datavorous/entrance-exam-dataset huggingface dataset:
A parallel plate capacitor is made of two square plates of side a, separated by a distance d d≪a. The lower triangular portion filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is:

Options:
A Kε0a22dK+1
B Kε0a2dK-1ln⁡K
C Kε0a2dln⁡K
D 12Kε0a2d

The correct answer is:
Kε0a2dK-1ln⁡K

Solution:
Consider a small element at a distance x. It is a two capacitors in series.
dc=dc1×dc2dc1+dc2
dc=ε0.adxxtanθk+d-xtanθ1
∴dc=ε0K.adx-xtan⁡θK-1+d.k
∴C=∫0aε0K.adx-xtan⁡θK-1+d.k
=ε0k.atan⁡θ(K-1)ln -atan⁡θK-1 +d.kd.K
=ε0Ka2dK-1ln K

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor is made of two square plates of side a, separated by a distance d d≪a. The lower triangular portion filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is:

Options:
A Kε0a22dK+1
B Kε0a2dK-1ln⁡K
C Kε0a2dln⁡K
D 12Kε0a2d

The correct answer is:
Kε0a2dK-1ln⁡K

Solution:
Consider a small element at a distance x. It is a two capacitors in series.
dc=dc1×dc2dc1+dc2
dc=ε0.adxxtanθk+d-xtanθ1
∴dc=ε0K.adx-xtan⁡θK-1+d.k
∴C=∫0aε0K.adx-xtan⁡θK-1+d.k
=ε0k.atan⁡θ(K-1)ln -atan⁡θK-1 +d.kd.K
=ε0Ka2dK-1ln K

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A parallel plate capacitor is of area 6 cm2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant K1=10, K2=12 and K3=14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be:

Options:
A 4
B 36
C 14
D 12

The correct answer is:
12

Solution:
C=∈0Ad
C1=K1∈0A3d=K1C3
C2=K2C3
C3=K3C3
C eq = K eq C= C 3 ( K 1 + K 2 + K 3 )
Keq=K1+K2+K33=363=12

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A parallel plate capacitor of \( 6 \mu \mathrm{F} \) is connected across \( 18 \mathrm{~V} \) battery and charged. The battery is then disconnected and oil of dielectric constant \( \mathrm{k}=2.1 \) is introduced between the plates. What will be the charge on capacitor?

Options:
A \( 51.4 \mu \mathrm{C} \)
B \( 108 \mu \mathrm{C} \)
C \( 8.5 \mu \mathrm{C} \)
D \( 92.5 \times 10^{2} \mathrm{C} \)

The correct answer is:
\( 108 \mu \mathrm{C} \)

Solution:
When the capacitor is charged and then disconnected, then form conservation of charge, the charge 'Q' across the plates remains constant, even when oil is introduced between the plates. So the initial and final charge on the capacitor remains same, i.e.,Qfinal=Qinitial=CV=6×18=108 μC.

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor of area $60 \mathrm{~cm}^2$ and separation $3 \mathrm{~mm}$ is charged initially to $90 \mu \mathrm{C}$. If the medium between the plate gets slightly conducting and the plate loses the charge initially at the rate of $2.5 \times 10^{-8} \mathrm{C} / \mathrm{s}$, then what is the magnetic field between the plates ?

Options:
A
$2.5 \times 10^{-8} \mathrm{~T}$

B
$2.0 \times 10^{-7} \mathrm{~T}$

C
$1.63 \times 10^{-11} \mathrm{~T}$

D
Zero

The correct answer is:
Zero

Solution:
Magnetic field between the plates in this case is zero.

