A capacitor of capacitance C = .015 F is connes
to parallel conducting rail and a conducting
rod of mass m= 100 g and length 1 = lm star
fall under gravity in vertical plane. A uniform
magnetic field of 2T exist in space directed
perpendicular to rod as shown in figure. Find
acceleration of rod.
C
х
Answers
1) Initially, Capacitor will be charged to chage q(i) = CV on one side of plate capacitor .
After that, on changing the polarity of Battery and of EMF 2V ,Final charge on same side of plate capacitor q(f) = -C (2V) .
2) Hence, magnitude of charge transferred by Battery of 2V is
| CV - (-2CV)| = 3CV
Now,
Work done by battery = Charge transferred * EMF
= (3CV) *(2V) = 6CV^2
Answer:
let at any time t, q = EC (1 – e^-t/CR) E = Energy stored = q^2/2c = E^2C^2/2c (1 – e^-t/CR)^2 = E^2C/2 (1 – e^-t/CR)^2 R = rate of energy stored = dE/dt = -E^2C/2 (-1/RC)^2 (1 – e^-t/RC)e^-t/RC = E^2/CR . e^-t/RC (1 – e^-t/CR) dR/dt = E^2/2R [-1/RC e^-t/CR . (1 – e^-t/CR)+(-) . e^-t/CR(1-/ RC) . e^-t/CR] E^2/2R = (-e^-t/CR/RC + e^-2t/CR/RC + 1/RC. E^-2t/CR) = E^2/2R(2/RC . e^-2t/CR – e^-t/CR/RC) ….(1) For R base max dR/dt = 0 ⇒ 2.e^-t/RC -1 = 0 ⇒ e^-t/CR = ½ ⇒ -t/RC = -In^2 ⇒ t = RC In 2 ∴ Putting t = RC In 2 in equation (1) We get dR/dt = E^2/4R .