21. ⇒ (MHT CET 2021 21th September Morning Shift )

An air filled parallel plate capacitor has a capacity $2 pF$ . The sepeartion of the plates is doubled and the interspace between the plates is filled with dielectric material, then the capacity is increased to $6 pF$ . The dielectric constant of the material is

A. 3

B. 6

C. 2

D. 4

Correct Option is (B)

$\begin{aligned} C = \frac{KA ε_{0}}{d} \\ ∴ \frac{C_{2}}{C_{1}} = \frac{k_{2}}{k_{1}} \cdot \frac{d_{1}}{d_{2}} \end{aligned}$

$\begin{aligned} ∴ \frac{6}{2} = \frac{k_{2}}{1} \cdot \frac{1}{2} \\ ∴ k_{2} = 6 \end{aligned}$

22. ⇒ (MHT CET 2021 20th September Evening Shift )

In the arrangement of the capacitors as shown in figure, each capacitor is of 6 $μ$ F, then equivalent capacity between points A and B is

MHT CET 2021 20th September Evening Shift Physics - Capacitor Question 23 English

A. 12 $μ$ F

B. 6 $μ$ F

C. 4 $μ$ F

D. 10 $μ$ F

Correct Option is (D)

MHT CET 2021 20th September Evening Shift Physics - Capacitor Question 23 English Explanation

C $_{1}$ = C $_{2}$ = C $_{3}$ = C $_{4}$ = 6 $μ$ F

C $_{2}$ and C $_{3}$ are in parallel.

Hence their equivalent capacitance, C $_{5}$ = 2 $\times$ 6 = 12 $μ$ F

C $_{5}$ and C $_{1}$ in series.

Hence their equivalent capacitance C $_{6}$ is given by

$\frac{1}{C_{6}} = \frac{1}{12} + \frac{1}{6} = \frac{3}{12} = \frac{1}{4}$

$∴$ C $_{6}$ = 4 $μ$ F

C $_{6}$ and C $_{4}$ are in parallel.

Hence their equivalent capacitance is C = 4 + 6 = 10 $μ$ F

23. ⇒ (MHT CET 2021 20th September Evening Shift )

A parallel plate air capacitor is charged upto 100 V. A plate 2 mm thick is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. The dielectric constant of the thick plate is

A. 4

B. 5

C. 2

D. 3

Correct Option is (B)

If d is the initial distance between the plates then capacitance is given by $C = \frac{kA \in_{0}}{d}$

When a plate of thickness t = 2 mm is inserted between the plates the capacitance,

$C_{1} = \frac{k A \in_{0}}{(d - t + \frac{t}{k})}$

If distance between the plates is increased by $x = 1.6 mm$

Capacitance becomes, $C_{2} = \frac{kA \in_{0}}{(d + x - t + \frac{t}{k})}$ But $C_{2} = C$

$\begin{aligned} ∴ d + x - t + \frac{t}{k} = d \\ ∴ t - x = \frac{t}{k} \\ ∴ 2 - 1.6 = \frac{2}{k} \\ ∴ 0.4 = \frac{2}{k} ∴ k = 5 \end{aligned}$

24. ⇒ (MHT CET 2021 20th September Morning Shift )

A parallel plate capacitor filled with oil of a dielectric constant 3 between the plates has capacitance 'C'. If the oil is removed, then the capacitance of the capacitor will be

A. $\frac{C}{\sqrt{3}}$

B. 3C

C. $\sqrt{3}$ C

D. $\frac{C}{3}$

Correct Option is (D)

If dielectric of dielectric constant k is placed between the plates of a capacitor its capacitance increase k times. If it is removed the capacitance will become $\frac{1}{k}$ times.

⇒Electrostatics