Charging a capacitor using a capacitor? I recently did an experiment using a fully charged capacitor (assumed) and charged a separate capacitor through fixed resistor. We measured the potential difference of the capacitor being charged. We produced a graph of the potential difference against time. I expected the the graph to follow a normal charging graph with a gradient similar to a logarithmic graph which is true however the gradient of the graph at the end will become negative which suggested there is potential difference lost in the capacitor. Is there any reasons for this?
I cant show the circuit but it is similar to a charging circuit but the cell is replaced with a fully charged capacitor.
 A: Comment on the question (v1): If you're confused by long-time behavior of your circuit, its behavior may be sensitive to details which are safe to neglect when describing short-time behavior. If you want a real answer, you should edit your question to include a complete circuit diagram, and perhaps a plot of your data.
For example: in one of my labs I have students charge a 200 µF capacitor through a 500 kΩ resistor, which has a time constant of 100 seconds.
However, the voltmeter they use to monitor the capacitor has an input resistance of 10 MΩ.
Students (and sometimes faculty, ahem) who disconnect the power supply while the capacitor is charged and monitor the "unconnected" voltage on the capacitor are surprised to see the charge on the capacitor trickle away; it trickles away through the voltmeter, with a time constant $\rm 10\,M\Omega \cdot 200\,\mu{}F \approx 30\,minutes$.
A: In the real world you can never rid yourself of resistance in an electrical circuit. There is resistance in the connecting wires and resistance internal to the capacitor itself. So once you close the circuit and current starts moving, some of that electrical energy will be lost to heat, dissipated by resistance. 
