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(a) A 900 pF capacitor is charged by 100 V battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system? |
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Answer» Solution :(a) The charge on the capacitor is Q = CV = `900 xx 10^(-12) F xx 100 V = 9 xx 10^(-8) C` The energy STORED by the capacitance is `=(1//2)CV^(2) = (1//2) QV = (1//2) xx 9 xx 10^(-8) C xx 100 V = 4.5 xx 10^(-6)` J (b) In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V′. The charge on each capacitance is then `Q. = CV.`. By charge conservation, `Q. = Q//2`. This implies `V. = V//2`. The total energy of the system is `=2 xx 1/2 Q.V. = 1/4 QV = 2.25 xx 10^(-6)` J Thus in going from (a) to (b), though no charge is lost, the final energy is only half the initial energy. Where has the remaining energy gone? There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of HEAT and electromagnetic RADIATION. |
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