Q: In the given circuit, the charge on \$4 \mu F\$ capacitor will be :

So total charge flow \$= q = 5.4 \,\mu F \times 10V\$ \$= 54 \,\mu C\$The charge will be distributed in the ratio of capacitance\$\Rightarrow \dfrac{q_1}{q_2} = \dfrac{2.4}{3} = \dfrac{4}{5}\$\$\therefore 9X = 54\mu C \Rightarrow X = 6\mu C\$\$\therefore\$ Charge on \$4 \mu F\$ capacitor will be \$= 4X = 4 \times 6\mu C\$ \$= 24 \mu C\$

Q: A capacitor of capacitance \$C_1 = 1\mu F\$ withstand the maximum voltage \$V_1 = 6.0 kV\$ while the capacitor of capacitance \$C_2 = 2\mu F\$, the maximum voltage \$V_2 = 4.0 kV\$. What voltage in Kilo Volt will the system of these two capacitors withstand if they are connected in series?

Maximum charge capacitance \$C_1\$ can withstand is \$Q_1=C_1V_1\$ while capacitance \$C_2\$ can withstand maximum charge of \$Q_2=C_2V_2\$When \$C_1\ and\ C_2\$ are connected in series the charge on each capacitor is equal and is equal to the maximum charge the combination can withstand, \$Q_{max} = C_1V_1\ \because C_1V_1< C_2V_2\$ ... for given valuesThe resultant capacitance of the combination is \$C_R=\dfrac{C_2C_1}{C_1+C_2}\$\$Q_{max} =6 \times10^{-3}\ C\$ and \$C_R=\dfrac{2}{3} \times10^{-6} \ F\$Maximum voltage that the combination of capacitance can withstand is \$V_{max}=\dfrac{Q_{max}}{C_R}\$\$V_{max}=9\ kV\$

Q: A parallel plate capacitor has \$1 \mu F\$ capacitance. One of its two plates is given \$+2 \mu C\$ charge and the other plate, \$+ 4 \mu C\$ charge. The potential difference developed across the capacitor is :-

\$\because\$ Electric fd. due to cap plate is \$\dfrac{\sigma}{2\varepsilon _0}\$Since both the plate has given positive charges.\$\boxed{E=\dfrac{\sigma}{2\varepsilon_0}}\$Net Electric .field. \$\dfrac{\sigma_2}{2\varepsilon_0}-\dfrac{\sigma_1}{2\varepsilon_0}\$ \$\boxed{\sigma=\dfrac{Q}{A}}\$\$E_{net}=\dfrac{Q_2}{2AE_0}-\dfrac{Q_1}{2AE_0}\$\$E_{net}=\dfrac{1}{1AE_0}(Q_2-Q_1)\$ here \$Q_2=4\mu c\$ Here \$Q_2=2 \mu c\$ \$\boxed{E_{net}=\dfrac{1}{2AE_0}(4-2)=\dfrac{1}{AE_0}}\$ Also \$V=ED\$ \$=E_{net}\times d=\dfrac{1}{AE_0}\times d=\dfrac{1}{c}\$ \$\dfrac{1}{1}\$ \$\because C=\dfrac{E_0A}{d}\$\$\because C=1\mu F\$\$\boxed {\therefore V=1V}\$ \$V = \dfrac{Q}{C} = \dfrac{1\mu C}{1\mu F} = 1\$ Volt

Q: Define capacitance of a capacitor and its SI unit.

The capacitance of any capacitor is its ability to store charge \$C=\dfrac{Q}{V}\$\$C=\$ capacitance \$Q=\$ Charge \$V=\$ Potential across capacitor.The SI unit of capacitance is farad(F).

Q: The earth has Volume 'V' and surface area 'A' then its capacitance would be

If we assume shape of earth be solid spherical.Then, the volume of earth\$=4/3\pi R^3\$where R is radius of earth.Given, volume of earth\$=V=4/3\pi R^3\$surface area of earth\$=A=4\pi R^2\$So, \$V/A=(4/3R^3)/4\pi R^2\$or \$V/A=R/3\$or, \$R=3V/A\$Now, capacitance of earth\$=4\pi \in_0R\$\$=4\pi \in_0\dfrac{3V}{A}\$\$=\dfrac{12\pi\in_0V}{A}\$.

