Question

An electron with a speed of $5\times {10}^6$ $\text{m/s}$ enters an electric field of magnitude ${10}^3$ $\text{N/C}$, travelling along the field lines in the direction that retards its motion. How far will the electron travel in the field before stopping momentarily ?

• A. $4.9$ $\text{cm}$
• B. $5.9$ $\text{cm}$
• C. $6.9$ $\text{cm}$
• D. $7.9$ $\text{cm}$

Answer 1

The correct option is C $6.9$ $\text{cm}$

Given, initial speed of electron, $u=5\times {10}^6\text{m/s}$
Electric field intensity, $E={10}^3\text{N/C}$
So, force,
$\Rightarrow F=qE$
$\Rightarrow F=1.6\times {10}^{-19}\times {10}^3=1.6\times {10}^{-16}\text{N}$
Further, retardation $a=\frac{F}{m}$
$\Rightarrow a=\frac{1.6\times {10}^{-16}}{9.1\times {10}^{-31}}=1.8\times {10}^{14}{\text{m/s}}^2$
On using
$v^2-u^2=2as$
$\Rightarrow 0^2-(5\times {10}^6{)}^2=-2\times 1.8\times {10}^{14}\times s$
$\Rightarrow s=0.069\text{m}=6.9\text{cm}$