Question

An electron is traveling through a uniform electric field. The field is constant and given by E = 2.00 x 10-11 N/C i + (-1.20 x 10^-11N/C j. At t = 0, the electron is at the origin and traveling in the x direction with a speed of 2.10 m/s. What is its position 1.80 s later? Express the position as coordinates (x,y).

Answer 1

Answer:

Explanation:

The electric field is given by

$\overrightarrow{E}=2\times 10^{-11}\widehat{i}-1.20\times 10^{-11}\widehat{j}$

mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration.

u = 2.10 m/s along X axis

time, t = 1.8 s

charge of electron, q = 1.6 x 10^-19 C

Force on electron, F = q E

$\overrightarrow{F}=1.6\times 10^{-19}\left (2\times 10^{-11}\widehat{i}-1.20\times 10^{-11}\widehat{j} \right )$

$\overrightarrow{F}=\left (3.2\times 10^{-30}\widehat{i}-1.92\times 10^{-30}\widehat{j} \right )$

a = F / m

$\overrightarrow{a}=\frac{\left (3.2\times 10^{-30}\widehat{i}-1.92\times 10^{-30}\widehat{j} \right )}{9.1\times 10^{-31}}$

$\overrightarrow{a} = 3.52\widehat{i}-2.12\widehat{j}$

Use second equation of motion

s = ut + 1/2 at²

$\overrightarrow{s} = 2.10\widehat{i}\times 1.8+0.5\times \left (3.52\widehat{i}-2.12\widehat{j} \right )\times 1.8^{2}$

$\overrightarrow{s} = 9.48\widehat{i}-3.43\widehat{j}$

s = (9.48, - 3.43) m