6. ⇒ (MHT CET 2023 11th May Evening Shift )

Three identical capacitors of capacitance ' $C$ ' each are connected in series and this connection is connected in parallel with one more such identical capacitor. Then the capacitance of whole combination is

A. $3 C$

B. $2 C$

C. $\frac{4}{3} C$

D. $\frac{3}{4} C$

Correct Option is (C)

The equivalent capacitance of three capacitances connected in series is

$\begin{aligned} \frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} \\ \frac{1}{C_{e q}} = \frac{3}{C} \\ C_{e q} = \frac{C}{3} \end{aligned}$

This capacitance is connected in parallel with another capacitance.

$\begin{aligned} \Rightarrow C_{total} = \frac{C}{3} + C \\ C_{total} = \frac{4 C}{3} \end{aligned}$

7. ⇒ (MHT CET 2023 11th May Evening Shift )

A parallel plate capacitor is charged by a battery and battery remains connected. The dielectric slab of constant ' $K$ ' is inserted between the plates and then taken out. Then electric field between the plates

A. remains the same

B. increases

C. decreases

D. becomes zero

Correct Option is (A)

Currently no explanation available

8. ⇒ (MHT CET 2023 11th May Morning Shift )

Two identical capacitors have the same capacitance ' $C$ '. One of them is charged to a potential $V_{1}$ and the other to $V_{2}$ . The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

A. $\frac{1}{4} C (V_{1}^{2} - V_{2}^{2})$

B. $\frac{1}{4} C (V_{1}^{2} + V_{2}^{2})$

C. $\frac{1}{4} C {(V_{1} - V_{2})}^{2}$

D. $\frac{1}{4} C {(V_{1} + V_{2})}^{2}$

Correct Option is (C)

Initial energy of the combined system

$U_{1} = \frac{1}{2} {CV}_{1}^{2} + \frac{1}{2} {CV}_{2}^{2} = \frac{C}{2} (V_{1}^{2} + V_{2}^{2})$

On joining the two condensers in parallel, common potential,

$V = \frac{V_{1} + V_{2}}{2}$

$∴$ Final energy of the combined system:

$U_{2} = \frac{1}{2} (C + C) {(\frac{V_{1} + V_{2}}{2})}^{2}$

$∴$ Decrease in energy will be:

$\begin{aligned} Δ U & = U_{1} - U_{2} \\ = \frac{C}{2} (V_{1}^{2} + V_{2}^{2}) - \frac{1}{2} (C + C) {(\frac{V_{1} + V_{2}}{2})}^{2} \\ = \frac{1}{4} C {(V_{f} - V_{2})}^{2} \end{aligned}$

9. ⇒ (MHT CET 2023 10th May Evening Shift )

If a capacitor of capacity $900 μ F$ is charged to $100 V$ and its total energy is transferred to a capacitor of capacity $100 μ F$ , then its potential will be

A. $30 V$

B. $200 V$

C. $300 V$

D. $400 V$

Correct Option is (C)

Let the unknown potential be $V_{u}$

Energy of a capacitor $= \frac{1}{2} {CV}^{2}$

$\begin{aligned} \Rightarrow \frac{1}{2} C_{1} V_{1}^{2} = \frac{1}{2} C_{2} {(V_{u})}^{2} \\ \frac{1}{2} \times 900 \times 10^{- 6} \times (100)^{2} = \frac{1}{2} \times 100 \times 10^{- 6} \times {(V_{u})}^{2} \\ {(V_{u})}^{2} = 9 \times 100^{2} \\ ∴ & V_{u} = 300 V \end{aligned}$

10. ⇒ (MHT CET 2023 10th May Evening Shift )

Two dielectric slabs having dielectric constant ' $K_{1}$ ' and ' $K_{2}$ ' of thickness $\frac{d}{4}$ and $\frac{3 d}{4}$ are inserted between the plates as shown in figure. The net capacitance between $A$ and $B$ is $[ε_{0}$ is permittivity of free space]

MHT CET 2023 10th May Evening Shift Physics - Capacitor Question 11 English

A. $\frac{2 A ε_{0}}{d} [\frac{K_{1} K_{2}}{3 K_{1} + K_{2}}]$

B. $\frac{3 A ε_{0}}{d} [\frac{K_{1} + K_{2}}{K_{1} K_{2}}]$

C. $\frac{3 A_{0}}{2 d} [\frac{K_{1} + K_{2}}{K_{1} K_{2}}]$

D. $\frac{4 A ε_{0}}{d} [\frac{K_{1} K_{2}}{3 K_{1} + K_{2}}]$

Correct Option is (D)

Capacity of $1^{st}$ Capacitor,

$C_{1} = \frac{K_{1} ε_{0} A}{d / 4} = \frac{4 K_{1} ε_{0} A}{d}$

Capacity of $2^{nd}$ Capacitor,

$C_{2} = \frac{K_{2} ε_{0} A}{3 d / 4} = \frac{4 K_{2} ε_{0} A}{3 d}$

Equivalent capacitance $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\begin{aligned} \frac{1}{C_{1}} & = \frac{d}{4 K_{1} ε_{0} A}; \frac{1}{C_{2}} = \frac{3 d}{4 K_{2} ε_{0} A} \\ \frac{1}{C} & = \frac{d}{4 K_{1} ε_{0} A} + \frac{3 d}{4 K_{2} ε_{0} A} \\ \frac{1}{C} & = \frac{d}{4 ε_{0} A} [\frac{1}{K_{1}} + \frac{3}{K_{2}}] \\ \frac{1}{C} & = \frac{d}{4 ε_{0} A} [\frac{K_{2} + 3 K_{1}}{K_{1} K_{2}}] \\ ∴ C & = \frac{4 ε_{0} A}{d} [\frac{K_{1} K_{2}}{3 K_{1} + K_{2}}] \end{aligned}$

⇒Electrostatics