Consider the circuit shown in the figure. Find the charge on capacitor C between A and D in a steady state.
\(C_\epsilon\)
\(\frac{C_\epsilon}{2}\)
\(\frac{C_\epsilon}{3}\)
zero
The correct answer is d. Zero
Explanation: A capacitor blocks the DC steady state i.e. a capacitor behaves as an open circuit in a steady state. Hence no current will flow through the capacitor. Therefore, the potential difference between A and D is, V = 0
The capacitance of a capacitor is given by, C = qV
⇒ C = 0
Two charges \( +q \) and \( -q \) are placed at points \( A \) and \( B \) respectively which are at a distance \( 2L \) apart. \( C \) is the midpoint of \( AB \). The work done in moving a charge \( +Q \) along the semicircle CSD (\( W_1 \)) and along the line CBD (\( W_2 \)) are 
Find work done in bringing charge q = 3nC from infinity to point A as shown in the figure : 