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Chapter 23: Q.51 (page 656)

A parallel-plate capacitor has 2.0cm脳2.0cmelectrodes with surface charge densities 卤1.0脳10鈭�6C/m2. A proton traveling parallel to the electrodes at 1.0脳106m/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.

Short Answer

Expert verified

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

a=0.108脳1014m2s

t=2脳10鈭�8sec

y=2.2mm

Step by step solution

01

Acceleration and Velocity

鈫�畏=1脳10鈭�6( charge densities)

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

E=畏系o=acceleration

a=qEmp

=q畏系omp

a=1.6脳10鈭�19脳10鈭�68.85脳10鈭�12脳1.67脳10鈭�27

a=0.108脳1014m2s

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

d=vt

t=dv

=2脳10鈭�2106

=2脳10鈭�8sec

02

Distance

With in vertical motion, ion sweep area

y=ut+12at2

y=12脳0.108脳1014脳2脳10鈭�82

y=2.2mm

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