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A parallel-plate capacitor has 2.0cm×2.0cmelectrodes with surface charge densities ±1.0×106C/m2. A proton traveling parallel to the electrodes at 1.0×106m/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=0.108×1014m2s

t=2×108sec

y=2.2mm

Step by step solution

01

Acceleration and Velocity

η=1×106( charge densities)

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

E=ηϵo=acceleration

a=qEmp

=qηϵomp

a=1.6×1019×1068.85×1012×1.67×1027

a=0.108×1014m2s

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

d=vt

t=dv

=2×102106

=2×108sec

02

Distance

With in vertical motion, ion sweep area

y=ut+12at2

y=12×0.108×1014×2×1082

y=2.2mm

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