Q: A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of \$2.5 \mu C\$. The space between the concentric spheres is filled with a liquid of dielectric constant 32.(a) Determine the capacitance of the capacitor.(b) What is the potential of the inner sphere?(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

Radius of the inner sphere, \$r_2\$= 12 cm = 0.12 mRadius of the outer sphere, \$r_1\$= 13 cm = 0.13 mCharge on the inner sphere, \$q=2.5 \times 10^{-6}\ C\$Dielectric constant of a liquid, \$\varepsilon_r=32\$(a)Capacitance, \$C= \dfrac{4 \pi \varepsilon_o \varepsilon_r r_1r_2}{r_1-r_2}\$where,\$\varepsilon_o\$= Permittivity of free space \$=8.85 \times 10^{-12}\ C^2N^{-1}m^2\$\$\therefore C \approx 5.5 \times 10^{-9}\ F\$(b) Potential of the inner sphere is given by, \$V=q/C\$ \$=4.5 \times 10^2 V\$(c) Radius of an isolated sphere, r= 12 cm Capacitance of the sphere is given by the relation, \$C'= 4 \pi \varepsilon_o r\$\$=1.33 \times 10^{-11}\ F\$ The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.

Q: In the circuit \$ C_1= 20 \mu F , C_2 = 40 \mu F\$ and \$C_3= 50 \mu F\$. If no capacitor can sustain more than \$50V\$, then maximumpotential difference between \$X\$ and \$Y\$ is

Let the charge on each capacitor be \$q\$thus the potential against each capacitor will be\$(q\$ is in micro coulomb\$)\$across \$C_1\$ capacitor \$= q/20V\$across \$C_2\$ capacitor \$= q/40V\$across \$C_3\$ capacitor \$= q/50V\$as we can see for same charge on each capacitor as they connected in series, the capacitor \$C_1\$has maximum potential drop, which according to the situation can be \$50V\$so \$q/20=50\$so \$q= 1000\$ micro coulombso the potential drop against \$C_2\$ capacitor \$= 1000/40 = 25\$ voltswhile the potential drop against \$C_3\$ capacitor will be \$1000/50 = 20\$ voltshence as all the capacitors are in series so the potential difference across \$X\$ and \$Y\$ \$=50 + 25 + 20 = 95\$ volts

Question 1

In the given circuit, the charge on $4 \mu F$ capacitor will be :

Accepted Answer

So total charge flow $= q = 5.4 \,\mu F \times 10V$                                $= 54 \,\mu C$The charge will be distributed in the ratio of capacitance$\Rightarrow \dfrac{q_1}{q_2} = \dfrac{2.4}{3} = \dfrac{4}{5}$$\therefore 9X = 54\mu C \Rightarrow X = 6\mu C$$\therefore$ Charge on $4 \mu F$ capacitor will be $= 4X = 4 \times 6\mu C$           $= 24 \mu C$

Question 2

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig.). Show that the capacitance of a spherical capacitor is given by
$c = \frac{4 π ϵ _{0} r _{1} r _{2}}{r _{1} - r _{2}}$
where $r_{1}$ and $r_{2}$ are the radii of outer and inner spheres, respectively.

Accepted Answer

Radius of the outer shell = r1Radius of the inner shell = r2The inner surface of the outer shell has charge +Q.The outer surface of the inner shell has induced charge &#8722;Q.Potential difference between the two shells is given by,$V=\cfrac{Q}{4 \pi \varepsilon_o r_2}-\cfrac{Q}{4 \pi \varepsilon_o r_1}$Where,$\varepsilon_o$= Permittivity of free space$V=\cfrac{Q}{4 \pi \varepsilon_o r_2}-\cfrac{Q}{4 \pi \varepsilon_o r_1}$$V=\cfrac{Q}{4 \pi \varepsilon_o}\left[\cfrac{1}{r_2}-\cfrac{1}{r_1}\right]$$V=\cfrac{Q(r_1-r_2)}{4 \pi \varepsilon_o r_1r_2}$Capacitance of the given system is given by,$C=\dfrac{Q}{V}$$C=\dfrac{4 \pi \varepsilon_o r_1 r_2}{r_1-r_2}$Hence, proved.

Question 3

A capacitor of capacitance $C_1 = 1\mu F$ withstand the maximum voltage $V_1 = 6.0 kV$ while the capacitor of capacitance $C_2 = 2\mu F$, the maximum voltage $V_2 = 4.0 kV$. What voltage in Kilo Volt will the system of these two capacitors withstand if they are connected in series?

