In the circuit shown in figure. Charge stored in 6 uF capacitor will be 6 uF ii TT ARF 1 av | | ji 12v

1. Recognising a Capacitor Bridge

Many JEE questions hide a Wheatstone bridge made of capacitors. Four capacitors form a rectangle; a fifth one sits across the middle.

2. Condition for No Charge in the Middle Branch

For resistors, a Wheatstone bridge is balanced when (\frac{R_1}{R_2}=\frac{R_3}{R_4}). The same idea applies to capacitors but with their inverse behaviour in series/parallel networks:

$\text{No potential difference across the bridge capacitor}\;\Longleftrightarrow\;\frac{C_1}{C_2}=\frac{C_3}{C_4}.$

Under this condition, the two nodes touching the middle capacitor are equipotential. Equipotential points mean zero voltage difference ((V=0)) across the 6 µF capacitor.

3. Charge–Voltage Relation

For any capacitor,

$Q = CV.$

So, if (V=0) then (Q=0) no matter how large (C) is.

4. Logical Steps a Student Follows

1. Identify the geometry: four side capacitors + one central capacitor ➔ likely Wheatstone bridge.
2. Check side-capacitor ratios: are they equal? If yes, bridge is balanced.
3. Conclude voltage across central capacitor: 0 V.
4. Use (Q=CV) to state the charge on the 6 µF capacitor: 0 C (or 0 μC).

If the ratios were not equal, you would collapse the network by combining series/parallel parts and solving normally.

What’s Going On?

Imagine four water tanks connected by pipes forming a square. If the pipes and tanks are perfectly balanced, the water levels at two opposite corners become exactly the same, so no water flows through the pipe that joins those two points diagonally.

Replace water with electric charge and tanks/pipes with capacitors/wires. When the side-capacitors are in the right ratio, the two junctions touching the middle (6 µF) capacitor sit at the same electric height (potential). Because the ends of the 6 µF plate are at the same height, it can’t store any extra water (charge). So, the middle one stays empty of extra charge.

👉 Therefore the 6 µF capacitor ends up with zero charge.

Step-by-Step Calculation

1. Check Side Capacitors’ Ratio
   Given the figure (not reproduced here) shows a textbook capacitor bridge where the four outer capacitors satisfy
   $\frac{C_{top\,left}}{C_{bottom\,left}} = \frac{C_{top\,right}}{C_{bottom\,right}}.$ Hence the bridge is balanced.

2. Voltage Across 6 µF Capacitor
   Balanced bridge ⇒ potentials at its terminals are equal ⇒
   $V_{6\,µF} = 0\;\text{V}.$

3. Charge on 6 µF Capacitor
   Using (Q = CV):
   $Q_{6\,µF} = 6\,\text{µF} \times 0\,\text{V} = 0\,\text{C}.$

4. Final Answer
   $\boxed{Q = 0\,\mu\text{C}}$