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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359173 A parallel plate condenser with oil between the plates (dielectric constant of oil $K =$ 2) has a capacitance $C$ . If the oil is removed, then capacitance of the capacitor becomes

1

2 C

2

\frac{C}{2}

3

\sqrt{2} C

4

\frac{C}{\sqrt{2}}

Explanation:

$C_{a i r} = \frac{C_{m e d i u m}}{K} = \frac{C}{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359174 A capacitor connected to a 10 $V$ battery collects a charge of $20 μ C$ with air as dielectric and $100 μ C$ with a given oil as dielectric. The dielectric constant of the oil is

1

1.5

2

2.0

3

5

4

3.0

Explanation:

$q = C V,$ As $V$ is constant, therefore $q \propto C$
$\frac{q_{1}}{q_{2}} = \frac{C_{0}}{K C_{0}} = \frac{1}{K}$
$\Rightarrow K = \frac{q_{1}}{q_{2}} = 5$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359175 A capacitor has air as dielectric medium and two conducting plates of area $12 c m^{2}$ and they are $0.6 c m$ apart. When a slab of dielectric constant $(K)$ having same area and thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 c m$ to keep the capacitance same as in previous case. The dielectric constant of the slab is (Given $ε_{0} = 8.834 \times 10^{- 12} F / m)$

1 1.50

2 1

3 0.66

4 1.33

Explanation:

The given condition is shown below :
supporting img

Area of plate, $A = 12 c m^{2}$
The capacitance is same for both cases.
So, $C_{1} = C_{2}$
$\frac{ε_{0} A}{d_{1}} = \frac{ε_{0} A}{d_{2} - t + \frac{t}{k}}$
$\Rightarrow \frac{1}{0.6} = \frac{1}{0.8 - 0.6 + \frac{0.6}{k}}$
$\Rightarrow \frac{1}{0.6} = \frac{1}{0.2 + \frac{0.6}{k}}$
On solving, we get $k = 1.5$

JEE - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359176 The capacity of a parallel plate condenser is $5 μ F$ without dielectric. Dielectric of constant 3 is used to fill half the distance between the plates, the new capacitance in $μ F$ is

1

10

2

5

3

15

4

7.5

Explanation:

Let $d$ be the distance between the plates. Before inserting the slab the capacitance is
$C_{1} = \frac{ε_{0} A}{d}$
After insering the slab of thickness $\frac{d}{2}$ the capacitance is
$\frac{1}{C_{2}} = \frac{d / 2}{ε_{0} A} + \frac{d / 2}{K ε_{0} A} = \frac{d}{2 ε_{0} A} [1 + \frac{1}{k}]$
Where $k$ is the dielectric constant of the medium
$\Rightarrow \frac{1}{C_{2}} = \frac{1}{2 C_{1}} [1 + \frac{1}{k}]$
$C_{2} = \frac{2 C_{1} k}{k + 1} = \frac{2 \times 5 \times 3}{3 + 1} = 7.5$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359173 A parallel plate condenser with oil between the plates (dielectric constant of oil $K =$ 2) has a capacitance $C$ . If the oil is removed, then capacitance of the capacitor becomes

1

2 C

2

\frac{C}{2}

3

\sqrt{2} C

4

\frac{C}{\sqrt{2}}

Explanation:

$C_{a i r} = \frac{C_{m e d i u m}}{K} = \frac{C}{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359174 A capacitor connected to a 10 $V$ battery collects a charge of $20 μ C$ with air as dielectric and $100 μ C$ with a given oil as dielectric. The dielectric constant of the oil is

1

1.5

2

2.0

3

5

4

3.0

Explanation:

