A parallel plate capacitor has plate area 40 cm$^2$ 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 :
Explanation:
Charge on $\mathrm{C}_2$ is $\mathrm{Q}_2=\mathrm{x} \times 10=10 \mathrm{x} \mu \mathrm{C}$ .....(ii)
Charge on $\mathrm{C}_3$ is $\mathrm{Q}_3=3 \times 10=30 \mu \mathrm{C}$ .....(iii)
Total charge $20+10 \mathrm{x}+30=100$ $\Rightarrow x=5$
In the circuit shown, the energy stored in the capacitor is $n ~\mu \mathrm{J}$. The value of $n$ is __________
Explanation:
$ \begin{aligned} & \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=3 \mathrm{I}_1=3 \mathrm{~V} \\\\ & \mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=2 \times 4=8 \mathrm{~V} \end{aligned} $
Subtracting eq. (1) from eq. (2)
$ \begin{aligned} & \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{D}}=5 \mathrm{~V} \Rightarrow \mathrm{V}=5 \mathrm{~V} \\\\ & \mathrm{U}=\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \times 6 \times 5^2=75 \mu \mathrm{J} \end{aligned} $
In the given circuit, $\mathrm{C}_{1}=2 \mu \mathrm{F}, \mathrm{C}_{2}=0.2 \mu \mathrm{F}, \mathrm{C}_{3}=2 \mu \mathrm{F}, \mathrm{C}_{4}=4 \mu \mathrm{F}, \mathrm{C}_{5}=2 \mu \mathrm{F}, \mathrm{C}_{6}=2 \mu \mathrm{F}$, The charge stored on capacitor $\mathrm{C}_{4}$ is ____________ $\mu \mathrm{C}$.
Explanation:
$ \begin{aligned} & \mathrm{C}_{\mathrm{eq}}=0.5 \mu \mathrm{F} \\\\ & \mathrm{Q}=0.5 \times 10=5 \mu \mathrm{C} \\\\ & \mathrm{Q}^{\prime}=\frac{5 \mu \mathrm{C} \times 0.8}{0.8+0.2}=4 \mu \mathrm{C} \end{aligned} $
A $600 ~\mathrm{pF}$ capacitor is charged by $200 \mathrm{~V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 ~\mathrm{pF}$ capacitor. Electrostatic energy lost in the process is ____________ $\mu \mathrm{J}$
Explanation:
The energy stored in a capacitor can be calculated using the formula:
$ U = \frac{1}{2} C V^2 $
where:
- (U) is the energy,
- (C) is the capacitance,
- (V) is the voltage.
Initially, the energy stored in the first capacitor is:
$ U_{\text{initial}} = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (200 \, \text{V})^2 = 0.012 \, \text{J} = 12 \, \mu\text{J}. $
When the charged capacitor is connected to the uncharged capacitor, the charge will distribute equally between them because they have the same capacitance. Therefore, the final voltage across each capacitor is half of the initial voltage, i.e., 100 V.
The energy in each capacitor after the redistribution is:
$ U_{\text{final each}} = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (100 \, \text{V})^2 = 0.003 \, \text{J} = 3 \, \mu\text{J}. $
As there are two capacitors, the total final energy is:
$ U_{\text{final total}} = 2 \times U_{\text{final each}} = 2 \times 3 \, \mu\text{J} = 6 \, \mu\text{J}. $
The energy loss is the difference between the initial energy and the final energy:
$ \Delta U = U_{\text{initial}} - U_{\text{final total}} = 12 \, \mu\text{J} - 6 \, \mu\text{J} = 6 \, \mu\text{J}. $
As shown in the figure, two parallel plate capacitors having equal plate area of $200 \mathrm{~cm}^{2}$ are joined in such a way that $a \neq b$. The equivalent capacitance of the combination is $x \in_{0} \mathrm{~F}$. The value of $x$ is ____________.
Explanation:
The situation is equivalent to a conducting slab placed between the plates
A parallel plate capacitor with plate area $\mathrm{A}$ and plate separation $\mathrm{d}$ is filled with a dielectric material of dielectric constant $K=4$. The thickness of the dielectric material is $x$, where $x < d$.
