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An electron is launched at a 45 degree angle and a speed of 5.0*10^6m/s from the positive plate of the parallel-plate capacitor shown in the figure(Figure 1) . The electron lands 4.0 cm away.

1- What is the electric field strength inside the capacitor?

2- What is the smallest possible spacing between the plates?

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electron follows projectile path whose range is 4cm.

Range = v^2sin(2*theta) / a

0.04 = (5x10^6) sin(2x45) / a

a =6.25 x 10^14 m/s2

and a = qE /m

6.25x 10^14 = 1.6x 10-19 xE / 9.109 x 10-31

E = 3558.20 N/C

b) that will be the max. height of the projectile.

h = u^2 sin^2 (theta) / 2a

= (5x10^6) ^2 sin^2(theta) / 2x 6.25x 10^14

h = 0.01 m = 1 cm

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