datavorous/entrance-exam-dataset huggingface dataset:
A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will

Options:
A increase by 50%
B decrease by 15%
C increase by 25%
D increase by 33%

The correct answer is:
increase by 50%

Solution:
When a dielectric of dielectric constant K is inserted between the plates of a capacitor C0, the capacitance becomes KC0.

Energy stored in a capacitor is given by, U=12CV2. Now,

Ui=12K1C0V2 and Uf=12K2C0V2

⇒ΔU=Uf-Ui=12K2-K1C0V2

Percentage change in the energy will be,

ΔUUi×100=12×5×C0212×10×C02×100=50%

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μC charge and the other plate, +4 μC charge. The potential difference developed across the capacitor is:

Options:
A 1 V
B 2 V
C 3 V
D 5 V

The correct answer is:
1 V

Solution:
Charge inside capacitor =1μC
V=QC=1×10-61×10-6=1 volt

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A parallel plate capacitor has a capacitance C=200 pF. It is connected to 230 V ac supply with an angular frequency 300 rad s-1. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

Options:
A 1.38μA and 1.38μA
B 14.3μA and 143 μA
C 13.8μA and 138 μA
D 13.8 μA and 13.8 μA

The correct answer is:
13.8 μA and 13.8 μA

Solution:
By Maxwell's equation for a time varying voltage (AC voltage), both conduction and displacement currents are the same. Therefore,

Idisplacement=Iconduction=VrmsXC=Vrms1ωC=230×300×200×10−12=13.8 μA

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A parallel plate capacitor has a capacity $80 \times 10^{-6} \mathrm{~F}$, when air is present between its plates. The space between the plates is filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of $30 \mathrm{~V}$ by wires. The dielectric slab is then removed. Then, the charge passing through the wire is

Options:
A $12 \times 10^{-3} \mathrm{C}$
B $25.3 \times 10^{-3} \mathrm{C}$
C $120 \times 10^{-3} \mathrm{C}$
D $45.6 \times 10^{-3} \mathrm{C}$

The correct answer is:
$45.6 \times 10^{-3} \mathrm{C}$

Solution:
$C_{\text {air }}=80 \mu \mathrm{F}$
$$
C_{\text {dielectric }}=\varepsilon_r C_{\text {air }}=1600 \mu \mathrm{F}
$$
Charge stored in the presence of air,
$$
q_{\text {air }}=C_{\text {air }} \times V=80 \times 30 \times 10^{-6}=2400 \mu \mathrm{C}
$$
Charge stored in presence of dielectric medium,
$$
\begin{aligned}
q_d=C_{\text {dielectric }} \times V & =1600 \times 30 \times 10^{-6} \\
& =48000 \mu \mathrm{C}
\end{aligned}
$$
When dielectric is removed, effective charge remained is
$$
\begin{aligned}
q & =q_d-q_{\text {air }} \\
& =(48-2.4) \times 10^{-3} \mathrm{C} \\
q & =45.6 \times 10^{-3} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Capacitance, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 1)
A parallel plate capacitor has a capacity $80 \times 10^{-6} \mathrm{~F}$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of $30 \mathrm{~V}$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

Options:
A $45.6 \times 10^{-3} \mathrm{C}$
B $25.3 \times 10^{-3} \mathrm{C}$
C $120 \times 10^{-3} \mathrm{C}$
D $125 \times 10^{-3} \mathrm{C}$

The correct answer is:
$45.6 \times 10^{-3} \mathrm{C}$

Solution:
Charge that through the wire
$\begin{aligned} \Delta q & =\Delta C V \\ & =\left(C^{\prime}-C\right) V \\ & =(k-1) \sigma V \\ & =(20-1)\left(80 \times 10^{-6}\right)(30) \\ & =4.56 \times 10^{-2} \\ & =45.6 \times 10^{-3} \mathrm{C}\end{aligned}$

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and area of each plate is $A$, the energy stored in the capacitor is

Options:
A $\frac{1}{2} \varepsilon_0 E^2$
B $E^2 A d / \varepsilon_0$
C $\frac{1}{2} \varepsilon_0 E^2 A d$
D $\varepsilon_0 E A d$

