Capacitors

JEE Main level, single correct, medium

114 of 180

Question 114 of 180

The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:
K(x)=Ko+λx(λ=a constant )\mathrm{K}(\mathrm{x})=\mathrm{K}_{\mathrm{o}}+\lambda \mathrm{x}(\lambda=\mathrm{a} \text { constant })
The capacitance C\mathrm{C}, of the capacitor, would be related to its vacuum capacitance Co\mathrm{C}_{\mathrm{o}} for the relation :

Show answer and solution

Answer: (C) C=λdln⁡(1+λd/KoCo)\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\lambda \mathrm{d} / \mathrm{K}_{\mathrm{o}}\right.} \mathrm{C}_{\mathrm{o}}\right)

The value of dielectric constant is given as,
K=K0+λx And, V=∫0dEdrV=∫0dσKdx\begin{aligned} &\mathrm{K}=\mathrm{K}_0+\lambda \mathrm{x} \\ &\text { And, } \mathrm{V}=\int_0^{\mathrm{d}} \mathrm{Edr} \\ &\mathrm{V}=\int_0^{\mathrm{d}} \frac{\sigma}{\mathrm{K}} \mathrm{dx} \end{aligned}

=σ∫0d1( K0+λxdx=σλ[ln⁡(K0+λd−ln⁡K0]=σλln⁡(1+λdK0)\begin{aligned} &=\sigma \int_0^{\mathrm{d}} \frac{1}{\left(\mathrm{~K}_0+\lambda \mathrm{x}\right.} \mathrm{dx} \\ &=\frac{\sigma}{\lambda}\left[\ln \left(\mathrm{K}_0+\lambda \mathrm{d}-\ln \mathrm{K}_0\right]\right. \\ &=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right) \end{aligned}
Now it is given that capacitance of vacuum =C0=\mathrm{C}_0.
Thus, C=QV\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}
=σ.sv( Let surface area of plates =s)=σ.sσλln⁡(1+λdK0)\begin{aligned} &=\frac{\sigma . \mathrm{s}}{\mathrm{v}}(\text { Let surface area of plates }=\mathrm{s}) \\ &=\frac{\sigma . \mathrm{s}}{\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)} \end{aligned}
=sλ⋅dd1ln⁡(1+λdK0)(∵=\mathrm{s} \lambda \cdot \frac{\mathrm{d}}{\mathrm{d}} \frac{1}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)}\left(\because\right. in vacuum ε0=1)\left.\varepsilon_0=1\right)
c=λdln⁡(1+λdK0)⋅C0(\mathrm{c}=\frac{\lambda \mathrm{d}}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)} \cdot \mathrm{C}_0 \quad\left(\right. here, C0=sd)\left.\mathrm{C}_0=\frac{\mathrm{s}}{\mathrm{d}}\right)

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Capacitors, dielectrics and RC circuits simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Capacitors (Capacitors with Dielectric), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.