Advantages of Using Dielectrics in Capacitors

Q. In the figure a capacitor is filled with dielectrics. The resultant capacitance is:

1. $\frac{2A{\epsilon }_0}{d}[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}]$
2. $\frac{2A{\epsilon }_0}{d}[k_1+k_2+k_3]$
3. $\frac{A{\epsilon }_0}{d}[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}]$
4. $\frac{A{\epsilon }_0}{d}[\frac{k_3}{2}+\frac{k_1{k}_2}{k_1+k_2}]$

Q. The amplifiers $X,Y$ and $Z$ are connected in series. If the voltage gains of $X,Y$ and $Z$ are $10,20$ and $30$, respectively and the input signal is $1mV$ peak value, then what is the output signal voltage (peak value)

(i) if $dc$ supply voltage is $10V$?

(ii) if $dc$ supply voltage is $5V$?

Q.

In an aluminum $\left(Al\right)$ bar of square cross section, a square hole is drilled and is filled with iron $\left(Fe\right)$ as shown in the figure. The electrical resistivities of Al and Fe are $2.7\times {10}^{-8}\Omega -m$ and $1.0\times {10}^{-7}\Omega -m$, respectively. The electrical resistance between the two faces $P$ and $Q$ of the composite bar is

1. 
   $\frac{2475}{64}\mu \Omega$
2. 
   $\frac{1875}{64}\mu \Omega$
3. 
   $\frac{1875}{49}\mu \Omega$
4. 
   $\frac{2475}{132}\mu \Omega$

Q. A metallic plate of thickness (t) and face area of one side (A) is inserted between the plates of a parallel plate air capacitor with a separation (d) and face area (A). Then the equivalent capacitance is :

1. $\frac{{ϵ}_{0}A}{d}$
2. $\frac{{ϵ}_{0}A}{(d\times t)}$
3. $\frac{{ϵ}_{0}A}{(d-t)}$
4. $\frac{{ϵ}_{0}A}{(d+t)}$

Q. A parallel plate capacitor has the space between its plates filled by two slabs of thickness $d/2$ each with dielectric constants $k_1$ and $k_2$. Given that $d$ is the plate separation of the capacitor, the effective capacitance is

1. $\frac{2{ϵ}_{0}A}{d}\frac{k_1{k}_2}{k_1+k_2}$
2. $\frac{{ϵ}_{0}A}{d}\frac{k_1{k}_2}{k_1-k_2}$
3. $\frac{{ϵ}_{0}A(k_1-k_2)}{d}$
4. $\frac{{ϵ}_{0}A(k_1+k_2)}{2d}$

Q.

How do you solve a capacitor problem?

Q.

The separation between the plates of a parallel-plate capacitor is 0.500 cm and its plate area is $100c{m}^2.$ A 0.400 cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.

Q. The advantages of using dielectrics are :

1. Increases capacitance.
2. Increased stability.
3. All of these.
4. Higher dielectric strength.

Q.

Between the plates of parallel plate capacitor of capacitance C, two parallel plates, of the same material and area same as per the plate of the original capacitor, are placed. If the thickness of these plates is equal to ${\frac{1}{5}}^{th}$ of the distance between the plates of the original capacitor, then the capacitance of the new capacitor is

1. $\frac{10}{3}$ C

2. $\frac{1}{5}$ C
   

3. $\frac{3}{5}$ C
   

4. $\frac{3}{10}$ C
   

Q.

Four resistors are connected in series such that their total resistance is $50\Omega$. if the resistance of three resistors is $5\Omega$, $10\Omega$ and $15\Omega$ respectively, find the resistance of the fourth resistor.

Q. A parallel plate capacitor of $20\mu \text{F}$ is connected to a battery of emf $8.85\text{V}$. If the plate area of each plate is $2{\text{cm}}^2$, with what magnitude of force the positively charged plate attracts the negative one?

1. $4.425\times {10}^6\text{N}$
2. $17.7\times {10}^6\text{N}$
3. $8.85\times {10}^6\text{N}$
4. $2.21\times {10}^6\text{N}$

Q. If a battery is placed between $\text{A}$ and $\text{B}$, with what capacitor you can replace the combination of capacitors?

1. $10/3\mu \text{F}$
2. $26\mu \text{F}$
3. $8\mu \text{F}$
4. $6\mu \text{F}$

Q.

