Q: When a dielectric material is introduced between the plates of a charged condenser, after disconnecting the battery, the electric field between the plates:

We know electric field between capacitor plates: \$E= \dfrac{\sigma}{\epsilon _0}\$\$= \dfrac{V}{d}\$, where \$d\$ is the distance between the plates.When battery is disconnected, the change present on capacitor is fixed, i.e., \$V\$ is constant.Now, when the dielectric is introduced, the electric field becomes: \$E' = \dfrac{V}{Kd}\$, where K is dielectric constant which is always > 1\$\therefore \$ electric field decreases.

Q: A condenser is charged and then battery is removed. A dielectric plate is put between the plates of condenser, then correct statement is

\$\textbf{Step 1 - Given data Refer Figure}\$let Cross-section of plate \$= A\$ distance between plate \$= d\$ surface charge density \$= \sigma\$Capacitance of capacitor \$= c\$\$\textbf{Step 2 - Explanation of condenser without slab}\$Since, we know that - when battery is removed from a charge capacitorthen charge is constant.So, electric field inside capacitor \$= E = \dfrac {\sigma}{\varepsilon_{0}}\$ Potential between plate \$= V = E \cdot d = \dfrac {\sigma d}{\varepsilon_{0}}\$ \$Capacitance = c = \dfrac {Q}{v} = \dfrac {\sigma A}{v} = \dfrac {\sigma A}{\left (\dfrac {\sigma d}{\varepsilon_{0}}\right )}\$ \$c = \dfrac {A\varepsilon_{0}}{d}\$energy store in capacitance \$= v = \dfrac {1}{2} \dfrac {\theta^{2}}{c}\$\$\textbf{Step 3 - Explanation of condenser with dielectric slab}\$Now, when we interduce a di-electric in capacitor then -electric field between plate \$= E' = \dfrac {E}{k} = \dfrac {\sigma}{k\varepsilon_{0}}\$ Potential between plate \$= v' = E'\cdot d = \dfrac {\sigma d}{k\varepsilon_{0}} = \dfrac {v}{k}\$ Capacitance of capacitor \$= c' = \dfrac {\theta}{v'} = \dfrac {\sigma A}{\left (\dfrac {\sigma d}{k\varepsilon_{0}}\right )}\$\$e' = \dfrac {k\varepsilon_{0}A}{d} = ck\$So, store energy -\$v' = \dfrac {1}{2} \dfrac {\theta^{2}}{c'} = \dfrac {1}{2} \dfrac {\theta^{2}}{ck} = \dfrac {v}{k}\$\$\textbf{Step 4 - Observation -}\$\$\bullet\$ Since, we can see that - after inducing the slab in capacitor is -\$Capacitance = c' = ck \rightarrow increase\$ \$Potential = v' = \dfrac {v}{k} \rightarrow decrease\$ \$energy = u' = \dfrac {u}{k} \rightarrow decrease\$\$\rightarrow\$ Hence option (A) is correct.

Q: Two charges \$q\$ and \$-q\$ are kept apart. Then at any point on the perpendicular bisector of line joining the two charges:

We know voltage is given by \$\dfrac{Kq}{r}\$All the points on perpendicular bisectore are at equidistant from the two points A and B.\$\therefore Potential = \dfrac{Kq}{r}+\dfrac{K(-q)}{r}\$\$= 0\$

Q: A parallel plate capacitor, filled with a material of dielectric constant \$K\$, is charged to a certain voltage and is isolated. The dielectric material is removed. Then: (a) capacitance decreases by a factor \$K\$(b) electric field reduces by a factor \$K\$(c) voltage across the capacitor increases by a factor \$K\$(d) charge stored in the capacitor increases by a factor \$K\$

Initial capacitance \$c= \dfrac {k\varepsilon_0 A}{d}\$When the system is isolated, charge (\$Q\$) present on it is constant.Now, when dielectric is removed \$C=\dfrac {\varepsilon_0 A}{d}\$\$\Rightarrow\$ \$C\$ decreases by a factor of \$K\$.We know that \$Q=C \times V\$As \$C\$ decreases by factor of \$K\$ and \$Q\$ remains constant, \$V\$ increases by a factor of \$K\$.

