Question
Can you please help me answer these two questions because myphysics book does not explain the concepts well. 1)Because the charges on the plates of a parallel-platecapacitor are opposite in sign, they attract each other. Hence, itwould take positive work to increase the plate separation. Whattype of energy in the system changes due to the external work donein this process? Assume that the capacitor is disconnected from thebattery. 2)Think what would be the work done and the energy change ofthe capacitor in the previous question if the capacitor isconnected to the battery while its plates are being pulledapart. ***Is the answer for question 1 electric potential energyand does it increase Can you please help me answer these two questions because myphysics book does not explain the concepts well. 1)Because the charges on the plates of a parallel-platecapacitor are opposite in sign, they attract each other. Hence, itwould take positive work to increase the plate separation. Whattype of energy in the system changes due to the external work donein this process? Assume that the capacitor is disconnected from thebattery. 2)Think what would be the work done and the energy change ofthe capacitor in the previous question if the capacitor isconnected to the battery while its plates are being pulledapart. ***Is the answer for question 1 electric potential energyand does it increase
Explanation / Answer
Part 1... yes, the electric potential energy of thecapacitor increases. When the plates are pulled apart, in eithersituation, the capacitance of the capacitor decreases. Thepotential energy stored in the cap can be written: . U = (1/2) Q V = (1/2) CV2 = (1/2) Q2 /C . If we use the last expression, since Q is heldconstant, we can see that U increases when Cdecreases. . Part 2... now if the plates remain connected to thebattery, V is held constant and the amount of charge onthe plates is allowed to change. As the plates are pulled apart,the amount of charge on each plate will decrease (it goes back tothe battery) which will cause the stored energy to decrease. Mathematically, this can be shown byusing U = (1/2) CV2 since V isnow held constant. A decrease in C now results in a decrease inU.