Let $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ be the capacitance of the system for $\chi=\frac{1}{3} d$ and $\mathcal{X}=\frac{2 d}{3}$, respectively. If $\mathrm{C}_{1}=2 \mu \mathrm{F}$ the value of $\mathrm{C}_{2}$ is __________ $\mu \mathrm{F}$
Explanation:
$\Rightarrow C_1=\frac{\varepsilon_0 A}{\left(\frac{\frac{d}{3}}{K}+\frac{2 d}{3}\right)}=\frac{\varepsilon_0 A}{\left(\frac{d}{3 \times 4}+\frac{2 d}{3}\right)}$
$ \Rightarrow 2 \mu F=\frac{\varepsilon_0 A \times 12}{9 d}=\frac{4}{3} \frac{\varepsilon_0 A}{d} $
$\Rightarrow 2 \mu F=\frac{4}{3} \frac{\varepsilon_0 A}{d} \quad\left(\right.$ for $\left.x=\frac{1}{3} d\right)$
$\Rightarrow \frac{\varepsilon_0 A}{d}=\frac{3}{2} \mu \mathrm{F}$
Now,
$ \begin{aligned} & C_2=\frac{\varepsilon_0 A}{\left(\frac{\frac{2 d}{3}}{K}+\frac{d}{3}\right)} \quad\left(\text { for } x=\frac{2 d}{3}\right) \\\\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\\\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\\\ & \Rightarrow C_2=\frac{12 \varepsilon_0 A}{2 d+4 d}=\frac{2 \varepsilon_0 A}{d}=\frac{3}{2} \times 2 \\\\ & =3 \mu \mathrm{F} \end{aligned} $
Explanation:
$\Rightarrow i_{A B}=\frac{6}{3}=2 \mathrm{~A} $
$ i_{A D}=\frac{6}{12}=0.5 \mathrm{~A}$
$\Rightarrow V_{B}+2 \times 2-10 \times 0.5=V_{D}$
$\Rightarrow V_{B}-V_{D}=1$ volt
(Assuming Dielectric constant $=10$ )
Explanation:
And charge on $\mathrm{C}_{2}=\mathrm{CV}$
When they are connected in parallel charge will be equally divided so charge on one capacitor is
$q=\frac{K+1}{2} \mathrm{CV}$
So, $V=\frac{q}{K C}=\frac{K+1}{2 K}=55 \mathrm{~V}$
A capacitor of capacitance $900 \mu \mathrm{F}$ is charged by a $100 \mathrm{~V}$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $x \times 10^{-} { }^{2} \mathrm{~J}$. The value of $x$ is _____________.
Explanation:
${U_i} = {1 \over 2}C{V^2} = {1 \over 2} \times 900 \times {10^{ - 6}} \times {100^2} = 4.5$ J
As the other capacitor is identical therefore charge is equally divided and potential difference across the capacitors becomes half. So
${U_f} = {1 \over 2}2C{\left( {{V \over 2}} \right)^2} = {1 \over 2} \times 2 \times 900 \times {10^{ - 6}}{\left( {{{100} \over 2}} \right)^2}$
$ = {9 \over 4}$ J = 2.25 J
So, loss in energy $\Delta {U_{loss}} = {U_i} - {U_f}$
= 2.25 J
= 225 $\times$ 10$^{-2}$ J
A capacitor has capacitance 5$\mu$F when it's parallel plates are separated by air medium of thickness d. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness $\frac{d}{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be __________ $\mu$F.
Explanation:
When completely air filled
$C=5 \mu \mathrm{F}=\frac{\varepsilon_{0} A}{d} \quad...(1)$
When half filled with $K=1.5$
$ \begin{gathered} \frac{1}{C_{\mathrm{eq}}}=\frac{\frac{d}{2}}{\varepsilon_{0} A}+\frac{\frac{d}{2}}{\varepsilon_{0} A K} \\\\ C_{\text {eq }}=\left(\frac{2 K}{K+1}\right) \frac{\varepsilon_{0} A}{d} \quad...(2) \end{gathered} $
From (1) & (2)
$C_{\mathrm{eq}}=\left(\frac{2 \times 1.5}{1.5+1}\right) 5 \mu \mathrm{F}=6 \mu \mathrm{F}$
A parallel plate capacitor with air between the plate has a capacitance of 15pF. 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 $\frac{x}{4}$ pF. The value of $x$ is ____________.