$q = C V,$ As $V$ is constant, therefore $q \propto C$
$\frac{q_{1}}{q_{2}} = \frac{C_{0}}{K C_{0}} = \frac{1}{K}$
$\Rightarrow K = \frac{q_{1}}{q_{2}} = 5$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359175 A capacitor has air as dielectric medium and two conducting plates of area $12 c m^{2}$ and they are $0.6 c m$ apart. When a slab of dielectric constant $(K)$ having same area and thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 c m$ to keep the capacitance same as in previous case. The dielectric constant of the slab is (Given $ε_{0} = 8.834 \times 10^{- 12} F / m)$

1 1.50

2 1

3 0.66

4 1.33

Explanation:

The given condition is shown below :
supporting img

Area of plate, $A = 12 c m^{2}$
The capacitance is same for both cases.
So, $C_{1} = C_{2}$
$\frac{ε_{0} A}{d_{1}} = \frac{ε_{0} A}{d_{2} - t + \frac{t}{k}}$
$\Rightarrow \frac{1}{0.6} = \frac{1}{0.8 - 0.6 + \frac{0.6}{k}}$
$\Rightarrow \frac{1}{0.6} = \frac{1}{0.2 + \frac{0.6}{k}}$
On solving, we get $k = 1.5$

JEE - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359176 The capacity of a parallel plate condenser is $5 μ F$ without dielectric. Dielectric of constant 3 is used to fill half the distance between the plates, the new capacitance in $μ F$ is

1

10

2

5

3

15

4

7.5

Explanation:

Let $d$ be the distance between the plates. Before inserting the slab the capacitance is
$C_{1} = \frac{ε_{0} A}{d}$
After insering the slab of thickness $\frac{d}{2}$ the capacitance is
$\frac{1}{C_{2}} = \frac{d / 2}{ε_{0} A} + \frac{d / 2}{K ε_{0} A} = \frac{d}{2 ε_{0} A} [1 + \frac{1}{k}]$
Where $k$ is the dielectric constant of the medium
$\Rightarrow \frac{1}{C_{2}} = \frac{1}{2 C_{1}} [1 + \frac{1}{k}]$
$C_{2} = \frac{2 C_{1} k}{k + 1} = \frac{2 \times 5 \times 3}{3 + 1} = 7.5$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359173 A parallel plate condenser with oil between the plates (dielectric constant of oil $K =$ 2) has a capacitance $C$ . If the oil is removed, then capacitance of the capacitor becomes

1

2 C

2

\frac{C}{2}

3

\sqrt{2} C

4

\frac{C}{\sqrt{2}}

Explanation:

$C_{a i r} = \frac{C_{m e d i u m}}{K} = \frac{C}{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359174 A capacitor connected to a 10 $V$ battery collects a charge of $20 μ C$ with air as dielectric and $100 μ C$ with a given oil as dielectric. The dielectric constant of the oil is

1

1.5

2

2.0

3

5

4

3.0

Explanation:

$q = C V,$ As $V$ is constant, therefore $q \propto C$
$\frac{q_{1}}{q_{2}} = \frac{C_{0}}{K C_{0}} = \frac{1}{K}$
$\Rightarrow K = \frac{q_{1}}{q_{2}} = 5$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359175 A capacitor has air as dielectric medium and two conducting plates of area $12 c m^{2}$ and they are $0.6 c m$ apart. When a slab of dielectric constant $(K)$ having same area and thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 c m$ to keep the capacitance same as in previous case. The dielectric constant of the slab is (Given $ε_{0} = 8.834 \times 10^{- 12} F / m)$

1 1.50

2 1

3 0.66

4 1.33

Explanation:

The given condition is shown below :
supporting img

Area of plate, $A = 12 c m^{2}$
The capacitance is same for both cases.
So, $C_{1} = C_{2}$
$\frac{ε_{0} A}{d_{1}} = \frac{ε_{0} A}{d_{2} - t + \frac{t}{k}}$
$\Rightarrow \frac{1}{0.6} = \frac{1}{0.8 - 0.6 + \frac{0.6}{k}}$
$\Rightarrow \frac{1}{0.6} = \frac{1}{0.2 + \frac{0.6}{k}}$
On solving, we get $k = 1.5$