Topic: Physics, Capacitance, JEE Main, JEE Main 2013 (23 Apr Online)
A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three dielectric materials having dielectric constants k1,k2 and k3 as shown. If a single dielectric material is to be used to have the same capacitance C in the capacitor, then its dielectric constant k is given by

Options:
A k=k1+k2+2k3
B k=k1k2k1+k2+2k3
C 1k=1k1+1k2+12k3
D 1k=1k1+k2+12k3

The correct answer is:
1k=1k1+k2+12k3

Solution:
The capacitances, C1=k1Îµ0A/2d/2=k1Îµ0Ad
C2=k2Îµ0A/2d/2=k0Îµ0Ad
And C3=k3Îµ0A/2d/2=2k3Îµ0Ad
The equivalent capacitance is given by
â 1Ceq=1C1+C2+1C3=1Îµ0Adk1+k2+1Îµ0AdÃ2k3
â 1Ceq=1C1+C2+1C3=1Îµ0Adk1+k2+1Îµ0AdÃ2k3
â 1Ceq=dÎµ0A1k1+k2+12k3
â Ceq=1k1+k2+12k3-1.Îµ.Ad
â´ 1k=1k1+k2+12k3 âµC=KÎµAd
k= k 1 k 3 k 1 + k 3 + k 2 k 3 k 2 + k 3
C eq = Îµ o A/2 d 2 k 1 + d 2 k 3 + Îµ o A/2 d 2 k 2 + d 2 k 3
= Îµ o A d ( k 1 . k 3+ k 1 + k 3 + k 2 . k 3 k 2 + k 3 )
â´ equivalent
k= k 1 k 3 k 1 + k 3 + k 2 . k 3 k 2 + k 3

Topic: Physics, Electrostatics, JEE Main
A parallel plate capacitor of capacitance 1 Î¼F has a charge of +2 Î¼C on one of the plates and a charge of +4 Î¼C on the other. The potential difference developed across the capacitor is

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

The correct answer is:
1 V

Solution:
V=QC=1μC1μF=1 Volt

Topic: Physics, Electrostatics, JEE Main
A parallel plate capacitor of capacitance 20 μF is being charged by a voltage source whose potential is changing at the rate of 3 V s-1. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor would be, respectively:

Options:
A zero, 60 μA
B 60 μA, 60 μA
C 60 μA, zero
D zero, zero

The correct answer is:
60 μA, 60 μA

Solution:
q=CV
dqdt=Cdvdt
=20×10-6×3
Conduction current =60 μA
Displacement current =ε0dϕedt
=ε0AddVdt
=60 μA

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor of capacitance \( 90 \mathrm{pF} \) is connected to a battery of emf \( 20 \mathrm{~V} \). If a dielectric material of dielectric constant \( K=\frac{5}{3} \) is inserted between the plates, the magnitude of the induced charge will be:

Options:
A \( 2.4 \mathrm{nC} \)
B \( 0.9 \mathrm{nC} \)
C \( 1.2 \mathrm{nC} \)
D \( 0.3 \mathrm{nC} \)

The correct answer is:
\( 1.2 \mathrm{nC} \)

Solution:
Capacitance of parallel plate capacitor with air between the plates C=Aε0d; where A is plate area and d is separation between the plates.

As we know that Q = CV

Initial charge

Qi = 90×10-12× 20=1800×10-12 C

⇒Qi=1.8 nC

Dielectric material of dielectric constant k=53 is inserted between the plates, then capacitance becomes

C'=Aε0kd

⇒C'=kC=53C

⇒C'=53×90×10-12=150×10-12 F

Now the final charge Qf=C'V

⇒Qf=150×10-12×20=3000×10-12 C

⇒Qf=3 nC

Induced charge,

Qinduced=Qf-Qi

Qinduced=3-1.8=1.2 nC.

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor of capacitance 90 pF is connected to a battery of EMF 20 V. If a dielectric material of dielectric constant K=53 is inserted between the plates, the magnitude of the induced charge will be:

Options:
A 0.9 nC
B 1.2 nC
C 0.3 nC
D 2.4 nC

The correct answer is:
1.2 nC

Solution:
Induced charge on the dielectric,
Qinduced=KCV1-1k=53×90×10-12×201-35=1.2 nC.

Topic: Physics, Capacitance, JEE Main, JEE Main 2018 (08 Apr)
A parallel plate capacitor of capacitance 200μF is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be __________ J.

