Answer to capacitor question

(a)

V across 200k resistor = IR

V= (2x10⁵) (2x 10^-5) = 4V

V_C = E - V_R = 6-4 = 2V

(b)

Q=VC

Q = (2.0) (560 x 10^-6) = 1.12 x 10^-3 C

 

(c)

W = ½ CV²

W = ½ (560 x10^-6) (2)² = 1.12 x 10^-3 J

 

(d)

 W = Emf x charge

W = 6V x 1.12 x 10^-3 C = 6.72 x 10^-3 J

 

(e) Work is done transferring the charge through the external resistance. This is transferred to heat energy in the resistor