Question

In an electron gun electron is accelerated through a potential difference V₀. The distance from the electron gun at which electron becomes stationary for the first time is - (e = charge of electron, m = mass of electron, gravity = 0, E = Electric field outside the electron gun)

• A. $\frac{eV_{0}^{2}m}{E}$
• B. $\frac{2eV_{0}^{2}m}{E}$
• C. $\frac{eV_{0}^{2}m}{2E}$
• D. 2Ee$V_{0}^{2}$ m

Answer 1

Answer: (B)

$\frac{2eV_{0}^{2}m}{E}$

Answer 2

Due q₁, electric feed at P(E₁) =

q₂ electric fed at P(E₂) =

At P

\ tanq = = = $\left( \frac { 1 } { 2 } \right)$× $\left( \frac { 2 } { 1 } \right) ^ { 2 }$ = 2