Question

A charge particle of mass 'm' and charge e has velocity u when at a distance d from a conducting plate crosses it at point B. The length of the plates AB is l. Then the charge density of the plate is

• A. $\frac{2d\varepsilon_{0}mu^{2}}{e\mathcal{l}^{2}}$
• B. $\frac{2d\varepsilon_{0}mu}{e\mathcal{l}}$
• C. $\frac{d\varepsilon_{0}mu^{2}}{e\mathcal{l}}$
• D. $\frac{d\varepsilon_{0}mu}{e\mathcal{l}}$

Answer 1

Answer: (A)

$\frac{2d\varepsilon_{0}mu^{2}}{e\mathcal{l}^{2}}$

Answer 2

For the motion of electron along Y axis of the electron d = $\frac{1}{2} ⥂ \mspace{6mu}\mspace{6mu}\frac{eE}{m}$. t² ………………….(1)

The electric field E due to the charge plate is given by

E = $\frac{\sigma}{\varepsilon_{0}}$, σ is the surface charge density of the

plate……………(2)

The time taken by the electron is t = $\frac{\mathcal{l}}{u}$ …………..(3)

From (1). (2) and (3) we obtain. d = $\frac{1}{2}\mspace{6mu}\frac{e\sigma}{\varepsilon_{0}m}\mspace{6mu}\mspace{6mu}\left( \frac{\mathcal{l}}{u} \right)^{2}\mspace{6mu}$

⇒ σ = $\frac{2d\varepsilon_{0}mu^{2}}{e\mathcal{l}^{2}}$