The correct answer is:
$\frac{1}{2} \varepsilon_0 E^2 A d$

Solution:
Energy density for a parallel plate capacitor $=\frac{1}{2} \varepsilon_0 E^2$ and volume $=A d$
Total energy $=$ energy density $\times$ volume
$$
\begin{aligned}
& =\left(\frac{1}{2} \varepsilon_0 E^2\right) \times(A d) \\
& =\frac{1}{2} \varepsilon_0 E^2 A d
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and area of each plate is $A$, the energy stored in the capacitor is

Options:
A $\frac{1}{2} \varepsilon_0 E^2$
B $E^2 A d / \varepsilon_0$
C $\frac{1}{2} \varepsilon_0 E^2 A d$
D $\varepsilon_0 E A d$

The correct answer is:
$\frac{1}{2} \varepsilon_0 E^2 A d$

Solution:
Energy density for a parallel plate capacitor $=\frac{1}{2} \varepsilon_0 E^2$ and volume $=A d$
Total energy $=$ energy density $\times$ volume
$$
\begin{aligned}
& =\left(\frac{1}{2} \varepsilon_0 E^2\right) \times(A d) \\
& =\frac{1}{2} \varepsilon_0 E^2 A d
\end{aligned}
$$

Topic: Physics, Capacitance, NEET, NEET 2012 (Mains)
A parallel plate capacitor has plate area 40 cm2 and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is:

Options:
A 24ε0 F
B 310ε0 F
C 103ε0 F
D 10ε0 F

The correct answer is:
103ε0 F

Solution:
With insertion of dielectric, the system can be considered as combination of two capacitors in series.

Thus, equivalent capacitance of the system is

1Ceq=1C1+1C2

=1Kε0 At+1ε0 A d-t

=tKε0 A+d-tε0 A

=1×10-35ε0×40×10-4+1×10-3ε0×40×10-4

⇒1Ceq=120ε0+14ε0

⇒Ceq=20×4ε024=10ε03 F.

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A parallel plate capacitor has plate area 100 m2 and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10 . The resultant capacitance of the system is x pF. The value of ε0=8.85×10-12 F m-1. The value of x to the nearest integer is ______.

Options:

The correct answer is:
161

Solution:
Here, Plate areaA=100 m2
Using, Capacity of parallel plate capacitor is,

C=kε0 A d; where d is separation between the plates and k is dielectric constant.

The capacitance with dielectric will be,
C1=10ϵ0(100)5
=200ϵ0

The capacitance without dielectric or air will be,
C2=ϵ0(100)5=20ϵ0
C1 & C2 are in series so,

Ceqv. =C1C2C1+C2

=4000ϵ0220
=160.9×10-12≃161pF

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A parallel plate capacitor has plate of length l, width w and separation of plates is d. It is connected to a battery of emf V. A dielectric slab of the same thickness d and of dielectric constant K=4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

Options:
A 2l3
B l3
C l4
D l2

The correct answer is:
l3

Solution:
Given,

Length of plate of parallel plate capacitor is l

Width of plate is w

Separation of plates is d.

Emf of battery is V

Thickness of dielectric slab is d

Dielectric constant of slab is K = 4

Capacitance of capacitor before the insertion of dielectric slab is

C0=Aε0d=wlε0d ...(1)

Energy stored in the capacitor before the insertion of dielectrics is given by,

U0=12C0V2=12wlε0dV2 ...(2)

Suppose that, the slab is inside the plate at y distance, therefore, it becomes the combination of two capacitors one with air with length (l-y) and another with dielectric with length y as shown in diagram.

