**·**2014-05-04 06:51:15

Let a charge Q flow from the 2nd capacitor to first capacitor.

At any time 't' the capacitors must be at same potential.

200-QAe10^{-4}+.001t=200+QAe(10^{-4}-.001t)

Simplifying and diff. w.r.t 't'

We get,i=2μA

- Aditya Agarwal But isnt the total charge on the 2 capacitors 200 and on each one? You've taken 200 to be on each initially....
- Niraj kumar Jha oh! yes it should be 100 on each initially.
- Aditya Agarwal But on putting 100 there, i was getting 1 mA not even 1 uA. I can't figure out where i'm going wrong. Can you please upload the solution.
- Akash Anand Excellent work Niraj ..keep it up
- Akshay Ginodia Sir can we do it this way that initially seperation was 0.1 mm and the plates of one capacitor are approaching each other with 1mm/s so after 0.1 sec the plates are joined and thus the pd becomes zero across both the capacitors...so in 0.1 sec 200 uC of charge will flow through them so in 1 sec 2000uC or 2 mA will have been flown..so current is 2mA
- Akash Anand @Akshaya I dont think it will be appropriate, think like if the speed will be different and in that case also if they touch each other then potential will be zero, but if they not then whole answer will be different.

·0·Reply·2014-05-04 08:23:20