Options:

The correct answer is:
4

Solution:
Initially
C=200μF
Ei=12CV2=12×200×10-6×2002
Finally
C'=KC=400μF
Ef=12C'V2=12×400×10-6×(200)2
ΔE=12×400-200×10-6×4×104=4 J

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A parallel plate capacitor of capacity $100 \mu \mathrm{F}$ is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joules is

Options:
A $\frac{125}{2} \times 10^{-3}$
B $125 \times 10^{-3}$
C $1.25 \times 10^{-3}$
D $0.0125 \times 10^{-3}$

The correct answer is:
$\frac{125}{2} \times 10^{-3}$

Solution:
$$
C=100 \mu \mathrm{F}, V=50 \text { volt }
$$
Capacitance of parallel plate capacitor
$$
C=\frac{\varepsilon_0 A}{d}
$$
$d=$ separation between the plates
Initial energy
$$
\begin{aligned}
E_1 & =\frac{1}{2} C V^2 \\
& =\frac{1}{2} \times\left(100 \times 10^{-6}\right) \times(50)^2 \\
& =50 \times 10^{-6} \times 2500 \\
& =125 \times 10^{-3} \mathrm{~J}
\end{aligned}
$$
When distance between the plates becomes double, then capacitance $C^{\prime}=\frac{C}{2}=\frac{100}{2}=50 \mu \mathrm{F}$
Final energy $E_2=\frac{1}{2} C^{\prime} V^2$
$\begin{aligned} & =\frac{1}{2} \times 50 \times 10^{-6} \times(50)^2 \\ & =25 \times 10^{-6} \times 2500 \\ & =625 \times 10^{-4} \\ & =625 \times 10^{-3} \mathrm{~J}\end{aligned}$
$\begin{aligned} & \text { Additional energy }=E_1-E_2 \\ & =125 \times 10^{-3}-625 \times 10^{-3} \\ & =62.5 \times 10^{-3} \\ & =\frac{125}{2} \times 10^{-3} \mathrm{~J} \\ & \end{aligned}$

Topic: Physics, Capacitance, AP EAMCET, AP EAMCET 2002
A parallel plate capacitor of capacity $100 \mu \mathrm{F}$ is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joules is

Options:
A $\frac{125}{2} \times 10^{-3}$
B $125 \times 10^{-3}$
C $1.25 \times 10^{-3}$
D $0.0125 \times 10^{-3}$

The correct answer is:
$\frac{125}{2} \times 10^{-3}$

Solution:
$$
C=100 \mu \mathrm{F}, V=50 \text { volt }
$$
Capacitance of parallel plate capacitor
$$
C=\frac{\varepsilon_0 A}{d}
$$
$d=$ separation between the plates
Initial energy
$$
\begin{aligned}
E_1 & =\frac{1}{2} C V^2 \\
& =\frac{1}{2} \times\left(100 \times 10^{-6}\right) \times(50)^2 \\
& =50 \times 10^{-6} \times 2500 \\
& =125 \times 10^{-3} \mathrm{~J}
\end{aligned}
$$
When distance between the plates becomes double, then capacitance $C^{\prime}=\frac{C}{2}=\frac{100}{2}=50 \mu \mathrm{F}$
Final energy $E_2=\frac{1}{2} C^{\prime} V^2$
$\begin{aligned} & =\frac{1}{2} \times 50 \times 10^{-6} \times(50)^2 \\ & =25 \times 10^{-6} \times 2500 \\ & =625 \times 10^{-4} \\ & =625 \times 10^{-3} \mathrm{~J}\end{aligned}$
$\begin{aligned} & \text { Additional energy }=E_1-E_2 \\ & =125 \times 10^{-3}-625 \times 10^{-3} \\ & =62.5 \times 10^{-3} \\ & =\frac{125}{2} \times 10^{-3} \mathrm{~J} \\ & \end{aligned}$

Topic: Physics, Capacitance, TS EAMCET, TS EAMCET 2002
A parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then, $E_1 / E_2$ value is:

Options:
A 4
B $3 / 2$
C 2
D $1 / 2$

The correct answer is:
4

Solution:
Capacitance of parallel plate capacitor,
$C_0=\frac{\varepsilon_0 A}{d}$
where, $A=$ area of the plates, $d=$ separation between the plates Charge stored in the capacitor,
$Q=C_0 V_0$
When battery is disconnected, then charge remains same.
So, energy, $E_1=\frac{1}{2} \frac{Q^2}{C}$
$C=$ capacitance when plate separation is doubled
So, $\quad C_1=\frac{C_0}{2}$
$E_1=\frac{1}{2} \frac{Q^2}{C_0 / 2}=\frac{Q^2}{C_0}=\frac{C_0^2 V_0^2}{C_0}=C_0 V_0^2$
When battery is connected, then
Energy, $E_2=\frac{1}{2} C V_0^2$
where, $E_2=\frac{1}{2} \frac{C_0}{2} V_0^2=\frac{1}{4}\left(C_0 V_0^2\right)$
$\therefore \quad \frac{E_1}{E_2}=\frac{C_0 V_0^2}{\frac{1}{4} C_0 V_0^2}=\frac{1}{4}$
$E_1: E_2=4: 1$

Topic: Physics, Capacitance, AP EAMCET, AP EAMCET 2003
A parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then, $E_1 / E_2$ value is:

Options:
A 4
B $3 / 2$
C 2
D $1 / 2$

The correct answer is:
4

Solution:
Capacitance of parallel plate capacitor,
$C_0=\frac{\varepsilon_0 A}{d}$
where, $A=$ area of the plates, $d=$ separation between the plates Charge stored in the capacitor,
$Q=C_0 V_0$
When battery is disconnected, then charge remains same.
So, energy, $E_1=\frac{1}{2} \frac{Q^2}{C}$
$C=$ capacitance when plate separation is doubled
So, $\quad C_1=\frac{C_0}{2}$
$E_1=\frac{1}{2} \frac{Q^2}{C_0 / 2}=\frac{Q^2}{C_0}=\frac{C_0^2 V_0^2}{C_0}=C_0 V_0^2$
When battery is connected, then
Energy, $E_2=\frac{1}{2} C V_0^2$
where, $E_2=\frac{1}{2} \frac{C_0}{2} V_0^2=\frac{1}{4}\left(C_0 V_0^2\right)$
$\therefore \quad \frac{E_1}{E_2}=\frac{C_0 V_0^2}{\frac{1}{4} C_0 V_0^2}=\frac{1}{4}$
$E_1: E_2=4: 1$

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then, $E_1 / E_2$ value is:

Options:
A 4
B $3 / 2$
C 2
D $1 / 2$

The correct answer is:
4

Solution:
Capacitance of parallel plate capacitor,
$C_0=\frac{\varepsilon_0 A}{d}$
where, $A=$ area of the plates, $d=$ separation between the plates Charge stored in the capacitor,
$Q=C_0 V_0$
When battery is disconnected, then charge remains same.
So, energy, $E_1=\frac{1}{2} \frac{Q^2}{C}$
$C=$ capacitance when plate separation is doubled
So, $\quad C_1=\frac{C_0}{2}$
$E_1=\frac{1}{2} \frac{Q^2}{C_0 / 2}=\frac{Q^2}{C_0}=\frac{C_0^2 V_0^2}{C_0}=C_0 V_0^2$
When battery is connected, then
Energy, $E_2=\frac{1}{2} C V_0^2$
where, $E_2=\frac{1}{2} \frac{C_0}{2} V_0^2=\frac{1}{4}\left(C_0 V_0^2\right)$
$\therefore \quad \frac{E_1}{E_2}=\frac{C_0 V_0^2}{\frac{1}{4} C_0 V_0^2}=\frac{1}{4}$
$E_1: E_2=4: 1$

Topic: Physics, Capacitance, TS EAMCET, TS EAMCET 2003
A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V=12 V between its plates. The charging battery is now disconnected and a porcelain plate with k=7 is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of _______ pJ.

