Q: \$100J\$ of work is done when \$2 \mu C\$ charge is moved in an electric field between two points. The p.d. between the points is

We know work done \$=q\Delta V \$Where \$\Delta V \$ is change in potential \$V_2-V_1\$\$100=\Delta V\times 2\times 10^{-6}\$\$50\times 10^6=\Delta V\$\$\therefore p\cdot d= 5\times 10^7 V\$

Q: There are \$10\$ condensers each of capacity \$5\; \mu F\$. The ratio of minimum to maximum capacity obtained from these condensers will be :

\$C = 5 \mu F\$minimum capacity is obtained when all are in a series,\$\therefore \dfrac{1}{c_{s}}=\dfrac{1}{c_1}+\dfrac{1}{c_2}.....+\dfrac{1}{c_{10}}\$\$ \dfrac{1}{c_{s}}=\dfrac{1}{5}+\dfrac{1}{5}+.....+\dfrac{1}{5}=\dfrac{10}{5}\$\$\therefore c_{s}=\dfrac{1}{2}\$maximum capacity is obtained when they are parallety connected\$\therefore c_{eff}=c_1+c_2+.......+c_{10}\$= 5 + 5 + ......+5\$c_{p}{=10(5)}\$\$c_{p}=50\$\$\therefore\$ Ratio of min to max capacity, \$\dfrac{c_s}{c_p}= \dfrac{0.5}{50}=1 : 100\$

Q: When air is replaced by a dielectric medium of constant \$K\$, the capacity of the condenser:

We know, capacity: \$C=\dfrac {\varepsilon_0 A}{d}\$Now when air is replaced by dielectric medium of constant \$k\$\$C'=\dfrac {K\varepsilon_0 A}{d}\$This implies that the capacitance increases \$K\$ times.

Q: A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,

\$\rightarrow \$ When the capacitor is charged and battery is removed the charge present on capacitor remains constant. (Law of conservation of charge)\$\rightarrow\$ We know \$C=\dfrac {\varepsilon_0 A}{d}\$When slab is introduced, \$C=\dfrac {k\varepsilon_0 A}{d}\$\$C\$ increases and \$Q\$ is constant.We know: \$Q=CV\$ \$\therefore \$ \$V\$ has to decrease.We know energy store in capacitor \$=\dfrac{1}{2} QV\$\$\therefore \$ as V decreases, energy decreases.

Q: A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is:

Initial capacitance \$=\frac {\varepsilon_0 A}{d}\$Now after metal plate is inserted it behaves as two capacitor plates in series with capacitance \$\frac {\varepsilon_0 A}{(\frac {d}{a})}\$We know when capacitance are in series\$\frac {1}{C_{eff}}=\frac {1}{C_1}+\frac {1}{C_2}\$\$\Rightarrow \frac {1}{C_{eff}}=\frac {d}{4\varepsilon_0 A}+\frac {d}{4\varepsilon_0 A}\$\$\Rightarrow C_{eff}=\frac {2\varepsilon_0 A}{d.}\$\$\therefore Now \ C_{eff}= 2C\$

Q: The equivalent capacitance between P and Q of the given figure is (the capacitance of each capacitor is\$1\ \mu F\$ ):

We know when capacitors are in series:\$ \dfrac{1}{C_{eff}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}\$and when in parallel:\$C_{eff}=C_{1}+C_{2}\$So, for the given combination, between P and Q there are 2 combinations of 2 parallely connected capacitors and these combinations come in series with a single capacitor.So, effective capacitance between P and Q may expressed as:\$\dfrac{1}{C_{eff}}=\dfrac{1}{2C}+\dfrac{1}{C}+\dfrac{1}{2C}\$or, \$C_{eff}=\dfrac{C}{2}\$\$=0.5 C\$\$\therefore C_{eff}=0.5\ \mu F\$

Question 1

$100J$ of work is done when $2 \mu C$ charge is moved in an electric field between two points. The p.d. between the points is

Accepted Answer

We know work done $=q\Delta V $Where $\Delta V $ is change in potential $V_2-V_1$$100=\Delta V\times 2\times 10^{-6}$$50\times 10^6=\Delta V$$\therefore p\cdot d= 5\times 10^7 V$

