Q: Two parallel infinite line charges with linear charge densities \$+ \lambda \$ C/m and \$-\lambda\$ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges ?

Electric field due to line charge (1)\$\vec{E}_1 = \dfrac{\lambda}{2 \pi \varepsilon_0 R} \hat{i} N/C\$Electric field due to line charge (2)\$\vec{E}_2 = \dfrac{\lambda}{2 \pi \varepsilon_0 R} \hat{i} N/C\$\$\vec{E}_{net} = \vec{E}_1 + \vec{E}_2\$\$= \dfrac{\lambda}{2 \pi \varepsilon_0 R} \hat{i} + \dfrac{\lambda}{2 \pi \varepsilon_0 R} \hat{i}\$\$= \dfrac{\lambda}{\pi \varepsilon_0 R} \hat{i} N/C\$

Q: A parallel-plate capacitor with plate area \$A\$ and separation between the plates \$d\$, is charged by a constant current \$i\$. Consider a plane surface of area \$A/2\$ parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Given that, Area \$A=\dfrac{A}{2}\$ Separation between the plates \$=d\$ Constant current \$=i\$ Now, the electric field between plates of the capacitor \$E=\dfrac{q}{{{\varepsilon }_{0}}A}\$ Now, the flux through the area \$ \phi E=\dfrac{q}{{{\varepsilon }_{0}}A}\times \dfrac{A}{2} \$ \$ \phi E=\dfrac{q}{2{{\varepsilon }_{0}}} \$ Now, the displacement current is \$ {{I}_{d}}={{\varepsilon }_{0}}\times \dfrac{d\phi E}{dt} \$ \$ {{I}_{d}}={{\varepsilon }_{0}}\times \dfrac{d\left( \dfrac{q}{2{{\varepsilon }_{0}}} \right)}{dt} \$ \$ {{I}_{d}}=\dfrac{1}{2}\left( \dfrac{dq}{dt} \right) \$ \$ {{I}_{d}}=\dfrac{i}{2} \$ Hence, the displacement current is \$\dfrac{i}{2}\$

Q: Two identical metal plates are given positive charge \${Q}_{1}\$ and \${Q}_{2}\$ \$\left( <{ Q }_{ 1 } \right)\$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance \$C\$, the potential difference between them is

The potential difference between the two identical metal plates is given as \$C = \dfrac{{{\varepsilon _{\rm{0}}}A}}{d}\$ Let the surface charge density is given as \${\sigma _1} = {\sigma _{21}} = \dfrac{Q}{A}\$ The net electric field is \${E_{net}} = \dfrac{{{\sigma _1} - {\sigma _2}}}{{2{\varepsilon _{\rm{0}}}}}\$ We know the potential difference is given as \$V = E.d\$ By substituting the above values we get \$V = \dfrac{{{Q_1} - {Q_2}}}{{2C}}\$

Q: A capacitor of capacitance \$C\$ is charged to a potential difference \$V\$ from a cell and then disconnected from it. A charge \$+Q\$ is now given to its positive plate. The potential difference across the capacitor is now

Let the initial charge on the capacitor be \$Q_o\$.Thus we have \$Q_o=CV\$Now a charge \$Q\$ is given to the positive plate and the new potential is given as \$V'\$.Thus we get \$V'=Ed=(E_1+E_2)d\$Here \$E_1\$ and \$E_2\$ are the field in the gap due to charges \$Q+Q_o\$ and \$-Q_o\$ respectively.Thus we have\$E_1=\dfrac{Q+Q_o}{2\epsilon_oA}\$ and \$E_2=\dfrac{Q_o}{2\epsilon_oA}\$Thus we get\$V'=\left(\dfrac{Q+Q_o}{2\epsilon_oA}+\dfrac{Q_o}{2\epsilon_oA}\right)d=\dfrac{Qd}{2\epsilon_oA}+\dfrac{Q_od}{\epsilon_oA}=\dfrac{Q}{2C}+V\$

Q: Mention any two factors on which the capacitance of a parallel plate capacitor depends.

The capacitance of a parallel plate capacitor depends on area of each plate, dielectric medium between the plates and distance between the plates.