Now the capacitance of capacitor in parallel combination is given by,

C=C1+C2 ...(3)

Where,

C1=w(l-y)ε0d ...(4)

C1=wyKε0d ...(5)

Now energy stored in this combination of capacitor is given by

U=12(C1+C2)V2 ...(6)

Now according to the given condition,

2U0=U

⇒212C0V2=12C1+C2V2

⇒2C0=C1+C2

⇒2ϵ0wld=ϵ0Kwyd+ϵ0w(l-y)d

⇒2l=Ky+l-y

⇒l=4y-y=3y

⇒y=l3

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (05 Sep Shift 2)
A parallel plate capacitor has plates of area A separated by distance d between them. It is filled with a dielectric which has a dielectric constant that varies as Kx=K01+αx where x is the distance measured from one of the plates. If αd<<1, the total capacitance of the system is best given by the expression:

Options:
A AK0ε0d1+αd2
B AK0ε0d1+αd22
C AK0ε0d1+α2d22
D AK0ε0d1+αd

The correct answer is:
AK0ε0d1+αd2

Solution:
Capacitance of element, dC=K01+αxε0Adx
∫1dC=∫0ddxK0ε0A1+αx
⇒1C=1K0ε0Aαln1+αd
Given αd≪1
⇒1C=1K0ε0Aααd-α2d22
⇒1C=dK0ε0A1-αd2
⇒C=K0ε0Ad1+αd2

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A parallel plate capacitor has the space between its plates filled by two slabs of thickness \( \frac{d}{2} \) each and dielectric constant \( K_{1} \) and \( K_{2} . d \) is the plate separation of the capacitor. The capacity of the capacitor is :

Options:
A \( \frac{2 \varepsilon_{0} d}{A}\left(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\right) \)
B \( \frac{2 \varepsilon_{0} A}{d}\left(\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right) \)
C \( \frac{2 \varepsilon_{0} A}{A}\left(K_{1}+K_{2}\right) \)
D \( \frac{2 \varepsilon_{0} A}{d}\left(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\right) \)

The correct answer is:
\( \frac{2 \varepsilon_{0} A}{d}\left(\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right) \)

Solution:
Let us understand this such that two capacitors are connected in series with K1 and K2 as dielectric constant of the dielectric mediums present in between the plates.
See the figure down below for better understanding.
C1=K1ε0A(d2)=2K1ε0Ad
C2=K2ε0A(d2)=2K2ε0Ad
Now since both C1 and C2 are arranged in series, we apply the formula for series combination of capacitor:
1Cs=1C1 +1C2=d2K1ε0A+d2K2ε0A
=d2ε0AK1+K2K1K2
Cs=2ε0AdK1K2K1+K2

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor having a plate separation of $2 \mathrm{~mm}$ is charged by connecting it to a $300 \mathrm{~V}$ supply. The energy density is

Options:
A $0.01 \mathrm{~J} / \mathrm{m}$
B $0.1 \mathrm{~J} / \mathrm{m}$
C $\quad 1.0 \mathrm{~J} / \mathrm{m}$ (d)
D $10 \mathrm{~J} / \mathrm{m}$

The correct answer is:
$0.1 \mathrm{~J} / \mathrm{m}$

Solution:
The energy density of parallel plate capacitor
is given by $U=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \varepsilon_0\left(\frac{V}{d}\right)^2$
$=\frac{1}{2} \times 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2 \times\left(\frac{300 \mathrm{volt}}{2 \times 10^{-3} \mathrm{~m}}\right)^2=0.1 \mathrm{~J} / \mathrm{m}^3$

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor having a separation between the plates $\mathrm{d}$, plate area $\mathrm{A}$ and material with dielectric constant $\mathrm{K}$ has capacitance $\mathrm{C}_0$. Now one-third of the material is replaced by another material with dielectric constant $2 \mathrm{~K}$, so that effectively there are two capacitors one with area $\frac{1}{3} \mathrm{~A}$, dielectric constant $2 \mathrm{~K}$ and another with area $\frac{2}{3} \mathrm{~A}$ and dielectric constant $\mathrm{K}$. If the capacitance of this new capacitor is $\mathrm{C}$ then $\frac{\mathrm{C}}{\mathrm{C}_0}$ is