Question 2

There are $10$ condensers each of capacity $5\; \mu F$. The ratio of minimum to maximum capacity obtained from these condensers will be :

Accepted Answer

$C = 5 \mu F$minimum capacity is obtained when all are in a series,$\therefore \dfrac{1}{c_{s}}=\dfrac{1}{c_1}+\dfrac{1}{c_2}.....+\dfrac{1}{c_{10}}$$ \dfrac{1}{c_{s}}=\dfrac{1}{5}+\dfrac{1}{5}+.....+\dfrac{1}{5}=\dfrac{10}{5}$$\therefore c_{s}=\dfrac{1}{2}$maximum capacity is obtained when they are parallety connected$\therefore c_{eff}=c_1+c_2+.......+c_{10}$= 5 + 5 + ......+5$c_{p}{=10(5)}$$c_{p}=50$$\therefore$ Ratio of min to max capacity,                                              $\dfrac{c_s}{c_p}= \dfrac{0.5}{50}=1 : 100$

Question 3

When air is replaced by a dielectric medium of constant $K$, the capacity of the condenser:

Accepted Answer

We know, capacity: $C=\dfrac {\varepsilon_0 A}{d}$Now when air is replaced by dielectric medium of constant $k$$C'=\dfrac {K\varepsilon_0 A}{d}$This implies that the capacitance increases $K$ times.

Question 4

The resultant capacity between the points P and Q of the given figure is:

Accepted Answer

As seen from the circuit, between P and Q there is a combination of three capacitors. In this combination two $4\ \mu F$ capacitors are connected in series and this series connection is connected in parallel to the $2\ \mu F$ capacitor.So, effective capacitance of the series combination:$\dfrac{1}{C_{eff}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$or, $\dfrac{1}{C_{eff}}=\dfrac{1}{4}+\dfrac{1}{4}$or, $C_{eff}=2\mu F$Now, this $2\ \mu F$ capacitance comes in parallel with the $2\ \mu F$ capacitance.So, effective capacitance of this parallel combination:$C'_{eff}=C_{1}+C_{2}$or, $C'_{eff}=2\ \mu F+2\ \mu F$or, $C'_{eff}=4\ \mu F$

Question 5

Space between the plates of a parallel plate capacitor is filled with a dielectric slab. The capacitor is charged and then the supply is disconnected to it. If the slab is now taken out , then:

Accepted Answer

When a capacitor is charged and then the supply is disconnected charge present on it becomes constant (law of conservation of charge)Initial capacitance $C=\frac {k\varepsilon_0 A}{d}$ (with slab)Final capacitance $C=\frac {\varepsilon_0 A}{d}$ (slab removed)We know $Q= CV$As C decreases and Q remains constant V has to increase.

Question 6

A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,

Accepted Answer

$\rightarrow $ When the capacitor is charged and battery is removed the charge present on capacitor remains constant. (Law of conservation of charge)$\rightarrow$ We know $C=\dfrac {\varepsilon_0 A}{d}$When slab is introduced, $C=\dfrac {k\varepsilon_0 A}{d}$$C$ increases and $Q$ is constant.We know: $Q=CV$ $\therefore $ $V$ has to decrease.We know energy store in capacitor $=\dfrac{1}{2} QV$$\therefore $ as V decreases, energy decreases.

Question 7

Fill in the blank:In order to increase the capacity of a parallel plate condenser,  one should introduce a sheet of  $\underline{\hspace{0.5in}}$ between the plates (assume that the space is completely filled).

Accepted Answer

Dielectric constant is the factor by which the capacitance increases from its value in air when a dielectric is introduced between the plates.Mica, among the materials mentioned above, is a dielectric material and hence to be used to increase the capacity.