Q: A capacitance of \$2\ \mu F\$ is required in an electrical circuit across a potential difference of \$1.0\ kV\$. A large number of \$1\ \mu F\$ capacitors are available which can withstand a potential difference of not more than \$300\, V\$. The minimum number of capacitors required to achieve this is :

\$m\times 300 > 1000\$\$m > \dfrac{10}{3}\$or \$m = 4\$For \$m = 4\$, capacitance of 1 branch is \$\dfrac{1\mu F}{4}=0.25 \mu F\$For overall capacitance of \$2\mu F\$, \$8\$ such branches or equivalentis \$C_{eq}=8\times 0.25 = 2\mu F\$Total capacitors required= \$8\times 4 = 32\$

Q: Two spherical conductors A and B of radii \$1\$ mm and \$2\$ mm separated by a distance of \$5\$ cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is:

After connection, \$V_1=V_2\$\$\Rightarrow K\dfrac {Q_1}{r_1}=K\dfrac {Q_2}{r_2}\$\$\Rightarrow \dfrac {Q_1}{r_1}=\dfrac {Q_2}{r_2}\$The ratio of electric fields\$\dfrac {E_1}{E_2}=\dfrac {K\dfrac {Q_1}{r_1^2}}{K\dfrac {Q_2}{r^2_2}}\$\$\Rightarrow \dfrac {E_1}{E_2}=\dfrac {Q_1}{r_1^2}\times \dfrac {r_2^2}{Q_2}\$\$\Rightarrow \dfrac {E_1}{E_2}=\dfrac {r_1\times r_2^2}{r_1^2\times r_2}\Rightarrow \dfrac {E_1}{E_2}=\dfrac {r_2}{r_1}=\dfrac {2}{1}\$Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.

Question 1

Derive the expression for capacitance of parallel plate capacitor.

Accepted Answer

Let the two plates are kept parallel to each other separated be a distance $d$ and cross-sectional area of each plate is $A$.  Electric field by a single thin plate $E' = \dfrac{\sigma}{2\epsilon_o}$Total electric field between the plates $E = \dfrac{\sigma}{2\epsilon_o} +\dfrac{\sigma}{2\epsilon_o}$Or $E = \dfrac{\sigma}{\epsilon_o}$Or $E = \dfrac{Q}{A\epsilon_o}$Potential difference between the plates $V = Ed$$\implies$ $V = \dfrac{Qd}{A\epsilon_o}$Capacitance  $C = \dfrac{Q}{V}$Thus we get capacitance of parallel plate capacitor  $C  =\dfrac{A\epsilon_o}{d}$

Question 2

Two parallel infinite line charges with linear charge densities $+ \lambda $ C/m and $-\lambda$ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges ?

Accepted Answer

Electric field due to line charge (1)$\vec{E}_1 = \dfrac{\lambda}{2 \pi \varepsilon_0 R} \hat{i} N/C$Electric field due to line charge (2)$\vec{E}_2 = \dfrac{\lambda}{2 \pi \varepsilon_0 R} \hat{i} N/C$$\vec{E}_{net} = \vec{E}_1 + \vec{E}_2$$= \dfrac{\lambda}{2 \pi \varepsilon_0 R} \hat{i} + \dfrac{\lambda}{2 \pi \varepsilon_0 R} \hat{i}$$= \dfrac{\lambda}{\pi \varepsilon_0 R} \hat{i} N/C$

Question 3

A parallel plate capacitor is charged an then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

Accepted Answer

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 $\dfrac{A\epsilon_0}{d}$.When the plates are separated, the distance between the plates $d$ increases and hence the capacitance decreases.Hence from $Q=CV$, the voltage across the capacitor increases in order to keep the charge constant.

Question 4

A conducting sphere of radius $10 cm$ is charged with $\displaystyle 10\mu C$. Another uncharged sphere of radius 20 cm is allowed to touch it for sometime, after that if the spheres are separated, then surface density of charges on the spheres will be in the ratio of

Accepted Answer

After  touch if the common potential is V, according to charge conservation $Q=C_1V+C_2V$    here $Q=10 \mu C$ and $C_1, C_2$ are the capacitances of two sphere.or $V=\dfrac{Q}{C_1+C_2}$After touch the charge on first sphere is $Q_1=C_1V=\dfrac{C_1Q}{C_1+C_2}$ and charge on second sphere is  $Q_2=C_2V=\dfrac{C_2Q}{C_1+C_2}$Charge densities are $\sigma_1=\dfrac{Q_1}{S_1}=\dfrac{C_1V}{4\pi r_1^2}$ and $\sigma_2=\dfrac{Q_2}{S_1}=\dfrac{C_2V}{4\pi r_2^2}$ Thus, $\dfrac{\sigma_1}{\sigma_2}=\dfrac{C_1}{C_2}\times \dfrac{r_2^2}{r_1^2}=\dfrac{4\pi\epsilon_0r_1}{4\pi\epsilon_0r_2}\times \dfrac{r_2^2}{r_1^2}=\dfrac{r_2}{r_1}=20/10=2/1$      as the capacitance of isolated sphere $=4\pi \epsilon_0R$