(Assume no friction)

Options:

The correct answer is:
864

Solution:
Ui=12×14×12×12 pJ ∵U=12CV2

=1008 pJ

Uf=10087pJ=144 pJ ∵Cm=kC0

Mechanical energy=ΔU

=1008-144

=864 pJ

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A parallel plate capacitor with air between the plate has a capacitance of 15 pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes x4 pF. The value of x is____.

Options:

The correct answer is:
105

Solution:
Capacitance of the parallel plate capacitor when air is present between the plates, C=ε0Ad=15 pF

Capacitance of the parallel plate capacitor when dielectric is filled in between the plates,

C'=kε0Ad'=3.5ε0A2d=3.52×15 pF=1054 pF

Therefore, x=105.

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}$. The separation between its plates is ' $d$ '. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_1=3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $\mathrm{k}_2=6$ and thickness $\frac{2 \mathrm{~d}}{3}$. Capacitance of the capacitor is now

Options:
A
$1.8 \mathrm{pF}$

B
$45 \mathrm{pF}$

C
$40.5 \mathrm{pF}$

D
$20.25 \mathrm{pF}$

The correct answer is:
$40.5 \mathrm{pF}$

Solution:
$$
\begin{aligned}
& C^{\prime}=\frac{A \varepsilon_0}{\frac{d_1}{3}+\frac{d_2}{6}}=\frac{A \varepsilon_0}{\frac{d}{9}+\frac{2 d}{18}}=\frac{18 A \varepsilon_0}{4 d} \\
& C^{\prime}=40.5 \mathrm{PF}
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main, JEE Main 2008
A parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu \mathrm{F}$. The area of capacitor is divided into two equal halves and filled with two media (as shown in figure) having dielectric constant $K_1=2$ and $\mathrm{K}_2=4$. The capacitance of the system will be

Options:
A $10 \mu \mathrm{F}$
B $20 \mu \mathrm{F}$
C $30 \mu \mathrm{F}$
D $40 \mu \mathrm{F}$

The correct answer is:
$30 \mu \mathrm{F}$

Solution:
$\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=10 \mu \mathrm{F}$
After dividing the area into two equal halves, the resultant capacitance is calculated as,
$\begin{aligned}
\mathrm{C}_{\text {eq }} & =\mathrm{C}_1+\mathrm{C}_2 \\
& =\frac{\mathrm{K}_1 \varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}+\frac{\mathrm{K}_1 \varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}=\frac{\frac{2 \varepsilon_0 \mathrm{~A}}{2}}{\mathrm{~d}}+\frac{\frac{4 \varepsilon_0 \mathrm{~A}}{2}}{\mathrm{~d}} \\
\therefore \quad \mathrm{C}_{\text {eq }} & =\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}+\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{3 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=3 \times 10=30 \mu \mathrm{F}
\end{aligned}$

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor with area $200 \mathrm{~cm}^2$ and separation between the plates $1.5 \mathrm{~cm}$, is connected across a battery of emf V. If the force of attraction between the plates is $25 \times 10^{-6} \mathrm{~N}$, the value of $V$ is approximately:
$$
\left.\left(\varepsilon_0=8.85 \times 10^{-12} \frac{\mathrm{C}^2}{\text { N.m }}\right)^2\right)
$$

