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Six identical square metallic plates are arranged as in the figure. The length of each plate is l (facing plates are a distance d apart). The capacitance of this arrangement would be ______.

Asked in GSEB Board July 2015 · Capacitor networks

• (1) (3ε₀l²)/d
• (2) (4ε₀l²)/(3d)
• (3) (3ε₀l²)/(2d)
• (4) (4ε₀l²)/d

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives (3ε₀l²)/d. Sanity check: the lower pair alone gives (ε₀l²)/d and the upper branch adds only a third of that, so the total is (4ε₀l²)/(3d).
• (3) Treats the two isolated middle plates of the upper branch as one conductor, so that branch has two gaps in series ((C₀)/2) instead of three: C₀ + (C₀)/2 = (3ε₀l²)/(2d). The two middle plates are separate, with a gap d between them.
• (4) Counts every gap as a capacitor joined straight across A and B, as for alternately connected plates: four gaps give (4ε₀l²)/d. But the middle plates of the upper branch are connected to nothing, so its three gaps are in series, not in parallel.

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