Q: Twenty seven identical mercury drops each charged to \$10V\$, are allowed to form a big drop. The potential of the big drop is

Voltage of each mercury drop \$= 10 V\$\$\Rightarrow \dfrac {kq}{r}= 10 V\$\$\Rightarrow q= \dfrac {10r}{k} \$ \$total \ charge =\dfrac {27\times 10r}{k}\$Equating volumes to find radius of bigger charge, \$\dfrac {4}{3}\pi (R^3)=27\cdot \dfrac {4}{3}\pi r^3\$\$R=3r\$final voltage\$ = \dfrac {kq}{R}=\dfrac {k \times 27\times 10r}{k\times 3r}\$\$=90 V\$

Q: A charge of \$2\ C\$ is moved from a point \$2\ m\$ away from a charge of \$1\ C\$ to a point \$1\ m\$ away from that charge. The work done is:

Voltage due to \$1\ C\$ at a distance \$r\$ is \$\dfrac {k \times 1}{r}\$.Total change in potential in moving \$2\ C\$ charge from \$2\ m\$ away to \$1\ m\$ away:\$V_2-V_1=\dfrac {k \times 1}{1}-\dfrac {k \times 1}{2}\$or, \$V_2-V_1=\dfrac{k \times 1}{2} = \Delta V\$So, work done \$=\Delta V Q\$\$=\dfrac {k \times 1}{2}\times 2=9\times 10^9\ J\$

Q: (1): The dielectric medium between the plates of a parallel plate capacitor lowers the potential difference between the plates without a battery. (2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.

Consider a capacitor with charge density \$\sigma\$.The potential between its two plates is given by \$\sigma d/ \epsilon_{0} \$When a dielectric is inserted, the electric field inside the capacitor decrease decreasing the potential between the two plates of capacitor.However, this is nothing to the dielectric strength of the dielectric.

Q: One plate of parallel plate capacitor is smaller than the other. The charge on the smaller plate:

Hint: apply gauss law for the two plates (conductors) where E=0.Step 1: imagine a Gaussian surface Imagine a Gaussian surface surrounding the two plates as shown in figure. Step 2: Apply gauss law Gauss law: electric flux through any closed surface is directly proportional to the charge enclosed in it. \$\oint{E.dA}\$ =\$\dfrac{q}{{{\varepsilon _o}}}\$ As both plates are conductor hence E=0 for the Gaussian surface \$\dfrac{q}{{{\varepsilon _o}}}\$= 0 \${\varepsilon _o} \ne 0\$ So q=0 Step 3: find charge on each plate As, total q=0 So charge on each plate will be equal and opposite Final step: The charge on the smaller plate will be equal to other plate to make total charge zero on both the plates irrespective of shape and size.

Question 1

When a dielectric material is introduced between the plates of a charged condenser, after disconnecting the battery, the electric field between the plates:

Accepted Answer

We know electric field between capacitor plates: $E= \dfrac{\sigma}{\epsilon _0}$$= \dfrac{V}{d}$, where $d$ is the distance between the plates.When battery is disconnected, the change present on capacitor is fixed, i.e., $V$ is constant.Now, when the dielectric is introduced, the electric field becomes: $E' = \dfrac{V}{Kd}$, where K is dielectric constant which is always > 1$\therefore $ electric field decreases.

Question 2

A condenser is charged and then battery is removed. A dielectric plate is put between the plates of condenser, then correct statement is