Question 5

A parallel-plate capacitor with plate area $A$ and separation between the plates $d$, is charged by a constant current $i$. Consider a plane surface of area $A/2$ parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Accepted Answer

Given that,&#10;&#10;Area $A=\dfrac{A}{2}$ &#10;&#10;Separation between the plates $=d$ &#10;&#10;Constant current $=i$ &#10;&#10;Now, the electric field between plates of the capacitor&#10;&#10;$E=\dfrac{q}{{{\varepsilon }_{0}}A}$&#10;&#10;Now, the flux through the area&#10;&#10;  $ \phi&#10;E=\dfrac{q}{{{\varepsilon }_{0}}A}\times \dfrac{A}{2} $ &#10;&#10; $ \phi&#10;E=\dfrac{q}{2{{\varepsilon }_{0}}} $ &#10;&#10;Now, the displacement current is&#10;&#10;  $&#10;{{I}_{d}}={{\varepsilon }_{0}}\times \dfrac{d\phi E}{dt} $ &#10;&#10; $&#10;{{I}_{d}}={{\varepsilon }_{0}}\times \dfrac{d\left( \dfrac{q}{2{{\varepsilon&#10;}_{0}}} \right)}{dt} $ &#10;&#10; $&#10;{{I}_{d}}=\dfrac{1}{2}\left( \dfrac{dq}{dt} \right) $ &#10;&#10; $&#10;{{I}_{d}}=\dfrac{i}{2} $ &#10;&#10;Hence, the displacement current is $\dfrac{i}{2}$

Question 6

Two identical metal plates are given positive charge ${Q}_{1}$ and ${Q}_{2}$ $\left( <{ Q }_{ 1 } \right)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$, the potential difference between them is

Accepted Answer

The potential difference between the two&#10;identical metal plates is given as&#10;&#10;$C = \dfrac{{{\varepsilon _{\rm{0}}}A}}{d}$&#10;&#10;Let the surface charge density is given as &#10;&#10;${\sigma _1} = {\sigma _{21}} = \dfrac{Q}{A}$&#10;&#10;The net electric field is &#10;&#10;${E_{net}} = \dfrac{{{\sigma _1} - {\sigma&#10;_2}}}{{2{\varepsilon _{\rm{0}}}}}$&#10;&#10;We know the potential difference is given as &#10;&#10;$V = E.d$&#10;&#10;By substituting the above values we get&#10;&#10;$V = \dfrac{{{Q_1} - {Q_2}}}{{2C}}$&#10;&#10;

Question 7

A capacitor of capacitance $C$ is charged to a potential difference $V$ from a cell and then disconnected from it. A charge $+ Q$ is now given to its positive plate. The potential difference across the capacitor is now

Accepted Answer

Let the initial charge on the capacitor be $Q_o$.Thus we have $Q_o=CV$Now a charge $Q$ is given to the positive plate and the new potential is given as $V'$.Thus we get $V'=Ed=(E_1+E_2)d$Here $E_1$ and $E_2$ are the field in the gap due to charges $Q+Q_o$ and $-Q_o$ respectively.Thus we have$E_1=\dfrac{Q+Q_o}{2\epsilon_oA}$ and $E_2=\dfrac{Q_o}{2\epsilon_oA}$Thus we get$V'=\left(\dfrac{Q+Q_o}{2\epsilon_oA}+\dfrac{Q_o}{2\epsilon_oA}\right)d=\dfrac{Qd}{2\epsilon_oA}+\dfrac{Q_od}{\epsilon_oA}=\dfrac{Q}{2C}+V$

Question 8

Mention any two factors on which the capacitance of a parallel plate capacitor depends.

Accepted Answer

The capacitance of a parallel plate capacitor depends on area of each plate, dielectric medium between the plates and distance between the plates.

Question 9

A capacitance of $2\ \mu F$ is required in an electrical circuit across a potential difference of $1.0\ kV$. A large number of $1\ \mu F$ capacitors are available which can withstand a potential difference of not more than $300\, V$. The minimum number of capacitors required to achieve this is :

Accepted Answer

$m\times 300 > 1000$$m > \dfrac{10}{3}$or $m = 4$For $m = 4$, capacitance of 1 branch is $\dfrac{1\mu F}{4}=0.25 \mu F$For overall capacitance of $2\mu F$, $8$ such branches or equivalentis $C_{eq}=8\times 0.25 = 2\mu F$Total capacitors required= $8\times 4 = 32$

Question 10

Two spherical conductors A and B of radii $1$ mm and $2$ mm separated by a distance of $5$ cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is:

