Question

Two 2.0-cm-diameter-disks spaced 2.0 mm apart form a parallel-plate capacitor. The plates have charge ±  8.8nC.

What is the voltage across the capacitor? deltaV_c =_______________ Dont forget unit

An electron is launched from the negative plate. It strikes the positive plate at a speed of 5.5×10⁷ m/s . What was the electron's speed as it left the negative plate?

v_{initial=_____________________Dont forget unit}

Answer 1

1.

Voltage across capacitor is given by:

V = Q/C

Q = Charge on plates = 8.8*10^-9 C

C = Capacitance of capacitor = e0*A/d

A = Area of plates = pi*D^2/4

D = diameter of plates = 2.0 cm = 0.02 m

d = plate separation = 2.0 mm = 2.0*10^-3 m

e0 = 8.854*10^-12

So,

V = Q/(e0*A/d) = 4*Q*d/(e0*pi*D^2)

V = 4*8.8*10^-9*2.0*10^-3/(8.854*10^-12*pi*0.02^2)

V = 6327.37 V

In two significant figures:

V = 6.3*10^3 V

2.

Using energy conservation:

dKE = dPE

KEf - KEi = q*dV

KEi = (1/2)*m*Vi^2,

where Vi = Initial speed of electron at negative plate = ?

KEf = (1/2)*m*Vf^2

Vi = final speed of electron at positive plate = 5.5*10^7 m/s

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

So,

(1/2)*m*Vf^2 - (1/2)*m*Vi^2 = q*dV

Vi = sqrt (Vf^2 - 2*q*dV/m)

Using known values:

Vi = sqrt ((5.5*10^7)^2 - 2*1.6*10^-19*6327.37/(9.1*10^-31)) = 2.828*10^7 m/s

In two significant figures

Vi = 2.8*10^7 m/s

Let me know if you've any query.