Accepted Answer

$\textbf{Step 1 - Given data Refer Figure}$let Cross-section of plate $= A$  distance between plate $= d$   surface charge density $= \sigma$Capacitance of capacitor $= c$$\textbf{Step 2 - Explanation of condenser without slab}$Since, we know that - when battery is removed from a charge capacitorthen charge is constant.So, electric field inside capacitor $= E = \dfrac {\sigma}{\varepsilon_{0}}$              Potential between plate $= V = E \cdot d = \dfrac {\sigma d}{\varepsilon_{0}}$                      $Capacitance = c = \dfrac {Q}{v} = \dfrac {\sigma A}{v} = \dfrac {\sigma A}{\left (\dfrac {\sigma d}{\varepsilon_{0}}\right )}$                                                   $c = \dfrac {A\varepsilon_{0}}{d}$energy store in capacitance $= v = \dfrac {1}{2} \dfrac {\theta^{2}}{c}$$\textbf{Step 3 - Explanation of condenser with dielectric slab}$Now, when we interduce a di-electric in capacitor then -electric field between plate $= E' = \dfrac {E}{k} = \dfrac {\sigma}{k\varepsilon_{0}}$      Potential between plate $= v' = E'\cdot d = \dfrac {\sigma d}{k\varepsilon_{0}} = \dfrac {v}{k}$    Capacitance of capacitor $= c' = \dfrac {\theta}{v'} = \dfrac {\sigma A}{\left (\dfrac {\sigma d}{k\varepsilon_{0}}\right )}$$e' = \dfrac {k\varepsilon_{0}A}{d} = ck$So, store energy -$v' = \dfrac {1}{2} \dfrac {\theta^{2}}{c'} = \dfrac {1}{2} \dfrac {\theta^{2}}{ck} = \dfrac {v}{k}$$\textbf{Step 4 - Observation -}$$\bullet$ Since, we can see that - after inducing the slab in capacitor is -$Capacitance = c' = ck \rightarrow increase$      $Potential = v' = \dfrac {v}{k} \rightarrow decrease$           $energy = u' = \dfrac {u}{k} \rightarrow decrease$$\rightarrow$ Hence option (A) is correct.

Question 3

Two charges $q$ and $-q$ are kept apart. Then at any point on the perpendicular bisector of line joining the two charges:

Accepted Answer

We know voltage is given by $\dfrac{Kq}{r}$All the points on perpendicular bisectore are at equidistant from the two points A and B.$\therefore Potential = \dfrac{Kq}{r}+\dfrac{K(-q)}{r}$$= 0$

Question 4

Statement(A): Negative charges always move from a higher potential to lower potential point&#10;Statement (B): Electric potential is vector.

Accepted Answer

Positive charges move from higher to lower potential whereas negative charges move from lower to higher potential. Hence, A is false.Electric potential does not have direction and is a scalar quantity. Hence, B is false.

Question 5

A parallel plate condenser is charged by connecting it to a battery. Without disconnecting the battery, the space between the plates is completely filled with a medium of dielectric constant k. Then:

Accepted Answer

Question 6

A parallel plate capacitor, filled with a material of dielectric constant $K$ , is charged to a certain voltage and is isolated. The dielectric material is removed. Then:
(a) capacitance decreases by a factor $K$
(b) electric field reduces by a factor $K$
(c) voltage across the capacitor increases by a factor $K$
(d) charge stored in the capacitor increases by a factor $K$

Accepted Answer

Initial capacitance $c= \dfrac {k\varepsilon_0 A}{d}$When the system is isolated, charge ($Q$) present on it is constant.Now, when dielectric is removed $C=\dfrac {\varepsilon_0 A}{d}$$\Rightarrow$ $C$ decreases by a factor of $K$.We know that $Q=C \times V$As $C$ decreases by factor of $K$ and $Q$ remains constant, $V$ increases by a factor of $K$.

Question 7

Twenty seven identical mercury drops each charged to $10V$, are allowed to form a big drop. The potential of the big drop is

Accepted Answer

Voltage of each mercury drop $= 10 V$$\Rightarrow \dfrac {kq}{r}= 10 V$$\Rightarrow q= \dfrac {10r}{k}  $ $total \  charge  =\dfrac {27\times 10r}{k}$Equating volumes to find radius of bigger charge, $\dfrac {4}{3}\pi (R^3)=27\cdot \dfrac {4}{3}\pi r^3$$R=3r$final voltage$ = \dfrac {kq}{R}=\dfrac {k \times 27\times 10r}{k\times 3r}$$=90 V$

Question 8

A charge of $2\ C$ is moved from a point $2\ m$ away from a charge of $1\ C$ to a point $1\ m$ away from that charge. The work done is:

Accepted Answer

Voltage due to $1\ C$ at a distance $r$ is $\dfrac {k \times 1}{r}$.Total change in potential in moving $2\ C$ charge from $2\ m$ away to $1\ m$ away:$V_2-V_1=\dfrac {k \times 1}{1}-\dfrac {k \times 1}{2}$or, $V_2-V_1=\dfrac{k \times 1}{2} = \Delta V$So, work done $=\Delta V Q$$=\dfrac {k \times 1}{2}\times 2=9\times 10^9\ J$

Question 9

(1): The dielectric medium between the plates of a parallel plate capacitor lowers the potential difference between the plates without a battery.

