(A) A dielectric slab of thickness t, is introduced between the pla...

Question

(A) A dielectric slab of thickness t, is introduced between the plates of parallel plate capacitor of area A and separation d (where t<d). Find an expression for the capacitance with the dielectric slab.

(B) A copper sphere of capacitor C is dropped in ocean. Will the capacitance of the sphere increase, decrease or remain same? Justify.

(C) A capacitor is connected across a source of potential difference V and then the separation ‘d’ between the plates is increased using insulating stick. Plot ‘V’ vs ‘d’ graph for the given capacitor.

Accepted Answer

📌 Part (A) — Capacitor with Dielectric Slab:

Setup: Plate area = A, plate separation = d, dielectric slab thickness = t, constant = K.

Effective air gap = (d − t)
Effective dielectric gap = t/K

Formula:
C = ε₀A / [(d − t) + t/K]
C = ε₀A / [d − t(1 − 1/K)]

∴ C increases compared to air capacitor.

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📌 Part (B) — Copper Sphere Dropped in Ocean:

Capacitance of isolated sphere: C = 4πε₀R

Ocean (seawater) is a conducting medium. When sphere is submerged:
• The surrounding conductor acts as the outer shell.
• Effective outer radius → very small gap (water around sphere).
• Permittivity of water (K ≈ 80) also increases capacitance.

∴ Capacitance increases significantly.

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📌 Part (C) — V vs d Graph (Battery connected):

• Capacitor is connected to source → V = constant.
• Charge Q = CV = (ε₀A/d)V → Q decreases as d increases.
• But V remains fixed by the source.

Graph: V vs d is a horizontal straight line (V = constant)

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