Explanation:
Initially
$ \frac{\varepsilon_{0} A}{d}=15 \times 10^{-12} \mathrm{~F} $
Finally
$ \begin{aligned} & \frac{3.5 \varepsilon_{0} A}{2 d}=\frac{x}{4} \times 10^{-12} \mathrm{~F} \\\\ & \therefore \frac{3.5}{2} \times 15=\frac{x}{4} \\\\ & \Rightarrow x=\frac{3.5 \times 15 \times 4}{2}=105 \end{aligned} $Two identical thin metal plates has charge $q_{1}$ and $q_{2}$ respectively such that $q_{1}>q_{2}$. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :
A slab of dielectric constant $\mathrm{K}$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} \mathrm{~d}$, where $\mathrm{d}$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :
(Given $\mathrm{C}_{0}$ = capacitance of capacitor with air as medium between plates.)
Two capacitors, each having capacitance $40 \,\mu \mathrm{F}$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $\mathrm{K}$ such that the equivalence capacitance of the system became $24 \,\mu \mathrm{F}$. The value of $\mathrm{K}$ will be :
A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key '$K$' is closed, the total energy stored across the combination is $E_{1}$. Now key '$K$' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E_{2}$. The ratio $E_{1} / E_{2}$ will be :
The total charge on the system of capacitors $C_{1}=1 \mu \mathrm{F}, C_{2}=2 \mu \mathrm{F}, \mathrm{C}_{3}=4 \mu \mathrm{F}$ and $\mathrm{C}_{4}=3 \mu \mathrm{F}$ connected in parallel is :
(Assume a battery of $20 \mathrm{~V}$ is connected to the combination)
Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii $\left( {{{{R_2}} \over {{R_1}}}} \right)$ is :
A condenser of $2 \,\mu \mathrm{F}$ capacitance is charged steadily from 0 to $5 \,\mathrm{C}$. Which of the following graph represents correctly the variation of potential difference $(\mathrm{V})$ across it's plates with respect to the charge $(Q)$ on the condenser?
Co is the capacitance of a parallel plate capacitor with air as a medium between the plates (as shown in Fig. 1). If half space between the plates is filled with a dielectric of relative permittivity $\varepsilon $r (as shown in Fig. 2), the new capacitance of the capacitor will be :
A capacitor is discharging through a resistor R. Consider in time t1, the energy stored in the capacitor reduces to half of its initial value and in time t2, the charge stored reduces to one eighth of its initial value. The ratio t1/t2 will be
A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :
A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.
The charge on capacitor of capacitance 15$\mu$F in the figure given below is :
A parallel plate capacitor with plate area A and plate separation d = 2 m has a capacitance of 4 $\mu$F. 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 :
Two capacitors having capacitance C1 and C2 respectively are connected as shown in figure. Initially, capacitor C1 is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C1 is now connected to uncharged capacitor C2 by closing the switch S. The amount of charge on the capacitor C2, after equilibrium, is :
Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness ${d \over 2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?
If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)
A parallel plate capacitor is formed by two plates each of area 30$\pi$ cm2 separated by 1 mm. A material of dielectric strength 3.6 $\times$ 107 Vm$-$1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 $\times$ 10$-$6C, the value of dielectric constant of the material is :
[Use ${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}$ Nm2 C$-$2]
As show in the figure, in steady state, the charge stored in the capacitor is ____________ $\times\, 10^{-6}$ C.
