A parallel-plate capacitor is formed of two 10 cm times 10 cm plates spaced 1.0

Question

A parallel-plate capacitor is formed of two 10 cm times 10 cm plates spaced 1.0 cm apart. The plates are charged to plusminus 1,0 nC An electron is shot through a very small hole halfway up the positive plate. What is the slowest speed the electron can have as it enters the capacitor if it is to reach the negative plate? What is the electric potential the electron experiences inside the capacitor when it is a distance of 0.3 cm away from the positive plate? Now, a second vertical electric with E = 2300 N/C. Where does the electron land? (Either the x-position on the top/bottom of the capacitor or the y-position on the negative plate.)

Explanation / Answer

a)

Let the initial velcoity of the electron be u

So, its initial Kinetic energy,

KEi = 0.5*mu^2

where m = mass of electron = 9.1*10^-31 kg

After stopping at the right side negative plate,

final Kinetic energy = 0

So, By work energy principle,

Work done by Electric field = Change of Kinetic energy

So, (V)q = 0.5*mv^2 - 0

q = charge of electron = 1.6*10^-19 C

Work done by electric field = Potential difference(V)*distance between plates(d)

Now, we know for a capacitor, V = Q/C

where Q = charge of capacitor = 1nC

C = capacitance = eA/d

e = epsilon = 8.854*10^-12

d = 1cm = 0.01 m

A = area = 0.1*0.1 m2 = 0.1^2 m2

So, V = Q/(eA/d) = Qd/eA

So, 0.5*mu^2 = Vq = qQd/eA

So, u = sqrt(2qQd/eAm)

So, u = sqrt(2*1.6*10^-19*(1*10^-9)*0.01/(8.854*10^-12*0.1^2*9.1*10^-31))

So, u = 6.3*10^7 m/s <-------answer

b)

Electric field inside the plates,

E = V/d = Q/eA

So, potential at distance x = 0.3 cm (or 0.003 m) from the positive plate,

Vx = -E*x = -Qx/eA

So, Vx = -(1*10^-9*0.003)/(8.854*10^-12*0.1^2) = -33.9 V <----------answer

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