Options:
A
$1$

B
$\frac{4}{3}$

C
$\frac{2}{3}$

D
$\frac{1}{3}$

The correct answer is:
$\frac{4}{3}$

Solution:
$\mathrm{C}_0=\frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}$
$\mathrm{C}=\frac{\mathrm{k} \in_0 2}{3 \mathrm{~d}}+\frac{2 \mathrm{k} \in_0 \mathrm{~A}}{3 \mathrm{~d}}=\frac{4}{3} \frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}$
$\therefore \frac{\mathrm{C}}{\mathrm{C}_0}=\frac{\frac{4}{3} \frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}}{\frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}}=\frac{4}{3}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2013 (25 Apr Online)
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is

Options:
A 560 pJ
B 600 pJ
C 508 pJ
D 692 pJ

The correct answer is:
508 pJ

Solution:
Initial energy of capacitor

Ui=12q2c
=12×120×12012=600 pJ

Since battery is disconnected so charge remain same. Final energy of capacitor

Uf=12q2c k

=12×120×12012×6.5=92 pJ

W+Uf=Ui
W=508 pJ

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of the same area but the thickness, d2, is inserted between the plates as shown in figure having dielectric constant K(=4). The ratio of new capacitance to its original capacitance will be,

Options:
A 2:1
B 8:5
C 6:5
D 4:1

The correct answer is:
8:5

Solution:
First, recall the formula of capacity of a parallel plate capacitor at free space in terms of cross-section area and separation between plates,C0=ϵ0Ad, when a dielectric slabs of thickness, t and dielectric constant k, inserted between plates then-new capacitance becomes,

Ck=ϵ0Ad-t+tk⇒Ck=ϵ0Ad-d2+d8, here t=d2.

⇒Ck=85ϵ0Ad⇒Ck=85C0⇒CkC0=85

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

Options:
A the energy stored in the capacitor decreases
B the capacitance of the capacitor increases
C the charge on the capacitor decreases
D the voltage across the capacitor increases

The correct answer is:
the voltage across the capacitor increases

Solution:
We know that
The parallel plate capacitor $C=\frac{\varepsilon_{0} \cdot A}{d}$
When $d$ (distance between two parallel plate) increase, then $C$ will be decrease
$\because$
$Q=C V \quad$ where $Q=$ constant
$\therefore$ The voltage across the capacitor increases.

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulting handles, then

Options:
A the capacitance of the capacitor increases
B the voltage across the capacitor increases
C the energy stored in the capacitor decreases
D the charge on the capacitor decreases

The correct answer is:
the voltage across the capacitor increases

Solution:
When the capacitor is charged and then disconnected, the charge across the plates remains a constant, even when the plates are separated.
The capacitance is related to the structure of parallel plate capacitor as $\frac{A \varepsilon_0}{d}$. When the plate is separated, the distance between the plate $d$ increases and hence the capacitance decreases.
Hence, according to the $Q=C V$ relation, the voltage across the capacitor increases in order to keep the charge constant.

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulting handles, then

Options:
A the capacitance of the capacitor increases
B the voltage across the capacitor increases
C the energy stored in the capacitor decreases
D the charge on the capacitor decreases

The correct answer is:
the voltage across the capacitor increases

Solution:
When the capacitor is charged and then disconnected, the charge across the plates remains a constant, even when the plates are separated.
The capacitance is related to the structure of parallel plate capacitor as $\frac{A \varepsilon_0}{d}$. When the plate is separated, the distance between the plate $d$ increases and hence the capacitance decreases.
Hence, according to the $Q=C V$ relation, the voltage across the capacitor increases in order to keep the charge constant.

Topic: Physics, Capacitance, MHT CET, MHT CET 2022 (10 Aug Shift 2)
A parallel plate capacitor is charged and then isolated. The effect of increasing the plate
separation on charge, potential and capacitance respectively are

Options:
A constant, decreases, decreases
B increases, decreases, decreases
C constant, decreases, increases
D constant, increases, decreases

The correct answer is:
constant, increases, decreases

Solution:
Given, parallel plate is charged and then isolated. Therefore, \( Q= \) constant.
Now, capacitance is given as \( C=\frac{\varepsilon_{0} A}{d} \Rightarrow C \propto \frac{1}{d} \)
So, if the separation between plates is increased then, C decreases.
Also, we know \( Q=C V \Rightarrow V=\frac{Q}{C} \Rightarrow V \propto \frac{1}{C} \)
So, if \( C \) decreases then potential increases.
Thus, on increasing the plate separation charge remains constant, potential increases and capacitance decreases.

