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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359169 A capacitor having capacitance $1 μ F$ with air, is filled with two dielectrics as shown below. The capacitance between $A$ and $B$ becomes times.
supporting img

1 12

2 6

3

8 / 3

4 3

Explanation:

Initially, the capacitance of capacitor
$C = \frac{ε_{0} A}{d}$
Where, $A$ is area of each plate and $d$ is the seperation between the plates.
supporting img
After filling with dielectrics, we have two capacitors of capacitance.
$C_{1} = \frac{K_{1} ε_{0} (A / 2)}{d} = \frac{8}{2} \frac{ε_{0} A}{d} = \frac{4 ε_{0} A}{d} = 4 C$ and $C_{2} = \frac{K_{2} ε_{0} (A / 2)}{d} = \frac{4}{2} \frac{ε_{0} A}{d} = \frac{2 ε_{0} A}{d} = 2 C$
Hence, their equivalent capacitance is
$C_{e q} = C_{1} + C_{2} = 4 C + 2 C = 6 C$
i.e., new capacitance will be six times the original.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359170 Two parallel metal plates having charges $+ Q$ and $- Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will

1 remain same

2 become zero

3 increases

4 decrease

Explanation:

Electric field $E = \frac{σ}{ε} = \frac{Q}{A ε}$
$ε$ of kerosine oil is more than that of air.
As $ε$ increases, $E$ decreases.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359171 A parallel plate capacitor is made of two square plates of side $a$ , separated by a distance $d$ $(d << a)$ . The lower triangular portion is filled with a dielectric constant $K$ , as shown in the figure. Capacitance of this capacitor is
supporting img

1

\frac{1}{2} \frac{K ε_{0} a^{2}}{d}

2

\frac{K ε_{0}^{} a^{2}}{d (K - 1)} \ln K

3

\frac{K ε_{0}^{} a^{2}}{2 d (K + 1)}

4

\frac{K ε_{0}^{} a^{2}}{d} \ln K

Explanation:

Consider a thin capacitor of width $d x$ as shown the figure.
supporting img
The capacitance of the thin capacitor is

$d C_{1} = \frac{ε_{o} b d x}{d - y} & d C_{2} = \frac{K ε_{o} b d x}{y}$
$\frac{1}{d C} = \frac{1}{d C_{1}} + \frac{1}{d C_{2}}$
$\Rightarrow \int d C = \int \frac{ε_{o} b K d x}{K d + y (1 - K)}$
$C = \frac{ε_{o} b K}{Tan θ} \int_{0}^{d} \frac{d y}{K d + y (1 - K)}$
$(Tan θ = \frac{y}{x} & d x = \frac{d y}{Tan θ})$
$C = \frac{ε_{o} K a^{2}}{d (1 - K)} \ln (\frac{1}{K})$
$(Tan θ = \frac{d}{a})$

JEE - 2019

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359172 Assertion :
When air between the plates of a parallel plate condenser is replaced by an insulating medium of dielectric constant its capacity increases.
Reason :
Electric field intensity between the plates with dielectric in between it is reduced.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$C^{'} = ϵ_{r} C$ and $E^{'} = \frac{E}{ϵ_{r}}$ The introduction of a dielectric with a specific constant between capacitor plates reduces electric field intensity, resulting in increased capacitance compared to when air is present. This is due to the dielectric's ability to polarize and weaken the electric field. So correct option is (1).

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359169 A capacitor having capacitance $1 μ F$ with air, is filled with two dielectrics as shown below. The capacitance between $A$ and $B$ becomes times.
supporting img

1 12

2 6

3

8 / 3

4 3

Explanation:

Initially, the capacitance of capacitor
$C = \frac{ε_{0} A}{d}$
Where, $A$ is area of each plate and $d$ is the seperation between the plates.
supporting img
After filling with dielectrics, we have two capacitors of capacitance.
$C_{1} = \frac{K_{1} ε_{0} (A / 2)}{d} = \frac{8}{2} \frac{ε_{0} A}{d} = \frac{4 ε_{0} A}{d} = 4 C$ and $C_{2} = \frac{K_{2} ε_{0} (A / 2)}{d} = \frac{4}{2} \frac{ε_{0} A}{d} = \frac{2 ε_{0} A}{d} = 2 C$
Hence, their equivalent capacitance is
$C_{e q} = C_{1} + C_{2} = 4 C + 2 C = 6 C$
i.e., new capacitance will be six times the original.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359170 Two parallel metal plates having charges $+ Q$ and $- Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will

