The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness 3/8d and d/2, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1 = 1.25 K2, the value of K1 is :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q20 – NEET-UG 2025

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Home Physics Electrostatics & CapacitanceQuestion #128

The plates of a parallel plate capacitor are separated by

d

. Two slabs of different dielectric constant

K_{1}

and

K_{2}

with thickness

\frac{3}{8} d

and

\frac{d}{2}

, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If

K_{1} = 1.25 K_{2}

, the value of

K_{1}

is :

Clue 1 of 3: Capacitance with Dielectrics

When dielectric slabs are inserted into a parallel plate capacitor, they effectively divide the capacitor into several capacitors in series.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify Initial Capacitance and Effective Capacitance Formula✦ Active

The initial capacitance of a parallel plate capacitor with vacuum between plates is $C_{0} = \frac{ϵ_{0} A}{d}$ . When multiple dielectric slabs are inserted, the effective capacitance $C$ is given by the formula for capacitors in series:

C = \frac{ϵ_{0} A}{\sum \frac{t_{i}}{K_{i}}}

Here, $t_{i}$ are the thicknesses of the dielectric slabs and $K_{i}$ are their respective dielectric constants. Any air gap is treated as a slab with $K = 1$ .

💡 Teacher's Secret Hint

The sum $\sum \frac{t_{i}}{K_{i}}$ must include all regions between the plates, ensuring their thicknesses add up to the total plate separation $d$ .

Step 2: Formulate New Capacitance and Set Up the Equation○ Expand

The total thickness of the dielectric slabs is $t_{1} + t_{2} = \frac{3}{8} d + \frac{d}{2} = \frac{3}{8} d + \frac{4}{8} d = \frac{7}{8} d$ . The remaining air gap (with $K_{a i r} = 1$ ) is $t_{a i r} = d - \frac{7}{8} d = \frac{1}{8} d$ . The new capacitance $C$ is:

C = \frac{ϵ_{0} A}{\frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}} + \frac{t_{a i r}}{1}} = \frac{ϵ_{0} A}{\frac{3 d / 8}{K_{1}} + \frac{d / 2}{K_{2}} + \frac{d / 8}{1}}

Given that the new capacitance $C$ is two times larger than $C_{0}$ , we have $C = 2 C_{0}$ . Substituting $C_{0} = \frac{ϵ_{0} A}{d}$ :

2 \frac{ϵ_{0} A}{d} = \frac{ϵ_{0} A}{d (\frac{3}{8 K_{1}} + \frac{1}{2 K_{2}} + \frac{1}{8})}

Simplifying the equation by canceling $\frac{ϵ_{0} A}{d}$ from both sides:

\frac{1}{2} = \frac{3}{8 K_{1}} + \frac{1}{2 K_{2}} + \frac{1}{8}

💡 Teacher's Secret Hint

Remember to include the air gap in the sum of effective thicknesses. It's a common mistake to overlook it.

Step 3: Solve for

K_{1}

using the given relation○ Expand

We are given the relation $K_{1} = 1.25 K_{2}$ . This implies $K_{2} = \frac{K_{1}}{1.25} = \frac{K_{1}}{5 / 4} = \frac{4 K_{1}}{5}$ . Substitute this into the equation from Step 2:

\frac{1}{2} = \frac{3}{8 K_{1}} + \frac{1}{2 (\frac{4 K_{1}}{5})} + \frac{1}{8}

Simplify the second term:

\frac{1}{2} = \frac{3}{8 K_{1}} + \frac{5}{8 K_{1}} + \frac{1}{8}

Combine the terms with $K_{1}$ :

\frac{1}{2} = \frac{3 + 5}{8 K_{1}} + \frac{1}{8}

\frac{1}{2} = \frac{8}{8 K_{1}} + \frac{1}{8}

\frac{1}{2} = \frac{1}{K_{1}} + \frac{1}{8}

Now, isolate $\frac{1}{K_{1}}$ and solve for $K_{1}$ :

\frac{1}{K_{1}} = \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}

K_{1} = \frac{8}{3} \approx 2.666 . . .

Comparing this value with the given options, $2.66$ is the closest.

💡 Teacher's Secret Hint

Pay close attention to algebraic manipulation, especially when dealing with fractions and substitutions to avoid calculation errors.

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Physics Question 20 – NEET-UG 2025

Clue 1 of 3: Capacitance with Dielectrics

🧠 Full Solution Path