Topic: Physics, Electrostatics, KCET, KCET 2015
A parallel plate capacitor is charged by a battery and battery remains connected. The dielectric slab of constant ' $\mathrm{K}$ ' is inserted between the plates and then taken out. Then electric field between the plates

Options:
A remains the same
B increases
C decreases
D becomes zero

The correct answer is:
remains the same

Solution:
Std.12|Ch-8| Subtopic-8.10

Topic: Physics, Capacitance, MHT CET, MHT CET 2023 (11 May Shift 2)
A parallel plate capacitor is charged to \( 60 \mu \mathrm{C} \). Due to a radioactive source, the plate loses charge at the rate of
\( 1.8 \times 10^{-8} \mathrm{C} \mathrm{s}^{-1} \). The magnitude of displacement current is

Options:
A \( 1.8 \times 10^{-8} \mathrm{C} \mathrm{s}^{-1} \)
B \( 3.6 \times 10^{-8} \mathrm{C} \mathrm{s}^{-1} \)
C \( 4.1 \times 10^{-11} \mathrm{C} \mathrm{s}^{-1} \)
D \( 5.7 \times 10^{-12} \mathrm{C} \mathrm{s}^{-1} \)

The correct answer is:
\( 1.8 \times 10^{-8} \mathrm{C} \mathrm{s}^{-1} \)

Solution:
Given that,
dqdt=1.8×10-8
the displacement current is
Id=dqdt Id=1.8×10-8 C s-1

Topic: Physics, Electromagnetic Waves, JEE Main
A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations.
A. Key $\mathrm{K}$ is kept closed and plates of capacitors are moved apart using insulating handle.
B. Key $\mathrm{K}$ is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).

Options:
A
In $\mathrm{A} Q$ remains same but $\mathrm{C}$ changes

B
In $\mathrm{B} \mathrm{V}$ remains same but $\mathrm{C}$ changes

C
In A V remains same and hence $Q$ changes

D
In B Q remains same and hence $\mathrm{V}$ changes.

The correct answers are:
In A V remains same and hence $Q$ changes
,
In B Q remains same and hence $\mathrm{V}$ changes.

Solution:
Case $\mathbf{A}$ When $\mathrm{K}$ is kept closed and plates of capacitors are moved apart using insulating handle the space between two plates increases then the capacitance decrease by relation $\left(\mathrm{C}=\frac{K \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)$ and hence, the charge stored decreases as $\mathrm{Q}=\mathrm{CV}$ (potential continue to be the same as capacitor is still connected with cell).
Case B When $K$ is opened and plates of capacitors are moved apart using insulating handle, charge stored by disconnected charged capacitor remains conserved and capacitance decrease by moving apart plates of capacitor. So by relation $\mathrm{Q}=\mathrm{CV}$, here $\mathrm{K}$ is open, so charge $\mathrm{Q}$ remains same in turn (V) will increase on decreasing the capacitance.

Topic: Physics, Electrostatic Potential and Capacitance
A parallel plate capacitor is formed by two plates each of area 30πcm2 separated by 1 mm. A material of dielectric strength 3.6×107 V m-1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7×10-6 C, the value of dielectric constant of the material is :

[Use 14πεo=9×109 N m2 C-2]

Options:
A 1.66
B 1.75
C 2.25
D 2.33

The correct answer is:
2.33

Solution:
The dielectric strength given is 3.6×107 V m-1.and the separation is 1 mm. Therefore, maximum potential difference that can be applied will be,

Vmax=3.6×107×d.