1 remain same

2 become zero

3 increases

4 decrease

Explanation:

Electric field $E = \frac{σ}{ε} = \frac{Q}{A ε}$
$ε$ of kerosine oil is more than that of air.
As $ε$ increases, $E$ decreases.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359171 A parallel plate capacitor is made of two square plates of side $a$ , separated by a distance $d$ $(d << a)$ . The lower triangular portion is filled with a dielectric constant $K$ , as shown in the figure. Capacitance of this capacitor is
supporting img

1

\frac{1}{2} \frac{K ε_{0} a^{2}}{d}

2

\frac{K ε_{0}^{} a^{2}}{d (K - 1)} \ln K

3

\frac{K ε_{0}^{} a^{2}}{2 d (K + 1)}

4

\frac{K ε_{0}^{} a^{2}}{d} \ln K

Explanation:

Consider a thin capacitor of width $d x$ as shown the figure.
supporting img
The capacitance of the thin capacitor is

$d C_{1} = \frac{ε_{o} b d x}{d - y} & d C_{2} = \frac{K ε_{o} b d x}{y}$
$\frac{1}{d C} = \frac{1}{d C_{1}} + \frac{1}{d C_{2}}$
$\Rightarrow \int d C = \int \frac{ε_{o} b K d x}{K d + y (1 - K)}$
$C = \frac{ε_{o} b K}{Tan θ} \int_{0}^{d} \frac{d y}{K d + y (1 - K)}$
$(Tan θ = \frac{y}{x} & d x = \frac{d y}{Tan θ})$
$C = \frac{ε_{o} K a^{2}}{d (1 - K)} \ln (\frac{1}{K})$
$(Tan θ = \frac{d}{a})$

JEE - 2019

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359172 Assertion :
When air between the plates of a parallel plate condenser is replaced by an insulating medium of dielectric constant its capacity increases.
Reason :
Electric field intensity between the plates with dielectric in between it is reduced.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$C^{'} = ϵ_{r} C$ and $E^{'} = \frac{E}{ϵ_{r}}$ The introduction of a dielectric with a specific constant between capacitor plates reduces electric field intensity, resulting in increased capacitance compared to when air is present. This is due to the dielectric's ability to polarize and weaken the electric field. So correct option is (1).

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359169 A capacitor having capacitance $1 μ F$ with air, is filled with two dielectrics as shown below. The capacitance between $A$ and $B$ becomes times.
supporting img

1 12

2 6

3

8 / 3

4 3

Explanation:

Initially, the capacitance of capacitor
$C = \frac{ε_{0} A}{d}$
Where, $A$ is area of each plate and $d$ is the seperation between the plates.
supporting img
After filling with dielectrics, we have two capacitors of capacitance.
$C_{1} = \frac{K_{1} ε_{0} (A / 2)}{d} = \frac{8}{2} \frac{ε_{0} A}{d} = \frac{4 ε_{0} A}{d} = 4 C$ and $C_{2} = \frac{K_{2} ε_{0} (A / 2)}{d} = \frac{4}{2} \frac{ε_{0} A}{d} = \frac{2 ε_{0} A}{d} = 2 C$
Hence, their equivalent capacitance is
$C_{e q} = C_{1} + C_{2} = 4 C + 2 C = 6 C$
i.e., new capacitance will be six times the original.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359170 Two parallel metal plates having charges $+ Q$ and $- Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will

1 remain same

2 become zero

3 increases

4 decrease

Explanation:

Electric field $E = \frac{σ}{ε} = \frac{Q}{A ε}$
$ε$ of kerosine oil is more than that of air.
As $ε$ increases, $E$ decreases.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359171 A parallel plate capacitor is made of two square plates of side $a$ , separated by a distance $d$ $(d << a)$ . The lower triangular portion is filled with a dielectric constant $K$ , as shown in the figure. Capacitance of this capacitor is
supporting img

1

\frac{1}{2} \frac{K ε_{0} a^{2}}{d}

2

\frac{K ε_{0}^{} a^{2}}{d (K - 1)} \ln K

3

\frac{K ε_{0}^{} a^{2}}{2 d (K + 1)}

4

\frac{K ε_{0}^{} a^{2}}{d} \ln K

Explanation:

Consider a thin capacitor of width $d x$ as shown the figure.
supporting img
The capacitance of the thin capacitor is