Options:
A
$150 \mathrm{~V}$

B
$100 \mathrm{~V}$

C
$250 \mathrm{~V}$

D
$300 \mathrm{~V}$

The correct answer is:
$250 \mathrm{~V}$

Solution:
Given area of Parallel plate capacitor, $A=$ $200 \mathrm{~cm}^2$
Separation between the plates, $d=1.5 \mathrm{~cm}$ Force of attraction between the plates, $F=$ $25 \times 10^{-6} \mathrm{~N}$ $F=Q E$
$F=\frac{Q^2}{2 A \epsilon_0} \quad$ (E due to parallel plate $\left.=\frac{\sigma}{2 \epsilon_0}=\frac{Q}{A 2 \epsilon_0}\right)$
But $Q=C V=\frac{\in_0 A(V)}{d}$
$$
\begin{gathered}
\left.\therefore F=\frac{\left(\epsilon_0 A V^2\right)}{\mathrm{d}^2 \times 2 \mathrm{~A} \epsilon_0}\right) \\
=\frac{\left(\epsilon_0 A\right)^2 \times V^2}{d^2 \times 2 \times\left(A \epsilon_0\right)}=\frac{\left(\epsilon_0 A\right) \times V^2}{d^2 \times 2} \\
\text { or, } 25 \times 10^{-6}=\frac{\left(8.85 \times 10^{-12}\right) \times\left(200 \times 10^{-4}\right) \times V^2}{2.25 \times 10^{-4} \times 2} \\
\Rightarrow V=\frac{25 \times 10^{-6} \times 2.25 \times 10^{-4} \times 2}{8.85 \times 10^{-12} \times 200 \times 10^{-4}} \approx 250 \mathrm{~V}
\end{gathered}
$$

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor with area $200 \mathrm{~cm}^2$ and separation between the plates $1.5 \mathrm{~cm}$, is connected across a battery of emf V. If the force of attraction between the plates is $25 \times 10^{-6} \mathrm{~N}$, the value of $V$ is approximately:
$$
\left.\left(\varepsilon_0=8.85 \times 10^{-12} \frac{\mathrm{C}^2}{\text { N.m }}\right)^2\right)
$$

Options:
A
$150 \mathrm{~V}$

B
$100 \mathrm{~V}$

C
$250 \mathrm{~V}$

D
$300 \mathrm{~V}$

The correct answer is:
$250 \mathrm{~V}$

Solution:
Given area of Parallel plate capacitor, $A=$ $200 \mathrm{~cm}^2$
Separation between the plates, $d=1.5 \mathrm{~cm}$ Force of attraction between the plates, $F=$ $25 \times 10^{-6} \mathrm{~N}$ $F=Q E$
$F=\frac{Q^2}{2 A \epsilon_0} \quad$ (E due to parallel plate $\left.=\frac{\sigma}{2 \epsilon_0}=\frac{Q}{A 2 \epsilon_0}\right)$
But $Q=C V=\frac{\in_0 A(V)}{d}$
$$
\begin{gathered}
\left.\therefore F=\frac{\left(\epsilon_0 A V^2\right)}{\mathrm{d}^2 \times 2 \mathrm{~A} \epsilon_0}\right) \\
=\frac{\left(\epsilon_0 A\right)^2 \times V^2}{d^2 \times 2 \times\left(A \epsilon_0\right)}=\frac{\left(\epsilon_0 A\right) \times V^2}{d^2 \times 2} \\
\text { or, } 25 \times 10^{-6}=\frac{\left(8.85 \times 10^{-12}\right) \times\left(200 \times 10^{-4}\right) \times V^2}{2.25 \times 10^{-4} \times 2} \\
\Rightarrow V=\frac{25 \times 10^{-6} \times 2.25 \times 10^{-4} \times 2}{8.85 \times 10^{-12} \times 200 \times 10^{-4}} \approx 250 \mathrm{~V}
\end{gathered}
$$

Topic: Physics, Capacitance, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as:

εx=ε0+kx, for 0<x≤d2

εx=ε0+kd-x, for d2≤x≤d

Options:
A ε0+kd22/kA
B kA2ln2ε0+kd2ε0
C 0
D kA2ln2ε02ε0-kd

The correct answer is:
kA2ln2ε0+kd2ε0

Solution:
Taking an element of width dx at a distance xx<d/2 from left plate

dc=(ε0+kx)Adx

Capacitance of half of the capacitor

1C=∫0d/21dc=1 A∫0d/2dxε0+kx

1C=1kAlnε0+kd/2ε0

Capacitance of second half will be same

Ceq=C2=kA2ln2ε0+kd2ε0

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (25 Jul Shift 1)
A parallel plate capacitor with plate area A and plate separation d=2 m has a capacitance of 4 μF. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K=3 (as shown in figure) will be

Options:
A 2 μF
B 32 μF
C 6 μF
D 8 μF

The correct answer is:
6 μF

Solution:
This parallel plate capacitor can be divided into two capacitors. One with dielectric C1 and other without dielectricC2. The two capacitors will be in series.

