Question

A battery with an emf of V=100 volts is connected in series with a resistor of R=600000 ohms and an air filled parallel plate capacitor of capacitance C=18 microfarads a. Determine the energy stored in the capacitor when it is fully charged. 0.09 J The spacing between the capacitor plates is suddenly increased (in a time short compared to the time constant of the circuit) to four times its original value b. Determine the work that must be done in increasing the spacing in this fashion. 0.0675JX c. Determine the current in the resistor immediately after the spacing is increased. 1.67e-4 A X After a long time. the circuit reaches a new static state. d. Determine the total charge that has passed through the battery. 4.50-4 с Х e. Determine the energy that has been added to the battery. 0.0337 J X

It's my fourth time asking this question. Please get it right this time.

Answer 1

Given data, V = 100 volts R = 600000 Ohm C = 18 microfarads. (a) the energy stored in the capacitor is given by V = 1/2 CV^2 = 0.5 times 18 times 10^-6 times 100^2 = 0.09 J (b) The capacitor is given by C = epsilon_0 A/d Therefore if the spacing is increased to 4 times the capacitor is decreased by 4 times. The voltage also changes but the charge on the capacitor does not and the potential energy is also given by V = 1/2 Q^2/C From the above equation we can see that as Q is unchanged if the capacitance decreases by \r\n 4 times the potential energy increases by 4 times. Thus new P.E is = 4 times 0.09 = 0.36 J thus work done = 0.36 - 0.09 = 0.27 J (c) The voltage in the capacitance is given by V = Q/C Thus the voltage also increases by 4 times