Demystifying Multi-Plate Capacitors
A Visual Guide
Multi-plate capacitor problems often look like a tangled mess of wires and metal sheets. However, they are incredibly logical once you learn to see past the physical layout and focus purely on the electrical nodes. Let's break down this classic JEE problem step by step.
The Setup
Breaking Down the Plates
We are given four identical metal plates, S1, S2, S3, and S4, placed parallel to each other. The space between any two adjacent plates forms a capacitor.
Since there are four plates, we get exactly three distinct capacitors:
1. C12 between plates S1 and S2
2. C23 between plates S2 and S3
3. C34 between plates S3 and S4
Because the plates are identical and equally spaced, each of these capacitors has the same base capacitance:
Case P
The Simple Series
In the first scenario, the terminals are connected to the outermost plates, S1 and S4. The inner plates, S2 and S3, are left floating (not connected to any external wire).
This means the charge has only one path to flow—straight through all three capacitors sequentially. Therefore, C12, C23, and C34 are in a perfect series combination.
For three identical capacitors in series, the equivalent capacitance is:
This matches with option (3).
Case Q
The Short Circuit Trap
Here, the terminals are still S1 and S4, but a wire connects plate S2 directly to plate S3.
When two plates of a capacitor are connected by a conducting wire, they are forced to be at the exact same electrical potential. Since there is no potential difference (ΔV=0), the capacitor C23 stores no charge. It is effectively short-circuited and bypassed by the current.
We are left with only C12 and C34 in series. Their equivalent capacitance is:
This matches with option (2).
Case R
Redrawing is the Key
Things get interesting here. The terminals are S1 (let's call it Node A) and S3 (Node B). Plate S2 is shorted to plate S4, meaning they form a single electrical node.
Let's trace the connections:
- C12 is connected between S1 (Node A) and S2.
- C23 is connected between S2 and S3 (Node B).
- C34 is connected between S3 (Node B) and S4. But wait, S4 is the exact same node as S2!
This means both C23 and C34 are connected between the S2/S4 node and Node B. They are in parallel! Their combined capacitance is C0+C0=2C0.
This parallel combination is in series with C12. The final equivalent capacitance is:
Ceq=C0+2C0C0⋅2C0=32C0
This matches with option (4).
Case S
The Ultimate Parallel Setup
In the final case, the terminals are S1 (Node A) and S2 (Node B). Plate S1 is shorted to S3 (so S3 is also Node A), and plate S2 is shorted to S4 (so S4 is also Node B).
Let's check where each capacitor lives:
- C12 is between S1 (A) and S2 (B).
- C23 is between S2 (B) and S3 (A).
- C34 is between S3 (A) and S4 (B).
Every single capacitor is connected directly across the terminals A and B! They are all perfectly in parallel. The total capacitance is simply the sum:
This matches with option (1).
Final Conclusion: The correct matching sequence is P → 3, Q → 2, R → 4, and S → 1.
