Question

An electron is moving in a uniform horizontal electric field. If the acceleration of electron is ${45}^{\circ }$ downward to the horizontal, then the field intensity in newton/coloumb will be

Answer 1

R.E.F image

Solution It $\theta ={45}^{\circ }\Rightarrow Fe=mg$

$F=\frac{mg}{e}$

For electron,

$F=\frac{(9.1\times {10}^{-31})\left(9.8\right)}{1.6\times {10}^{-19}}$

$F=55.737\times {10}^{-12}n/c$

Note : Direction of ${\overrightarrow{F}}_{net}$ and

$\overrightarrow{a}$ net are same according to

Newton's second low of motion.