Chapter 23: Q.51 (page 656)

A parallel-plate capacitor has $2.0 cm \times 2.0 cm$ electrodes with surface charge densities $\pm 1.0 \times 10^{- 6} C / m^{2}$ . A proton traveling parallel to the electrodes at $1.0 \times 10^{6} m / s$ enters the center of the gap between them. By what distance has the proton been deflected sideways when it reaches the far edge of the capacitor? Assume the field is uniform inside the capacitor and zero outside the capacitor.

Expert verified

The proton been deflected sideways when it reaches the far edge of the capacitor is

$a = 0.108 \times 10^{14} \frac{m^{2}}{s}$

$t = 2 \times 10^{- 8} \sec$

$y = 2.2 mm$

Step by step solution

$\to η = 1 \times 10^{- 6}$ ( charge densities)

From inside battery, there must be an electric current due to just a paired surface.

$E = \frac{η}{ϵ_{o}} = acceleration$

$a = \frac{q E}{m_{p}}$

$= \frac{q η}{ϵ_{o} m_{p}}$

$a = \frac{1.6 \times 10^{- 19} \times 10^{- 6}}{8.85 \times 10^{- 12} \times 1.67 \times 10^{- 27}}$

$a = 0.108 \times 10^{14} \frac{m^{2}}{s}$

So there is no force as in downward motion, the pace is same:

$d = v t$

$t = \frac{d}{v}$

$= \frac{2 \times 10^{- 2}}{10^{6}}$

$= 2 \times 10^{- 8} \sec$

With in vertical motion, ion sweep area

$y = u t + \frac{1}{2} a t^{2}$

$y = \frac{1}{2} \times 0.108 \times 10^{14} \times {(2 \times 10^{- 8})}^{2}$

$y = 2.2 mm$

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