$d C_{1} = \frac{ε_{o} b d x}{d - y} & d C_{2} = \frac{K ε_{o} b d x}{y}$
$\frac{1}{d C} = \frac{1}{d C_{1}} + \frac{1}{d C_{2}}$
$\Rightarrow \int d C = \int \frac{ε_{o} b K d x}{K d + y (1 - K)}$
$C = \frac{ε_{o} b K}{Tan θ} \int_{0}^{d} \frac{d y}{K d + y (1 - K)}$
$(Tan θ = \frac{y}{x} & d x = \frac{d y}{Tan θ})$
$C = \frac{ε_{o} K a^{2}}{d (1 - K)} \ln (\frac{1}{K})$
$(Tan θ = \frac{d}{a})$

JEE - 2019

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359172 Assertion :
When air between the plates of a parallel plate condenser is replaced by an insulating medium of dielectric constant its capacity increases.
Reason :
Electric field intensity between the plates with dielectric in between it is reduced.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$C^{'} = ϵ_{r} C$ and $E^{'} = \frac{E}{ϵ_{r}}$ The introduction of a dielectric with a specific constant between capacitor plates reduces electric field intensity, resulting in increased capacitance compared to when air is present. This is due to the dielectric's ability to polarize and weaken the electric field. So correct option is (1).

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359169 A capacitor having capacitance $1 μ F$ with air, is filled with two dielectrics as shown below. The capacitance between $A$ and $B$ becomes times.
supporting img

1 12

2 6

3

8 / 3

4 3

Explanation:

Initially, the capacitance of capacitor
$C = \frac{ε_{0} A}{d}$
Where, $A$ is area of each plate and $d$ is the seperation between the plates.
supporting img
After filling with dielectrics, we have two capacitors of capacitance.
$C_{1} = \frac{K_{1} ε_{0} (A / 2)}{d} = \frac{8}{2} \frac{ε_{0} A}{d} = \frac{4 ε_{0} A}{d} = 4 C$ and $C_{2} = \frac{K_{2} ε_{0} (A / 2)}{d} = \frac{4}{2} \frac{ε_{0} A}{d} = \frac{2 ε_{0} A}{d} = 2 C$
Hence, their equivalent capacitance is
$C_{e q} = C_{1} + C_{2} = 4 C + 2 C = 6 C$
i.e., new capacitance will be six times the original.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359170 Two parallel metal plates having charges $+ Q$ and $- Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will

1 remain same

2 become zero

3 increases

4 decrease

Explanation:

Electric field $E = \frac{σ}{ε} = \frac{Q}{A ε}$
$ε$ of kerosine oil is more than that of air.
As $ε$ increases, $E$ decreases.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359171 A parallel plate capacitor is made of two square plates of side $a$ , separated by a distance $d$ $(d << a)$ . The lower triangular portion is filled with a dielectric constant $K$ , as shown in the figure. Capacitance of this capacitor is
supporting img

1

\frac{1}{2} \frac{K ε_{0} a^{2}}{d}

2

\frac{K ε_{0}^{} a^{2}}{d (K - 1)} \ln K

3

\frac{K ε_{0}^{} a^{2}}{2 d (K + 1)}

4

\frac{K ε_{0}^{} a^{2}}{d} \ln K

Explanation:

Consider a thin capacitor of width $d x$ as shown the figure.
supporting img
The capacitance of the thin capacitor is

$d C_{1} = \frac{ε_{o} b d x}{d - y} & d C_{2} = \frac{K ε_{o} b d x}{y}$
$\frac{1}{d C} = \frac{1}{d C_{1}} + \frac{1}{d C_{2}}$
$\Rightarrow \int d C = \int \frac{ε_{o} b K d x}{K d + y (1 - K)}$
$C = \frac{ε_{o} b K}{Tan θ} \int_{0}^{d} \frac{d y}{K d + y (1 - K)}$
$(Tan θ = \frac{y}{x} & d x = \frac{d y}{Tan θ})$
$C = \frac{ε_{o} K a^{2}}{d (1 - K)} \ln (\frac{1}{K})$
$(Tan θ = \frac{d}{a})$

JEE - 2019

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359172 Assertion :
When air between the plates of a parallel plate condenser is replaced by an insulating medium of dielectric constant its capacity increases.
Reason :
Electric field intensity between the plates with dielectric in between it is reduced.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$C^{'} = ϵ_{r} C$ and $E^{'} = \frac{E}{ϵ_{r}}$ The introduction of a dielectric with a specific constant between capacitor plates reduces electric field intensity, resulting in increased capacitance compared to when air is present. This is due to the dielectric's ability to polarize and weaken the electric field. So correct option is (1).