Accepted Answer

Maximum charge capacitance $C_1$ can withstand is $Q_1=C_1V_1$ while capacitance $C_2$ can withstand maximum charge of $Q_2=C_2V_2$When $C_1\ and\ C_2$ are connected in series the charge on each capacitor is equal and is equal to the maximum charge the combination can withstand, $Q_{max} = C_1V_1\ \because C_1V_1< C_2V_2$ ... for given valuesThe resultant capacitance of the combination is $C_R=\dfrac{C_2C_1}{C_1+C_2}$$Q_{max} =6 \times10^{-3}\ C$ and $C_R=\dfrac{2}{3} \times10^{-6} \ F$Maximum voltage that the combination of capacitance can withstand is $V_{max}=\dfrac{Q_{max}}{C_R}$$V_{max}=9\ kV$

Question 4

A parallel plate capacitor has $1 \mu F$ capacitance. One of its two plates is given $+2 \mu C$ charge and the other plate, $+ 4 \mu C$ charge. The potential difference developed across the capacitor is :-

Accepted Answer

$\because$ Electric fd. due to cap plate is $\dfrac{\sigma}{2\varepsilon _0}$Since both the plate has given positive charges.$\boxed{E=\dfrac{\sigma}{2\varepsilon_0}}$Net Electric .field. $\dfrac{\sigma_2}{2\varepsilon_0}-\dfrac{\sigma_1}{2\varepsilon_0}$   $\boxed{\sigma=\dfrac{Q}{A}}$$E_{net}=\dfrac{Q_2}{2AE_0}-\dfrac{Q_1}{2AE_0}$$E_{net}=\dfrac{1}{1AE_0}(Q_2-Q_1)$ here $Q_2=4\mu c$ Here                                                     $Q_2=2 \mu c$       $\boxed{E_{net}=\dfrac{1}{2AE_0}(4-2)=\dfrac{1}{AE_0}}$   Also $V=ED$         $=E_{net}\times d=\dfrac{1}{AE_0}\times d=\dfrac{1}{c}$               $\dfrac{1}{1}$         $\because C=\dfrac{E_0A}{d}$$\because C=1\mu F$$\boxed {\therefore V=1V}$                                             $V = \dfrac{Q}{C} = \dfrac{1\mu C}{1\mu F} = 1$ Volt

Question 5

Define capacitance of a capacitor and its SI unit.

Accepted Answer

The capacitance of any capacitor is its ability to store charge $C=\dfrac{Q}{V}$$C=$ capacitance $Q=$ Charge $V=$ Potential across capacitor.The SI unit of capacitance is farad(F).

Question 6

The earth has Volume 'V' and surface area 'A' then its capacitance would be

Accepted Answer

If we assume shape of earth be solid spherical.Then, the volume of earth$=4/3\pi R^3$where R is radius of earth.Given, volume of earth$=V=4/3\pi R^3$surface area of earth$=A=4\pi R^2$So, $V/A=(4/3R^3)/4\pi R^2$or $V/A=R/3$or, $R=3V/A$Now, capacitance of earth$=4\pi \in_0R$$=4\pi \in_0\dfrac{3V}{A}$$=\dfrac{12\pi\in_0V}{A}$.

Question 7

A parallel plate capacitor with air between the plates has a capacitance of $8pF$. the separation between the plates is now reduced by its half and the space between them is filled with a medium of dielectric constant $5$. Calculate the value of capacitance of the capacitor in the second case.

Accepted Answer

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Question 8

A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of $2.5 μ C$ . The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

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Accepted Answer

Radius of the inner sphere, $r_2$= 12 cm = 0.12 mRadius of the outer sphere, $r_1$= 13 cm = 0.13 mCharge on the inner sphere, $q=2.5 \times 10^{-6}\ C$Dielectric constant of a liquid, $\varepsilon_r=32$(a)Capacitance, $C= \dfrac{4 \pi \varepsilon_o \varepsilon_r r_1r_2}{r_1-r_2}$where,$\varepsilon_o$= Permittivity of free space $=8.85 \times 10^{-12}\ C^2N^{-1}m^2$$\therefore C \approx 5.5 \times 10^{-9}\ F$(b)&#10;Potential&#10;of the inner sphere is given by,&#10;$V=q/C$&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;&#10;&#10;&#10;$=4.5 \times 10^2 V$(c)&#10;Radius&#10;of an isolated sphere, r=&#10;12 cm&#10;Capacitance&#10;of the sphere is given by the relation,&#10;$C'= 4 \pi \varepsilon_o r$$=1.33 \times 10^{-11}\ F$&#10;The&#10;capacitance of the isolated sphere is less in comparison to the&#10;concentric spheres. This is because the outer sphere of the&#10;concentric spheres is earthed. Hence, the potential difference is&#10;less and the capacitance is more than the isolated sphere.