Question 8

A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is:

Accepted Answer

Initial capacitance $=\frac {\varepsilon_0 A}{d}$Now after metal plate is inserted it behaves as two capacitor plates in series with capacitance $\frac {\varepsilon_0 A}{(\frac {d}{a})}$We know when capacitance are in series$\frac {1}{C_{eff}}=\frac {1}{C_1}+\frac {1}{C_2}$$\Rightarrow \frac {1}{C_{eff}}=\frac {d}{4\varepsilon_0 A}+\frac {d}{4\varepsilon_0 A}$$\Rightarrow C_{eff}=\frac {2\varepsilon_0 A}{d.}$$\therefore Now  \ C_{eff}= 2C$

Question 9

The equivalent capacitance between P and Q of the given figure is (the capacitance of each capacitor is$1\ \mu F$ ):

Accepted Answer

We know when capacitors are in series:$ \dfrac{1}{C_{eff}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$and when in parallel:$C_{eff}=C_{1}+C_{2}$So, for the given combination, between P and Q there are 2 combinations of 2 parallely connected capacitors and these combinations come in series with a single capacitor.So, effective capacitance between P and Q may expressed as:$\dfrac{1}{C_{eff}}=\dfrac{1}{2C}+\dfrac{1}{C}+\dfrac{1}{2C}$or, $C_{eff}=\dfrac{C}{2}$$=0.5 C$$\therefore C_{eff}=0.5\ \mu F$

Question 10

A parallel plate capacitor is first charged and then isolated , and a dielectric slab is introduced between the plates. The quantity that remains unchanged is:

Accepted Answer

When the capacitor is kept at a voltage, it gains charge.Now when the system is isolated, the charge present on capacitor cannot change because of law of conservation of charge.$\therefore $ Charge always remains constant in isolated systems.

Electrostatic Potential And Capacitance

Physics
18069 Views

$100 J$ of work is done when $2 μ C$ charge is moved in an electric field between two points. The p.d. between the points is

There are $10$ condensers each of capacity $5 μ F$ . The ratio of minimum to maximum capacity obtained from these condensers will be :

When air is replaced by a dielectric medium of constant $K$ , the capacity of the condenser:

The resultant capacity between the points P and Q of the given figure is:

Space between the plates of a parallel plate capacitor is filled with a dielectric slab. The capacitor is charged and then the supply is disconnected to it. If the slab is now taken out , then:

A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,

Fill in the blank:

In order to increase the capacity of a parallel plate condenser, one should introduce a sheet of between the plates (assume that the space is completely filled).

A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is:

The equivalent capacitance between P and Q of the given figure is (the capacitance of each capacitor is $1 μ F$ ):

A parallel plate capacitor is first charged and then isolated , and a dielectric slab is introduced between the plates. The quantity that remains unchanged is:

Electrostatic Potential And Capacitance

Physics 18069 Views

100J of work is done when 2μC charge is moved in an electric field between two points. The p.d. between the points is

There are 10 condensers each of capacity 5μF. The ratio of minimum to maximum capacity obtained from these condensers will be :

When air is replaced by a dielectric medium of constant K, the capacity of the condenser:

The resultant capacity between the points P and Q of the given figure is:

Space between the plates of a parallel plate capacitor is filled with a dielectric slab. The capacitor is charged and then the supply is disconnected to it. If the slab is now taken out , then:

A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,

Fill in the blank: In order to increase the capacity of a parallel plate condenser, one should introduce a sheet of ​ between the plates (assume that the space is completely filled).

A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is:

The equivalent capacitance between P and Q of the given figure is (the capacitance of each capacitor is1 μF ):

A parallel plate capacitor is first charged and then isolated , and a dielectric slab is introduced between the plates. The quantity that remains unchanged is:

Physics
18069 Views

$100 J$ of work is done when $2 μ C$ charge is moved in an electric field between two points. The p.d. between the points is

There are $10$ condensers each of capacity $5 μ F$ . The ratio of minimum to maximum capacity obtained from these condensers will be :

When air is replaced by a dielectric medium of constant $K$ , the capacity of the condenser:

Fill in the blank:

In order to increase the capacity of a parallel plate condenser, one should introduce a sheet of between the plates (assume that the space is completely filled).

The equivalent capacitance between P and Q of the given figure is (the capacitance of each capacitor is $1 μ F$ ):