Find how the voltage across the capacitor `C` varies with time `t` after closing of the switch `S_(w)` at the moment `t=0`.
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Answer

`V=(1)/(2) xi(1-e^(-2t//RC))`.

Explanation

`V=(1)/(2) xi(1-e^(-2t//RC))`
Let, at any moment of time, charge on the plates be `+q` and `-q` respectively, then voltage
across the capacitor, `varphi=(q)/(C) rarr(1)`
Now, from charge conservation
`i=i_(1) +i_(2)`, where `i_(2) =(dq)/(dt) rarr (2)`
In the loop `65146`, using `- Delta varphi =0`
`(q)/(C)+(i_(1)+(dq)/(dt))R- xi =0 rarr (3)`
Using eq `(1)` and `(2)`
In the loop `25632`, using `- Delta varphi =0`
`-(q)/(C)+i_(1)R=0 or i_(1)R =(q)/(C) rarr (4)`

From `(1)` and `(2)`,
`(dq)/(dt)R = xi_(1)-(2q)/(C) or, (dq)/(xi-(2q)/(C))=(dt)/(R) rarr (5)`
On integrating the expression `(5)` between suitable limits,
`underset(q) overset(q) int (dq)/(xi-(2q)/(C))=(1)/(R) underset(0) overset(t) int dt or, -(C)/(2) i n(xi-(2q)/(C))/(xi)=(t)/(R)`
thus `(q)/(C)=V=(1)/(2) xi(1-e^(-2t//RC))`.