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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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Most popular questions from this chapter

A problem of practical interest is to make a beam of electrons turn a 90°corner. This can be done with the parallel-plate capacitor shown in FIGURE. An electron with kinetic energy 3.0×10-17Jenters through a small hole in the bottom plate of the capacitor.

a. Should the bottom plate be charged positive or negative relative to the top plate if you want the electron to turn to the right? Explain.

b. What strength electric field is needed if the electron is to emerge from an exit hole 1.0cmaway from the entrance hole, traveling at right angles to its original direction?

Hint: The difficulty of this problem depends on how you choose your coordinate system.

c. What minimum separation dminmust the capacitor plates have?

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aAt what distance along the z-axis is the electric field strength a maximum?

bWhat is the electric field strength at this point?

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