(2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.

Accepted Answer

Consider a capacitor with charge density $\sigma$.The potential between its two plates is given by $\sigma d/ \epsilon_{0} $When a dielectric is inserted, the electric field inside the capacitor decrease decreasing the potential between the two plates of capacitor.However, this is nothing to the dielectric strength of the dielectric.

Question 10

One plate of parallel plate capacitor is smaller than the other. The charge on the smaller plate:

Accepted Answer

Hint:&#10;apply gauss law for the two plates (conductors) where E=0.Step&#10;1: imagine a Gaussian surface Imagine a Gaussian surface&#10;surrounding the two plates as shown in&#10;figure.  Step&#10;2: Apply gauss law              Gauss law:&#10;electric flux through any closed surface is directly             proportional to&#10;the charge enclosed in it.                     $\oint{E.dA}$  =$\dfrac{q}{{{\varepsilon _o}}}$             As both plates&#10;are conductor hence E=0 for the Gaussian surface                            $\dfrac{q}{{{\varepsilon _o}}}$= 0                             ${\varepsilon _o} \ne 0$                           So q=0 Step&#10;3: find charge on each plate               As, total q=0             So charge on&#10;each plate will be equal and opposite &#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Final step: The charge on the smaller&#10;plate will be equal to other plate to make total charge zero on both the plates&#10;irrespective of shape and size.

Electrostatic Potential And Capacitance

Physics
14721 Views

When a dielectric material is introduced between the plates of a charged condenser, after disconnecting the battery, the electric field between the plates:

A condenser is charged and then battery is removed. A dielectric plate is put between the plates of condenser, then correct statement is

Two charges $q$ and $- q$ are kept apart. Then at any point on the perpendicular bisector of line joining the two charges:

Statement(A): Negative charges always move from a higher potential to lower potential point

Statement (B): Electric potential is vector.

A parallel plate condenser is charged by connecting it to a battery. Without disconnecting the battery, the space between the plates is completely filled with a medium of dielectric constant k. Then:

Twenty seven identical mercury drops each charged to $10 V$ , are allowed to form a big drop. The potential of the big drop is

A charge of $2 C$ is moved from a point $2 m$ away from a charge of $1 C$ to a point $1 m$ away from that charge. The work done is:

(1): The dielectric medium between the plates of a parallel plate capacitor lowers the potential difference between the plates without a battery.
(2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.

One plate of parallel plate capacitor is smaller than the other. The charge on the smaller plate:

Electrostatic Potential And Capacitance

Physics 14721 Views

When a dielectric material is introduced between the plates of a charged condenser, after disconnecting the battery, the electric field between the plates:

A condenser is charged and then battery is removed. A dielectric plate is put between the plates of condenser, then correct statement is

Two charges q and −q are kept apart. Then at any point on the perpendicular bisector of line joining the two charges:

Statement(A): Negative charges always move from a higher potential to lower potential point Statement (B): Electric potential is vector.

A parallel plate condenser is charged by connecting it to a battery. Without disconnecting the battery, the space between the plates is completely filled with a medium of dielectric constant k. Then:

Twenty seven identical mercury drops each charged to 10V, are allowed to form a big drop. The potential of the big drop is

A charge of 2 C is moved from a point 2 m away from a charge of 1 C to a point 1 m away from that charge. The work done is:

(1): The dielectric medium between the plates of a parallel plate capacitor lowers the potential difference between the plates without a battery. (2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.

One plate of parallel plate capacitor is smaller than the other. The charge on the smaller plate:

Physics
14721 Views

Two charges $q$ and $- q$ are kept apart. Then at any point on the perpendicular bisector of line joining the two charges:

Statement(A): Negative charges always move from a higher potential to lower potential point

Statement (B): Electric potential is vector.

Twenty seven identical mercury drops each charged to $10 V$ , are allowed to form a big drop. The potential of the big drop is

A charge of $2 C$ is moved from a point $2 m$ away from a charge of $1 C$ to a point $1 m$ away from that charge. The work done is:

(1): The dielectric medium between the plates of a parallel plate capacitor lowers the potential difference between the plates without a battery.
(2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.