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: 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: Three capacitors of \$3 \mu F, 2 \mu F \$ and \$6 \mu F\$ are connected in series. When a battery of 10V is connected to this combination then charge on \$3 \mu F\$capacitor will be :

\$q=CV\ \ \ \dfrac{1}{C_{eff}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\$\$\dfrac{1}{C_{ef}}=\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{6}\$\$\therefore C_{ef}=1 \mu F\$\$\therefore q=C_{ef} V = 10 \mu C\$q is same for all capacitors\$\therefore\$ q on any capacitor is \$10 \mu C\$

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: The capacity of a parallel plate air condenser is \$2\ \mu F\$. If a dielectric of dielectric constant \$4\$ is introduced between the plates, its new capacity is:

We know \$C=\dfrac{\epsilon_{0}A}{d}\$When dielectric is added:\$C'=\dfrac{K \epsilon_{0}A}{d}\$\$=4\times 2\ \mu F\$\$=8\mu F\$

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\$

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

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 3

Three capacitors of $3 \mu F, 2 \mu F $ and $6 \mu F$ are connected in series. When a battery of 10V is connected to this combination then charge on $3 \mu F$capacitor will be :

Accepted Answer

$q=CV\ \ \ \dfrac{1}{C_{eff}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$$\dfrac{1}{C_{ef}}=\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{6}$$\therefore C_{ef}=1 \mu F$$\therefore q=C_{ef} V = 10 \mu C$q is same for all capacitors$\therefore$ q on any capacitor is $10 \mu C$

Question 4

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 5

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 6

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 7

The capacity of a parallel plate air condenser is $2\ \mu F$. If a dielectric of dielectric constant $4$ is introduced between the plates, its new capacity is:

Accepted Answer

We know $C=\dfrac{\epsilon_{0}A}{d}$When dielectric is added:$C'=\dfrac{K \epsilon_{0}A}{d}$$=4\times 2\ \mu F$$=8\mu F$

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

In an electrical circuit, which of the following quantities is analogous to temperature?

Accepted Answer

When there is a temperature difference, heat flows from high temperature to low temperature. Similarly, when there is a potential difference electric current flows from high potential to low potential. Hence, potential is analogous to temperature.

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
1796 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

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

Three capacitors of $3 μ F, 2 μ F$ and $6 μ F$ are connected in series. When a battery of 10V is connected to this combination then charge on $3 μ F$ capacitor will be :

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).

The capacity of a parallel plate air condenser is $2 μ F$ . If a dielectric of dielectric constant $4$ is introduced between the plates, its new capacity is:

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

In an electrical circuit, which of the following quantities is analogous to temperature?

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 1796 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

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

Three capacitors of 3μF,2μF and 6μF are connected in series. When a battery of 10V is connected to this combination then charge on 3μFcapacitor will be :

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).

The capacity of a parallel plate air condenser is 2 μF. If a dielectric of dielectric constant 4 is introduced between the plates, its new capacity is:

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

In an electrical circuit, which of the following quantities is analogous to temperature?

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
1796 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

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

Three capacitors of $3 μ F, 2 μ F$ and $6 μ F$ are connected in series. When a battery of 10V is connected to this combination then charge on $3 μ F$ capacitor will be :

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 capacity of a parallel plate air condenser is $2 μ F$ . If a dielectric of dielectric constant $4$ is introduced between the plates, its new capacity is: