(Please see the attachment for detailed problem)
1) In a circuit shown below it took capacitor 0.138 seconds to discharge from its maximum charge Qmax to half of that maximum charge once the battery was removed.
a) How much time would it take that capacitor to discharge from its maximum charge to 1/16th of that maximum charge?
b) What is the time constant of this circuit?
c) What is the capacitance of the capacitor?
d) The battery is reconnected and the capacitor charges up again. When the capacitor is fully charged what is the voltage drop across the capacitor?
e) What is the charge Qmax on the capacitor?
f) How far apart are capacitors plates if each is a circle with 2 cm diameter?
g) With the battery positioned as shown in the figure above is the left plate of the capacitor charged positively or negatively?
h) What are the magnitude and the direction of the electric field inside this capacitor
i) A proton is placed in the space between the capacitor plates. What are the direction and magnitude of the force acting on that proton inside the capacitor?
j) If initially proton had velocity straight down the page will it be deflected from its straight path by the electric field of the capacitor? If so which way?
k) What is the magnitude and direction of that proton's acceleration (proton's mass mp =1.6726×10^-27 kg)
SOLUTION This solution is FREE courtesy of BrainMass!
Please refer to the attachment.
a) While discharging, charge Q on the capacitor at any instant t is given by :
Q = Qmax e-t/RC where Q0 = The initial charge and RC = τ = Time constant of the circuit
For Q = Qmax/2, t = 0.138 sec. Hence, Qmax/2 = Qmax e-(0.138)/τ
Or e(0.138)/τ = 2 or loge2 = 0.138/τ or 0.693 = 0.138/τ
Or τ = 0.2 sec
For Q = Qmax/16 we have : Qmax/16 = Qmax e-t/0.2
Or et/0.2 = 16 or loge16 = t/0.2 or 2.77 = t/0.2 or t = 0.554 sec
b) Time constant of the circuit = τ = 0.2 sec
c) τ = RC or C = τ/R = 0.2/100x109 = 2x10-12 F (or 2 pF)
C = 2 pF
d) When the capacitor is fully charged, current in the circuit becomes zero. Hence, full battery voltage comes across the capacitor.
Hence, voltage across capacitor when it is fully charged = 12V.
e) Max charge on the capacitor = Qmax = CV = 2x10-12 x 12 = 24x10-12 C
f) Capacitance of a parallel plate capacitor = C = ε0A/d
Here, ε0 = 8.85x10-12 C2N-1m-2, A = Πr2 = 3.14x(1/100)2 = 3.14 x 10-4 m2
Hence, d = ε0A/C = 8.85x10-12 x 3.14 x 10-4/2x10-12 = 13.89 x 10-4 m (or 1.39 mm)
g) Left plate being connected to +ve terminal of the battery is charged +ve.
h) Magnetic field inside the capacitor = E = σ/ε0 where σ = Charge density on a plate = Total charge/Area = 24x10-12/3.14 x 10-4 = 7.64x10-8 C/m2
Hence, E = 7.64x10-8/8.85x10-12 = 0.86x104 N/C from +ve plate towards -ve plate.
i) Force on proton = Charge on proton x Electric field intensity = 1.6x10-19x0.86x104 =
1.376x10-15 N in a direction from +ve plate towards the negative plate.
j) It will be deflected towards the negative plate.
k) Proton's acceleration = Force/Mass = 1.376x10-15/1.6726×10-27= 0.823x1012 m/s2 towards the negative plate.© BrainMass Inc. brainmass.com September 29, 2022, 1:06 pm ad1c9bdddf>