The charge on each of the capacitors, 0 . 20   ms after the switch S is closed in figure is

Step-by-step solution

Here, both the capacitors are in parallel so that, we can replace it with their equivalent. So, this circuit can be drawn like this. As the capacitor is being charged, Q = C V ( 1 - e - t RC ) R =   25   Ω C = 4   μF R C = ( 25 ) ( 4 ) × 10 - 6 R C = 10 - 4   s R C = ( 0 · 1 )   ms . After time, t = 0 . 2   ms , Q = ( 6 ) ( 4 ) × 10 - 6 1 - e - 0 . 2 0 . 1 Q = 24 × 10 - 6 ( 1 - ( 0 . 37 ) ( 0 . 37 ) ) Q = 24 × 10 - 6 ( 1 - 0 . 13 ) Q = 20 . 74