A parallel plate capacitor is having separation between plates 0.885 mm. It has a capacitance of 1 μF when the space between the plates is filled with an insulating material of resistivity 1 x 10^13 Ωm and resistance 17.7 x 10^14 Ω. Relative permittivity of the insulating material is α x 10^7. The value of α is _______. (Take permittivity of free space = 8.85 x 10^-12 F/m)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q46 – JEE-MAIN 2026

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Home Physics Electrostatics & CapacitanceQuestion #127

A parallel plate capacitor is having separation between plates

0.885

mm. It has a capacitance of

1 μ

F when the space between the plates is filled with an insulating material of resistivity

1 \times 10^{13} Ω

m and resistance

17.7 \times 10^{14} Ω

. Relative permittivity of the insulating material is

α \times 10^{7}

. The value of

α

is _______. (Take permittivity of free space

= 8.85 \times 10^{- 12}

F/m)

Enter Calculated Numerical Value:

Clue 1 of 3: Fundamental Capacitor and Resistor Relations

Recall the formulas for capacitance of a parallel plate capacitor with a dielectric and the resistance of a material.

🧠 Full Solution Path

Step 1: Relate Capacitance and Resistance to Area✦ Active

The capacitance of a parallel plate capacitor with a dielectric is given by $C = \frac{ϵ_{r} ϵ_{0} A}{d}$ , where $A$ is the plate area, $d$ is the separation, $ϵ_{r}$ is the relative permittivity, and $ϵ_{0}$ is the permittivity of free space.

The resistance of the insulating material between the plates is given by $R = \frac{ρ d}{A}$ , where $ρ$ is the resistivity.

Step 2: Derive a Relationship for Relative Permittivity○ Expand

From the capacitance formula, we can express the area as $A = \frac{C d}{ϵ_{r} ϵ_{0}}$ . Substitute this expression for $A$ into the resistance formula:

R = \frac{ρ d}{\frac{C d}{ϵ_{r} ϵ_{0}}} = \frac{ρ d ϵ_{r} ϵ_{0}}{C d} = \frac{ρ ϵ_{r} ϵ_{0}}{C}

Rearranging to solve for $ϵ_{r}$ : $ϵ_{r} = \frac{R C}{ρ ϵ_{0}}$ .

💡 Teacher's Secret Hint

Ensure correct algebraic manipulation to isolate $ϵ_{r}$ .

Step 3: Calculate the Value of

α

○ Expand

Substitute the given values: $R = 17.7 \times 10^{14} Ω$ , $C = 1 \times 10^{- 6} F$ , $ρ = 1 \times 10^{13} Ω m$ , and $ϵ_{0} = 8.85 \times 10^{- 12} F/m$ .

ϵ_{r} = \frac{(17.7 \times 10^{14}) \times (1 \times 10^{- 6})}{(1 \times 10^{13}) \times (8.85 \times 10^{- 12})} = \frac{17.7 \times 10^{8}}{8.85 \times 10^{1}} = 2 \times 10^{7}

Given that the relative permittivity is $α \times 10^{7}$ , comparing this with our calculated value, we find $α = 2$ .

💡 Teacher's Secret Hint

Pay attention to the powers of 10 during calculation.

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Physics Question 46 – JEE-MAIN 2026

Clue 1 of 3: Fundamental Capacitor and Resistor Relations

🧠 Full Solution Path