Now,

C=qmaxVmax⇒kε0Ad=7×10-73.6×107×dk=dε0×30π×10-4×7×10-73.6×107×d=2.33

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (24 Jun Shift 1)
A parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternately. If the capacitance between any two plates is $C$ then the resultant capacitance is

Options:
A $C$
B $n C$
C $(n-1) C$
D $(n+1) C$

The correct answer is:
$(n-1) C$

Solution:
For \(n\) plates joined alternately, there will be \((n-1)\) capacitors connected in parallel Fig. Their

equivalent capacitance will be \(\mathrm{C}_{\mathrm{eq}}=(\mathrm{n}-1) \mathrm{C}\).

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C then the resultant capacitance is

Options:
A
$(n-1) C$

B
$(n+1) C$

C
$C$

D
$nC$

The correct answer is:
$(n-1) C$

Solution:
$C_{e q}=(n-1) C(\because$ all capacitors are in parallel $)$

Topic: Physics, Capacitance, JEE Main, JEE Main 2005
A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 1 0 4 V/m , the charge density of the positive plate will be close to :

Options:
A 6 Ã 1 0 - 7 C / m 2
B 3 × 1 0 - 7 C / m 2
C 3 × 1 0 4 C / m 2
D 6 × 1 0 4 C / m 2

The correct answer is:
6 Ã 1 0 - 7 C / m 2

Solution:
E = 3 × 1 0 4 V/m
We know that
Q = CV
=kε0Ad·E·d
=kε0AE
QA=kε0E
=2.2×8.85×10-12×3×104
=58.41×10-8
=5.84×10-7
≈6×10-7C/m2

Topic: Physics, Capacitance, JEE Main, JEE Main 2014 (06 Apr)
A parallel plate capacitor is made of two plates of length 1 , width $w$ and separated by distance $d$. A dielectric slab (dielectric constant $\mathrm{K}$ ) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $\mathrm{F}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}$ where $\mathrm{U}$ is the energy of the capacitor when dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is Q then the force on the dielectric when it is near the edge is:

Options:
A
$\frac{\mathrm{Q}^2 \mathrm{~d}}{2 \mathrm{wl}^2 \varepsilon_{\mathrm{o}}} \mathrm{K}$

B
$\frac{\mathrm{Q}^2 \mathrm{~W}}{2 \mathrm{dl}^2 \varepsilon_0}(\mathrm{~K}-1)$

C
$\frac{\mathrm{Q}^2 \mathrm{~d}}{2 \mathrm{wl}^2 \varepsilon_{\mathrm{o}}}(\mathrm{K}-1)$

D
$\frac{\mathrm{Q}^2 \mathrm{w}}{2 \mathrm{dl}^2 \varepsilon_{\mathrm{o}}} \mathrm{K}$

The correct answer is:
$\frac{\mathrm{Q}^2 \mathrm{~d}}{2 \mathrm{wl}^2 \varepsilon_{\mathrm{o}}}(\mathrm{K}-1)$

Solution:
No Solution Available

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor is made of two plates of length 1 , width $w$ and separated by distance $d$. A dielectric slab (dielectric constant $\mathrm{K}$ ) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $\mathrm{F}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}$ where $\mathrm{U}$ is the energy of the capacitor when dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is Q then the force on the dielectric when it is near the edge is:

Options:
A
$\frac{\mathrm{Q}^2 \mathrm{~d}}{2 \mathrm{wl}^2 \varepsilon_{\mathrm{o}}} \mathrm{K}$

B
$\frac{\mathrm{Q}^2 \mathrm{~W}}{2 \mathrm{dl}^2 \varepsilon_0}(\mathrm{~K}-1)$

C
$\frac{\mathrm{Q}^2 \mathrm{~d}}{2 \mathrm{wl}^2 \varepsilon_{\mathrm{o}}}(\mathrm{K}-1)$

D
$\frac{\mathrm{Q}^2 \mathrm{w}}{2 \mathrm{dl}^2 \varepsilon_{\mathrm{o}}} \mathrm{K}$

The correct answer is:
$\frac{\mathrm{Q}^2 \mathrm{~d}}{2 \mathrm{wl}^2 \varepsilon_{\mathrm{o}}}(\mathrm{K}-1)$

Solution:
No Solution Available

Topic: Physics, Capacitance, JEE Main, JEE Main 2014 (11 Apr Online)