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An electron is launched at a 45angle and a speed of 5.0×106m/sfrom the positive plate of the parallel-plate capacitor shown in FIGURE P23.52. The electron lands 4.0cmaway.

a. What is the electric field strength inside the capacitor?

b. What is the smallest possible spacing between the plates?

Short Answer

Expert verified

a. The electric field strength inside the capacitor isa=6.3×1014m/s2

b. The smallest possible spacing between the plates is

E=3.6×103N/C

H=1cm

Step by step solution

01

Velocity (part a)

The pace as in downward motion Vocos45.

And there is no stress with in downward motion

so,

d=Vocos45×t.

t=dVocos45

localid="1650219090644" =4×102m×25×106m/s×1

=1.13×108s

At its highest point (vertical direction).

vf=0

v=uatt=t2=0.56×108s(a opposite to Force

a=ua

=Vosinθt

localid="1650219122312" =5×106m/s2×0.56×108s

a=6.3×1014m/s2

02

Force (part b)

Force is,

eE=ma

ma=eE~E

=mae

localid="1650219349035" role="math" =9.11×1031kg×6.3×1014m/s21.6×1019C

E=3.6×103N/C

MaximumH

v2=uv22aH

H=uv22a

localid="1650219390403" =52×10622×6.3×1014m/s2

=1×102m

H=1cm

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