JEE - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359176 The capacity of a parallel plate condenser is $5 μ F$ without dielectric. Dielectric of constant 3 is used to fill half the distance between the plates, the new capacitance in $μ F$ is

1

10

2

5

3

15

4

7.5

Explanation:

Let $d$ be the distance between the plates. Before inserting the slab the capacitance is
$C_{1} = \frac{ε_{0} A}{d}$
After insering the slab of thickness $\frac{d}{2}$ the capacitance is
$\frac{1}{C_{2}} = \frac{d / 2}{ε_{0} A} + \frac{d / 2}{K ε_{0} A} = \frac{d}{2 ε_{0} A} [1 + \frac{1}{k}]$
Where $k$ is the dielectric constant of the medium
$\Rightarrow \frac{1}{C_{2}} = \frac{1}{2 C_{1}} [1 + \frac{1}{k}]$
$C_{2} = \frac{2 C_{1} k}{k + 1} = \frac{2 \times 5 \times 3}{3 + 1} = 7.5$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359173 A parallel plate condenser with oil between the plates (dielectric constant of oil $K =$ 2) has a capacitance $C$ . If the oil is removed, then capacitance of the capacitor becomes

1

2 C

2

\frac{C}{2}

3

\sqrt{2} C

4

\frac{C}{\sqrt{2}}

Explanation:

$C_{a i r} = \frac{C_{m e d i u m}}{K} = \frac{C}{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359174 A capacitor connected to a 10 $V$ battery collects a charge of $20 μ C$ with air as dielectric and $100 μ C$ with a given oil as dielectric. The dielectric constant of the oil is

1

1.5

2

2.0

3

5

4

3.0

Explanation:

$q = C V,$ As $V$ is constant, therefore $q \propto C$
$\frac{q_{1}}{q_{2}} = \frac{C_{0}}{K C_{0}} = \frac{1}{K}$
$\Rightarrow K = \frac{q_{1}}{q_{2}} = 5$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359175 A capacitor has air as dielectric medium and two conducting plates of area $12 c m^{2}$ and they are $0.6 c m$ apart. When a slab of dielectric constant $(K)$ having same area and thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 c m$ to keep the capacitance same as in previous case. The dielectric constant of the slab is (Given $ε_{0} = 8.834 \times 10^{- 12} F / m)$

1 1.50

2 1

3 0.66

4 1.33

Explanation:

The given condition is shown below :
supporting img

Area of plate, $A = 12 c m^{2}$
The capacitance is same for both cases.
So, $C_{1} = C_{2}$
$\frac{ε_{0} A}{d_{1}} = \frac{ε_{0} A}{d_{2} - t + \frac{t}{k}}$
$\Rightarrow \frac{1}{0.6} = \frac{1}{0.8 - 0.6 + \frac{0.6}{k}}$
$\Rightarrow \frac{1}{0.6} = \frac{1}{0.2 + \frac{0.6}{k}}$
On solving, we get $k = 1.5$

JEE - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359176 The capacity of a parallel plate condenser is $5 μ F$ without dielectric. Dielectric of constant 3 is used to fill half the distance between the plates, the new capacitance in $μ F$ is

1

10

2

5

3

15

4

7.5

Explanation:

Let $d$ be the distance between the plates. Before inserting the slab the capacitance is
$C_{1} = \frac{ε_{0} A}{d}$
After insering the slab of thickness $\frac{d}{2}$ the capacitance is
$\frac{1}{C_{2}} = \frac{d / 2}{ε_{0} A} + \frac{d / 2}{K ε_{0} A} = \frac{d}{2 ε_{0} A} [1 + \frac{1}{k}]$
Where $k$ is the dielectric constant of the medium
$\Rightarrow \frac{1}{C_{2}} = \frac{1}{2 C_{1}} [1 + \frac{1}{k}]$
$C_{2} = \frac{2 C_{1} k}{k + 1} = \frac{2 \times 5 \times 3}{3 + 1} = 7.5$