A capacitor of 50 μ F is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is _________ μ C .

Step-by-step solution

At steady state current through branch of capacitor is zero. Potential difference across capacitor is 2   V . So, q = 50 × 2 = 100 μ C i = 6 ( 2 + 2 + 2 ) × 10 3 = 1 × 10 - 3 Amp V A , B = potential across 50 μ F V A , B = i × 2 × 10 3 = 1 × 10 - 3 × 2 × 10 3 = 2 volt Charge on 50 μ F q = C V = 50 × 10 - 6 × 2 ⇒ 100 × 10 - 6 C = 100 μ C