Initially C=ε0Ad=4 μF

Now, C1=kε0Ad2=2×3×ε0Ad=24 μF and C2=ε0Ad2=2×ε0Ad=8 μF

Finally C'=C1C2C1+C2=24×824+8=6 μF

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A parallel plate capacitor with plate area A and separation between the plates d is filled with different dielectrics as shown in the figure. If the equivalent capacitance between a & b is mAÎµ0nd then find the value of m+n, where m and n are least positive integers.

Options:

The correct answer is:
214

Solution:
C1=ε0A3 × d2 × K1=2C0
C2=ε0A3 × d2 × K2=4C0
C3=ε0A3 × d2 × K3=6C0
C4=ε0Ad2 × K4=4C0

Ceq=15955C0=159 Aε055d

m+n=159+55=214

Topic: Physics, Electrostatics, JEE Main
A parallel plate capacitor with plate separation 5 mm is charged up by a battery. It is found that on introducing a dielectric sheet of thickness 2 mm, while keeping the battery connections intact, the capacitor draws 25% more charge from the battery than before. The dielectric constant of the sheet is ____.

Options:

The correct answer is:
2

Solution:
Without dielectric, capacitance will be C=Aε0d.

Therefore, charge on it, Q=Aε0dV

Now, with dielectric capacitance will be C'=Aε0d-t+tK.

Hence, charge on it will be Q'=Aε0Vd-t+tK

Given:

Aε0Vd-t+tK=1.25Aε0Vd

⇒1.253+2K=5

⇒K=2

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (31 Jan Shift 1)
A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100  N/C, the magnitude of charge on each plate is: ( Take   ∈ 0 =8.85× 10 −12 C 2 N- m 2 )

Options:
A 8.85x10-10C
B 6.85x10-10C
C 9.85x10-10C
D 7.85x10-10C

The correct answer is:
8.85x10-10C

Solution:
E=ÏÎµ0=QAÎµ0=100 V/m
Q=100ÃA Îµ0
âQ=100Ã1Ã8.85Ã10-12C
=8.85Ã10-10C

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be:

Options:
A K=K1K2K1+K2+K3.K4K3+K4
B K=K1+K2K3+K42K1+K2+K3+K4
C K=(K1+K4)(K2+K3) 2K1+K2+K3+K4
D K=K1+K2(K3+K4)K1+K2+K3+K4

The correct answer is:
K=K1K2K1+K2+K3.K4K3+K4

Solution:
C1 and C2 are in series.
The equivalent capacitance of C1 and C2 is
C12=∈0A2d2K1+d2K2=∈0AdK1K2K1+K2

C3 and C4 are in series
C34=∈0A2d2K3+d2K4=∈0AdK3K4K3+K4
Ceq=C12+C34=∈0AdK1K2K1+K2+K3K4K3+K4
∴Keq=K1K2K1+K2+K3.K4K3+K4

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A parallel plate capacitor with width 4 cm, length 8 cm and separation between the plates of 4 mm is connected to a battery of 20 V. A dielectric slab of dielectric constant 5 having length 1 cm, width 4 cm and thickness 4 mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be _____ ϵ0 J. (Where ϵ0 is the permittivity of free space)

Options:

The correct answer is:
240

Solution:
The figure below shows the parallel plate capacitor with dielectric slab placed in between the plates.

Now, the effective capacitance of the combination is

Ceff=ε07×4410+5ε01×4410×10-2

=70ε0+50ε0×10-2=120ε0×10-2

⇒Ceff =1.2ε0

Electrostatics energy of this system is =12CeffV2

=121.2ε02020=240ε0

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (27 Jul Shift 2)