Accepted Answer

After connection, $V_1=V_2$$\Rightarrow K\dfrac {Q_1}{r_1}=K\dfrac {Q_2}{r_2}$$\Rightarrow \dfrac {Q_1}{r_1}=\dfrac {Q_2}{r_2}$The ratio of electric fields$\dfrac {E_1}{E_2}=\dfrac {K\dfrac {Q_1}{r_1^2}}{K\dfrac {Q_2}{r^2_2}}$$\Rightarrow \dfrac {E_1}{E_2}=\dfrac {Q_1}{r_1^2}\times \dfrac {r_2^2}{Q_2}$$\Rightarrow \dfrac {E_1}{E_2}=\dfrac {r_1\times r_2^2}{r_1^2\times r_2}\Rightarrow \dfrac {E_1}{E_2}=\dfrac {r_2}{r_1}=\dfrac {2}{1}$Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.

Revise with Concepts

Properties of Parallel Plate Capacitor

Capacitance of Parallel Plate Capacitor

Charge Distribution Between Plates of Parallel Plate Capacitors

Quick Summary With Stories

Capacitance of Parallel Plate Capacitor

Charge Distribution in Parrallel Plate Capacitor-I

Derive the expression for capacitance of parallel plate capacitor.

Two parallel infinite line charges with linear charge densities $+ λ$ C/m and $- λ$ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges ?

A parallel plate capacitor is charged an then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

A conducting sphere of radius $10 c m$ is charged with $10 μ C$ . Another uncharged sphere of radius 20 cm is allowed to touch it for sometime, after that if the spheres are separated, then surface density of charges on the spheres will be in the ratio of

A parallel-plate capacitor with plate area $A$ and separation between the plates $d$ , is charged by a constant current $i$ . Consider a plane surface of area $A / 2$ parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Two identical metal plates are given positive charge $Q_{1}$ and $Q_{2}$ $(< Q_{1})$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$ , the potential difference between them is

A capacitor of capacitance $C$ is charged to a potential difference $V$ from a cell and then disconnected from it. A charge $+ Q$ is now given to its positive plate. The potential difference across the capacitor is now

Mention any two factors on which the capacitance of a parallel plate capacitor depends.

A capacitance of $2 μ F$ is required in an electrical circuit across a potential difference of $1.0 k V$ . A large number of $1 μ F$ capacitors are available which can withstand a potential difference of not more than $300 V$ . The minimum number of capacitors required to achieve this is :

Two spherical conductors A and B of radii $1$ mm and $2$ mm separated by a distance of $5$ cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is:

Revise with Concepts

Properties of Parallel Plate Capacitor

Capacitance of Parallel Plate Capacitor

Charge Distribution Between Plates of Parallel Plate Capacitors

Quick Summary With Stories

Capacitance of Parallel Plate Capacitor

Charge Distribution in Parrallel Plate Capacitor-I

Derive the expression for capacitance of parallel plate capacitor.

Two parallel infinite line charges with linear charge densities +λ C/m and −λ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges ?

A parallel plate capacitor is charged an then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

A conducting sphere of radius 10cm is charged with 10μC. Another uncharged sphere of radius 20 cm is allowed to touch it for sometime, after that if the spheres are separated, then surface density of charges on the spheres will be in the ratio of

A parallel-plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Two identical metal plates are given positive charge Q1​ and Q2​ (<Q1​) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now

Mention any two factors on which the capacitance of a parallel plate capacitor depends.

A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300V. The minimum number of capacitors required to achieve this is :

Two spherical conductors A and B of radii 1 mm and 2 mm separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is:

Two parallel infinite line charges with linear charge densities $+ λ$ C/m and $- λ$ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges ?

A conducting sphere of radius $10 c m$ is charged with $10 μ C$ . Another uncharged sphere of radius 20 cm is allowed to touch it for sometime, after that if the spheres are separated, then surface density of charges on the spheres will be in the ratio of

A parallel-plate capacitor with plate area $A$ and separation between the plates $d$ , is charged by a constant current $i$ . Consider a plane surface of area $A / 2$ parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Two identical metal plates are given positive charge $Q_{1}$ and $Q_{2}$ $(< Q_{1})$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$ , the potential difference between them is

A capacitor of capacitance $C$ is charged to a potential difference $V$ from a cell and then disconnected from it. A charge $+ Q$ is now given to its positive plate. The potential difference across the capacitor is now

A capacitance of $2 μ F$ is required in an electrical circuit across a potential difference of $1.0 k V$ . A large number of $1 μ F$ capacitors are available which can withstand a potential difference of not more than $300 V$ . The minimum number of capacitors required to achieve this is :

Two spherical conductors A and B of radii $1$ mm and $2$ mm separated by a distance of $5$ cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is: