New Capacitance in Dielectric

Q. A parallel-plate capacitor of area $A$, plate separation 𝑑, and capacitance $C$ is filled with four dielectric material having dielectric constants $k_1,k_2,k_3$ and $k_4$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by

1. $k=k_1+k_2+k_3+3k_4$
2. $K=\frac{2}{3}(k_1+k_2+k_3)+2k_4$
3. $\frac{2}{k}=\frac{3}{k_1+k_2+k_3}+\frac{1}{k_4}$
4. $\frac{1}{k_4}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\frac{3}{2k_4}$

Q. A parallel plate capacitor has plates of area $A$ separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant which varies as $k\left(x\right)=k(1+\alpha x)$, where `$x$' is the distance measured from one of the plates. If $(\alpha d\ll 1)$, the total capacitance of the system is
best given by the expression:

1. $\frac{A{\epsilon }_{0}k}{d}[1+{\left(\frac{\alpha d}{2}\right)}^2]$
2. $\frac{Ak{\epsilon }_0}{d}[1+\left(\frac{\alpha d}{2}\right)]$
3. $\frac{A{\epsilon }_{0}k}{d}[1+\left(\frac{{\alpha }^{2}d}{2}\right)]$
4. $\frac{Ak{\epsilon }_0}{d}[1+\alpha d]$

Q. The plates of a parallel plate capacitor are charged up to $100\text{V}$. Now, after removing the battery, a $2\text{mm}$ thick plate is inserted between the plates . Then, to maintain the same potential difference, the distance between the capacitor plates is increased by $1.6\text{mm}$. The dielectric constant of the plate is

1. $5$
2. $1.25$
3. $4$
4. $2.5$

Q. Two parallel-plate capacitors of capacitance $C$ and $2C$ are connected in parallel and charged to a potential difference $V$. The battery is then disconnected and the region between the plates of capacitor $C$ is filled completely with a material of dielectric constant $K$. The common potential difference across the combination becomes

1. $\frac{2V}{K+2}$
2. $\frac{2V}{K+2}$
3. $\frac{3V}{K+3}$
4. $\frac{3V}{K+2}$

Q.

What happens when 2 capacitors are connected with:opposite polarity plates connected together and same polarity plates are connected together

Q. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Q.

A parallel plate capacitor has a plate area of $100m^2$ and plate separation of $10m$. The space between the plates is filled up to a thickness of $5m$ with a material of dielectric constant $10$. The resultant capacitance of the system is $xpF$. The value of ${\epsilon }_0=8.85\times {10}^{-12}F{m}^{-1}$. The value of $x$ to the nearest integer is _______.

Q.

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in:

1. None of the above

2. Reduction of charge on the plates and increase of potential across the plates

3. Decrease in the potential difference across the plates reduction in stored energy, but no change in the charge on the plates
   

4. Increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates

   
   

Q.

Why does increasing voltage decrease capacitance?

Q. Two identical parallel-plate capacitors are connected in series and then joined across a battery of 100 V. A slab of dielectric constant K = 3 is inserted between the plates of the first capacitor. Then, the potential difference across the capacitors will respectively be

1. 25 V, 75 V
2. 75 V, 25 V
3. 20 V, 80 V
4. 50 V, 50 V

Q. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.

Q. How can we increase capacitance

Q. Derive the expression for the capacitance of a parallel plate capacitor with no dielectric medium between the plates

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

1. $32$
2. $2$
3. $16$
4. $24$

Q. A parallel plate capacitor has capacitance C when no dielectric between the plates. Now a slab of dielectric constant K, having same thickness as the separation between the plates is introduced so as to fill one- fourth of the capacitor as shown in the figure, the new capacitance will be

1. $(K+3)\frac{C}{4}$
2. $(K+2)\frac{C}{4}$
3. $(K+1)\frac{C}{4}$
4. $\frac{KC}{4}$

Q. A capacitor is half filled with a dielectric (K = 2) as shown in the figure A. If the same capacitor is to be filled with the same dielectric as shown in figure B, what should be the thickness $t$ of the dielectric so that capacitance still has same value?

Q. Show the capacitance of the spherical conductor is 4pi epsilon times the radius of the spherical conductor

Q.

When a dielectric slab is introduced between the plates of an isolated charged capacitor, it

1. All of the above

2. Increase the capacitance of the capacitor

3. Decreases the electric filed between the plates

4. Decreases the amount of energy stored in the capacitor

Q. A parallel plate condenser has a capacitance 50μF in air and 110μF when immersed in an oil. The dielectric constant of the oil is?

1. 0.55

2. 0.45

3. 1.10

4. 2.20

Q.

Find the ratio of resistances of two copper rods X and Y of lengths 30 cm and 10 cm respectively and having radii 2 cm and 1 cm respectively.

Q.

The dielectric constant of air is

Q. Two parallel plate air filled capacitors, each of capacitance $C$, are joined in series to a battery of emf $V$. The space between the plates of one of the capacitors is then completely filled up with a uniform dielectric having dielectric constant $K$. The quantity of charge which flows through the battery is

1. $\frac{CV}{2}\left(\frac{K+1}{K-1}\right)$
2. $\frac{CV}{2}\left(\frac{K-1}{K+1}\right)$
3. $\frac{2}{CV}\left(\frac{K-1}{K+1}\right)$
4. $\frac{2}{CV}\left(\frac{K+1}{K-1}\right)$

Q.

The capacity of a parallel plate condenser is . When a glass plate is

placed between the plates of the conductor, its potential becomes 1/8thof

the original value. The value of dielectric constant will be

1. 1.6

2. 5

3. 8

4. 40

Q. There are two concentric conducting spherical shells of radius R and 2R respectively. Inner shell is given charge +Q and outer shell is neutral. If both shells are connected by conducting wire, then heat loss is?

Q. Two parallel-plate capacitors, each of capacitance $40\mu F$, are connected in series. The space between the plates of one capacitor is filled with a dielectric of dielectric constant $K=3$, then the equivalent capacitance of the combination is

1. $30\mu F$
2. $120\mu F$
3. $40\mu F$
4. $160\mu F$

Q.

A potential difference is set up between the plates of a parallel plate capacitor by a battery and then the battery is removed. If the distance between the plates is decreased, then how the (a) charge (b) potential difference (c) electric field (d) energy and (e) energy density will change?

Q.

A parallel plate capacitor is charged by connecting its plates to the terminals of a battery. The battery is then disconnected and a glass plate is interposed between the plates of the capacitor, then

1. The capacity increases while the potential difference increases

2. The capacity increases while the potential difference decreases

3. capacity decreases while the potential difference decreases

4. The capacity decreases while the potential difference increases

Q.

When a dielectric slab is inserted between the plates of one of the two identical capacitors shown in the figure, then

match the following.

$\begin{array}{cc}Column-I& Column-II\\ \left(A\right)ChargeonA& \left(p\right)increases\\ \left(B\right)PotentialdifferenceacrossA& \left(q\right)decreases\\ \left(C\right)PotentialdifferenceacrossB& \left(r\right)remainsconstant\\ \left(D\right)ChargeonB& \left(s\right)willchange\end{array}$

1. undefined
2. undefined
3. undefined
4. undefined

Q.

If the potential difference between the plates of a charged and isolated parallel-plate condenser of capacitance 1 μF changes from 12 V to 4 V when a dielectric slab of dielectric constant K is introduced to fill the entire space between the plates, then

1. The change in the charge on the capacitor is 8 µC

2. The electric field in the capacitor increases by 100 %

3. The change in the capacity is 3 µF

4. The value of K is 3

Q. A parallel plate air condenser has a capacity ‘C’. When the plate separation is halved and a dielectric of constant k is inserted, new capacitance is found to be 12C. The value of dielectric constant ‘k’ is?

1. 12
2. 24
3. 6
4. 3