Parallel plate capacitor... separation 5 mm... mica sheet 2 mm... draws 25 more charge. Dielectric constant is.

Given: A parallel plate capacitor has plate separation d = 5 mm. A mica sheet of thickness t = 2 mm is inserted, and the capacitor draws 25 more charge. Find: The dielectric constant K of mica. For a capacitor connected to the same potential difference, charge is proportional to capacitance. Therefore, drawing 25 more charge means C new = 1.25 C0 The initial capacitance is C0 = 0 Ad = 0 A5 mm When a dielectric slab of thickness t is inserted in a capacitor of plate separation d, the new capacitance is C new = 0 Ad - t (1 - 1K) Using d = 5 mm and t = 2 mm, C new = 0 A5 - 2 (1 - 1K) Now use C newC0 = 1.25 So, 0 A5 - 2 (1 - 1K) 0 A5 = 1.25 which gives 55 - 2 + 2K = 1.25 53 + 2K = 1.25 3 + 2K = 4 2K = 1 K = 2 Therefore, the dielectric constant is 2.0 and the correct option is A.

Parallel plate capacitor... separation 5 mm... mica sheet 2 mm...… — Solution | JEE Physics

MCQMediumJEE 2026Capacitors & Dielectrics

Parallel plate capacitor... separation $5 \, \text{mm}$ ... mica sheet $2 \, \text{mm}$ ... draws $25\%$ more charge. Dielectric constant is _____.

A
$2.0$
B
$1.0$
C
$1.5$
D
$2.5$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A parallel plate capacitor has plate separation $d = 5 \, \text{mm}$ . A mica sheet of thickness $t = 2 \, \text{mm}$ is inserted, and the capacitor draws $25\%$ more charge.

Find: The dielectric constant $K$ of mica.

For a capacitor connected to the same potential difference, charge is proportional to capacitance. Therefore, drawing $25\%$ more charge means

C_{\text{new}} = 1.25 \, C_0

Using effective separation

The initial capacitance is

C_0 = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 A}{5 \, \text{mm}}

When a dielectric slab of thickness $t$ is inserted in a capacitor of plate separation $d$ , the new capacitance is

C_{\text{new}} = \frac{\varepsilon_0 A}{d - t\left(1 - \frac{1}{K}\right)}

Using $d = 5 \, \text{mm}$ and $t = 2 \, \text{mm}$ ,

C_{\text{new}} = \frac{\varepsilon_0 A}{5 - 2\left(1 - \frac{1}{K}\right)}

Ratio approach

Now use

\frac{C_{\text{new}}}{C_0} = 1.25

So,

\frac{\dfrac{\varepsilon_0 A}{5 - 2\left(1 - \frac{1}{K}\right)}}{\dfrac{\varepsilon_0 A}{5}} = 1.25

which gives

\frac{5}{5 - 2 + \dfrac{2}{K}} = 1.25

\frac{5}{3 + \dfrac{2}{K}} = 1.25

3 + \frac{2}{K} = 4

\frac{2}{K} = 1

K = 2

Therefore, the dielectric constant is $2.0$ and the correct option is A.

Common mistakes

Using the air and mica parts as if their capacitances add directly in parallel is incorrect because the layers lie along the field direction and behave like capacitors in series. Use effective separation or series combination instead.
Taking $25\%$ more charge as $C_{\text{new}} = 0.25C_0$ is wrong. Since the potential difference remains the same, $25\%$ more charge means $C_{\text{new}} = 1.25C_0$ .
Forgetting to subtract the dielectric correction from the full separation leads to an incorrect denominator. The correct effective gap is $d - t\left(1 - \frac{1}{K}\right)$ , not just $d - t$ .

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