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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Explanation

Let, at any moment of time charge on the plates be `+q` and `-q` respectively, then voltage across the capacitor, `varphi = q//C` ....(1)
Now, from charge conservation,
`i = i_(1) + i_(2)`, where `i_(2) = (dq)/(dt)` ....(2)
In the loop `65146`, using `-Delta varphi = 0`,
`(q)/(C) + (i_(1) + (dq)/(dt))R - xi = 0` .....(3)
[using (1) and (2)]
In the loop `25632` using `-Delta varphi = 0`
`- (q)/(C) + i_(1) R = 0` or `i_(1) R = (q)/(C)` ....(4)
From (1) and (2)
`(dq)/(dt) R = xi_(1) = - (2q)/(C)` or, `(dq)/(xi - (2q)/(C)) = (dt)/(R)` ....(5)
On intergating the expression (5) between suitable limits,
`int_(n)^(q) (dq)/(xi - (2q)/(C)) = (1)/(R) int_(0)^(t) dt` or, `- (C)/(2) In (xi - (2q)/(C))/(xi) = (t)/(R)`
Thus `(q)/(C) = V = (1)/(2) xi (1 - e^(-2t//R C))`