Explanation:
At steady state potential difference across capacitor
${V_c} = {{10 \times 100} \over {110}}\,V$
$Q = C{V_c}$
$ = {{1.1 \times {{10}^{ - 6}} \times 10 \times 100} \over {110}}\,C = 10\,\mu C$
A parallel plate capacitor with width $4 \mathrm{~cm}$, length $8 \mathrm{~cm}$ and separation between the plates of $4 \mathrm{~mm}$ is connected to a battery of $20 \mathrm{~V}$. A dielectric slab of dielectric constant 5 having length $1 \mathrm{~cm}$, width $4 \mathrm{~cm}$ and thickness $4 \mathrm{~mm}$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ____________ $\epsilon_{0}$ J. (Where $\epsilon_{0}$ is the permittivity of free space)
Explanation:
${d_1} = 4 \times {10^{ - 3}}$
${A_1} = 8 \times 4 \times {10^{ - 4}}\,{m^2}$
$V = 20\,V$
${d_2} = 4 \times {10^{ - 3}},$
${A_2} = 4 \times 1 \times {10^{ - 4}}\,{m^2}$
${C_{eq}} = {{({A_1} + 5{A_2} - {A_2}){\varepsilon _0}} \over d} = {{3(16) \times {{10}^{ - 4}}} \over {4 \times {{10}^{ - 3}}}}{\varepsilon _0}$
$\varepsilon = {1 \over 2}{C_{eq}}{V^2} = {3 \over 2}\left( {{4 \over {10}}} \right)(400){\varepsilon _0} = 240{\varepsilon _0}$
A composite parallel plate capacitor is made up of two different dielectric materials with different thickness $\left(t_{1}\right.$ and $\left.t_{2}\right)$ as shown in figure. The two different dielectric materials are separated by a conducting foil $\mathrm{F}$. The voltage of the conducting foil is V.
Explanation:
${{{C_1}} \over {{C_2}}} = {{3 \times {t_2}} \over {{t_1} \times 4}} = {3 \over 2}$
${q \over {{C_1}}} = {v_1},\,{q \over {{C_2}}} = {v_2}$
${{{v_1}} \over {{v_2}}} = {{{C_2}} \over {{C_1}}} = {2 \over 3}$
Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 V. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be ___________ V.
Explanation:
${V_{common}} = {{18CV + 54CV} \over {3C + 9C}} = 6\,V$
A capacitor C1 of capacitance 5 $\mu$F is charged to a potential of 30 V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C2 of capacitance 10 $\mu$F as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C2 is __________ $\mu$C.
Explanation:
Let the charge q is flown in the circuit.
So using Kirchhoff's law,
${q \over {10}} = {{150 - q} \over 5}$
$q = 100\,\mu C$
A parallel plate capacitor is made up of stair like structure with a plate area A of each stair and that is connected with a wire of length b, as shown in the figure. The capacitance of the arrangement is ${x \over {15}}{{{ \in _0}A} \over b}$. The value of x is ____________.
Explanation:
The circuit is equivalent to 3 capacitors in parallel as shown
${C_{eq}} = {{{\varepsilon _0}A} \over b}\left( {1 + {1 \over 3} + {1 \over 5}} \right) = {{23} \over {15}}{{{\varepsilon _0}A} \over b}$
$ \Rightarrow x = 23$
A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is ___________ nJ.
Explanation:
Electrical energy lost $ = {1 \over 2}\left( {{1 \over 2}C{V^2}} \right)$
$ = {1 \over 2} \times {1 \over 2} \times 50 \times {10^{ - 12}} \times {(100)^2}$
$ = {{500} \over 4}$ nJ
$ = 125$ nJ
The equivalent capacitance between points A and B in below shown figure will be __________ $\mu$F.
Explanation:
${C_{eq}} = {{(3 \times 8) \times 8} \over {(3 \times 8) + 8}}$
$ = {{24 \times 8} \over {32}}$
$ = 6\,\mu F$
| List - I | List - II | ||
|---|---|---|---|
| (a) | Capacitance, C | (i) | ${M^1}{L^1}{T^{ - 3}}{A^{ - 1}}$ |
| (b) | Permittivity of free space, ${\varepsilon _0}$ | (ii) | ${M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$ |
| (c) | Permeability of free space, ${\mu _0}$ | (iii) | ${M^{ - 1}}{L^{ - 2}}{T^4}{A^2}$ |
| (d) | Electric field, E | (iv) | ${M^1}{L^1}{T^{ - 2}}{A^{ - 2}}$ |
Choose the correct answer from the options given below
A capacitor of capacitance C = 1 $\mu$F is suddenly connected to a battery of 100 volt through a resistance R = 100 $\Omega$. The time taken for the capacitor to be charged to get 50 V is :
[Take ln 2 = 0.69]
(Given area of plate = A)
$\varepsilon (x) = {\varepsilon _0} + kx$, for $\left( {0 < x \le {d \over 2}} \right)$
$\varepsilon (x) = {\varepsilon _0} + k(d - x)$, for $\left( {{d \over 2} \le x \le d} \right)$
