JEE Advanced · Physics · 18. Capacitance

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is $C_{1}$ . When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_{1}$ and the rest of the area gets charge $Q_{2}$ . The electric field in the dielectric is $E_{1}$ and that in the other portion is $E_{2}$ . Choose the correct option/options, ignoring edge effects.

A $\frac{E_{1}}{E_{2}} = 1$
B $\frac{E_{1}}{E_{2}} = \frac{1}{K}$
C $\frac{Q_{1}}{Q_{2}} = \frac{3}{K}$
D $\frac{C}{C_{1}} = \frac{2 + K}{K}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{C}{C_{1}} = \frac{2 + K}{K}$

Step-by-step Solution

Detailed explanation

E = \frac{v}{d}

E_{1} = \frac{v}{d}

E_{2} = \frac{v}{d}

(As both parts have some P.D.) upper capacitor have capacitance

C_{1} = K ε_{0} A / d

Lower capacitor

C_{2} = 2 ε_{0} A / d

∴

equivalent capacitance

C = (k + 2) ϵ_{0} A / d

∴ \frac{C}{C_{1}} = \frac{k + 2}{k}

And

E_{1} : E_{2} = \frac{v}{d} : \frac{v}{d} = 1

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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	LIST -I		LIST-II
A)	$\frac{v_{1}}{v_{2}}$	P)	$\frac{1}{8}$
B)	$\frac{L_{1}}{L_{2}}$	Q)	1
C)	$\frac{K_{1}}{K_{2}}$	R)	2
D)	$\frac{T_{1}}{T_{2}}$	S)	8