Question 9

A metal plate is introduces between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor

Accepted Answer

By introducing the metal plate between the plates of charged capacitor, the capacitance of capacitor increasesReason: It is thickness of metal plate, then$ C= \dfrac{\varepsilon oA}{d-t(1-\dfrac{1}{k})} $For metal plate K= $\infty $,  $ C=\dfrac{\varepsilon oA}{(d-t)} $

Question 10

In the circuit $ C_1= 20 \mu F ,  C_2 = 40 \mu F$ and $C_3= 50 \mu F$. If no capacitor can sustain more than $50V$, then maximumpotential difference between $X$ and $Y$ is

Accepted Answer

Let the charge on each capacitor be $q$thus the potential against each capacitor will be$(q$ is in micro coulomb$)$across $C_1$ capacitor $= q/20V$across $C_2$ capacitor $= q/40V$across $C_3$ capacitor $= q/50V$as we can see for same charge on each capacitor as they connected in series, the capacitor $C_1$has maximum  potential drop, which according to the situation can be $50V$so $q/20=50$so $q= 1000$ micro coulombso the potential drop against $C_2$ capacitor $= 1000/40 = 25$ voltswhile the potential drop against $C_3$ capacitor will be $1000/50 = 20$ voltshence as all the capacitors are in series so the potential difference across $X$ and $Y$ $=50 + 25 + 20 = 95$ volts

Revise with Concepts

Capacitor and Capacitance

Quick Summary With Stories

Capacitance and Capacitor

In the given circuit, the charge on $4 μ F$ capacitor will be :

A capacitor of capacitance $C_{1} = 1 μ F$ withstand the maximum voltage $V_{1} = 6.0 k V$ while the capacitor of capacitance $C_{2} = 2 μ F$ , the maximum voltage $V_{2} = 4.0 k V$ . What voltage in Kilo Volt will the system of these two capacitors withstand if they are connected in series?

A parallel plate capacitor has $1 μ F$ capacitance. One of its two plates is given $+ 2 μ C$ charge and the other plate, $+ 4 μ C$ charge. The potential difference developed across the capacitor is :-

Define capacitance of a capacitor and its SI unit.

The earth has Volume 'V' and surface area 'A' then its capacitance would be

A parallel plate capacitor with air between the plates has a capacitance of $8 p F$ . the separation between the plates is now reduced by its half and the space between them is filled with a medium of dielectric constant $5$ . Calculate the value of capacitance of the capacitor in the second case.

A metal plate is introduces between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor

In the circuit $C_{1} = 20 μ F, C_{2} = 40 μ F$ and $C_{3} = 50 μ F$ . If no capacitor can sustain more than $50 V$ , then maximum
potential difference between $X$ and $Y$ is

Revise with Concepts

Capacitor and Capacitance

Quick Summary With Stories

Capacitance and Capacitor

In the given circuit, the charge on 4μF capacitor will be :

A capacitor of capacitance C1​=1μF withstand the maximum voltage V1​=6.0kV while the capacitor of capacitance C2​=2μF, the maximum voltage V2​=4.0kV. What voltage in Kilo Volt will the system of these two capacitors withstand if they are connected in series?

A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μC charge and the other plate, +4μC charge. The potential difference developed across the capacitor is :-

Define capacitance of a capacitor and its SI unit.

The earth has Volume 'V' and surface area 'A' then its capacitance would be

A parallel plate capacitor with air between the plates has a capacitance of 8pF. the separation between the plates is now reduced by its half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in the second case.

A metal plate is introduces between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor

In the circuit C1​=20μF,C2​=40μF and C3​=50μF. If no capacitor can sustain more than 50V, then maximum potential difference between X and Y is

In the given circuit, the charge on $4 μ F$ capacitor will be :

A capacitor of capacitance $C_{1} = 1 μ F$ withstand the maximum voltage $V_{1} = 6.0 k V$ while the capacitor of capacitance $C_{2} = 2 μ F$ , the maximum voltage $V_{2} = 4.0 k V$ . What voltage in Kilo Volt will the system of these two capacitors withstand if they are connected in series?

A parallel plate capacitor has $1 μ F$ capacitance. One of its two plates is given $+ 2 μ C$ charge and the other plate, $+ 4 μ C$ charge. The potential difference developed across the capacitor is :-

A parallel plate capacitor with air between the plates has a capacitance of $8 p F$ . the separation between the plates is now reduced by its half and the space between them is filled with a medium of dielectric constant $5$ . Calculate the value of capacitance of the capacitor in the second case.

In the circuit $C_{1} = 20 μ F, C_{2} = 40 μ F$ and $C_{3} = 50 μ F$ . If no capacitor can sustain more than $50 V$ , then maximum
potential difference between $X$ and $Y$ is