Find the co-ordinates of center of mass of the lamina shown in the figure below

1. $(0.75m,1.75m)$

2. $(0.75m,0.75m)$

3. $(1.25m,1.5m)$

4. $(1m,1.75m)$

Q. The equivalent capacitance between $\text{A}$ and $\text{B}$ is

1. $2\mu \text{F}$
2. $3\mu \text{F}$
3. $5\mu \text{F}$
4. $0.5\mu \text{F}$

Q. An air capacitor of capacity $10\mu \text{F}$ is connected to a battery of $12\text{V}$. Now the space between the plates is filled with a liquid of dielectric constant $5$. The additional charge that flows to the capacitor is

1. $120\mu \text{C}$
2. $600\mu \text{C}$
3. $480\mu \text{C}$
4. $24\mu \text{C}$

Q.

What is dielectric brake in topic atmospheric electricity

Q. How does inserting a dielectric between plates increase the capacitor’s stability?

1. The attractive force on plates makes the capacitor unstable. Dielectric gives support for the plates.
2. The dielectrics have higher relative permittivity than air.
3. The dielectric gives more voltage capacity.
4. None of these

Q. A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will
(a) increase
(b) decrease
(c) remain unchanged
(d) become zero.

Q. Two inductance coils, of same self inductance, are corrected in parallel and the distance between them is large. The resultant self inductance of the coil will be

1. $\frac{L}{4}$
2. $L$
3. $\frac{L}{2}$
4. $2L$

Q.

Between the plates of parallel plate capacitor of capacitance C, two parallel plates, of the same material and area same as per the plate of the original capacitor, are placed. If the thickness of these plates is equal to ${\frac{1}{5}}^{th}$ of the distance between the plates of the original capacitor, then the capacitance of the new capacitor is

1. $\frac{1}{5}$ C
   

2. $\frac{3}{5}$ C
   

3. $\frac{3}{10}$ C
   

4. $\frac{10}{3}$ C

Q.

An electric oven of resistance $125\Omega$, a toaster of resistance $50\Omega$ and an electric geyser of $250\Omega$ are connected in parallel to a $220V$ source. What is the resistance of an electric iron connected to the same source that takes as much as all the three appliances and what is the current through it?

Q.

The magnetic susceptibility of a rod is $499$. The absolute permeability of vacuum is $4\pi \times {10}^{-7}H/m$. The absolute permeability of the material of the rod is

1. $2\pi \times {10}^{-4}H/m$
2. $3\pi \times {10}^{-4}H/m$
3. $4\pi \times {10}^{-4}H/m$
4. $\pi \times {10}^{-4}H/m$

Q. A parallel plate capacitor has area $20c{m}^2$ and separation between the plates is $0.1mm$. The dielectric break down strength is $3\times {10}^6$$v/m$. The maximum r.m.s voltage which can be safely applied is :

1. $210V$
2. $300V$
3. $100V$
4. $200$

Q.

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

1. C
2. $\frac{C}{2}$
3. zero
4. 2C

Q. A parallel plate capacitor has $91$ plates, all are identical and arranged with same spacing between them. If the capacitance between adjacent plates is $3pF$. What will be the resultant capacitance?

1. $273pF$
2. $94pF$
3. $30pF$
4. $270pF$

Q. A magnetic wire of dipole moment $4\pi$ $A{m}^2$ is bent in the form of semi-circle. The new magnetic moment is?

1. $4\pi$ $A{m}^2$
2. $8\pi$ $A{m}^2$
3. $4$ $A{m}^2$
4. $8A{m}^2$

Q. A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d=5cm. The coil is connected to a battery producing a current of 4.0 A in the wire. (a) What is the magnitude of the magnetic dipole moment of this device? (b) At what axial distance $z>>d$ will the magnetic field have the magnitude $5.0\mu T$ (approximately one-tenth that ofEarth’s magnetic field)?

Q. Capacitance depends on the

1. Charge on capacitor
2. Potential difference across capacitor
3. Dielectric constant of the substance
4. Area of plates

Q.

A capacitor is made of a flat plate of area A and second plate having a stair-like structure as shown in figure. The width of each plate is a and the height is b. The capacitance of the capacitor is :

1. $\frac{2A{ϵ}_0}{3(d+b)}$
2. $\frac{A{ϵ}_0(3d^2+6bd+2b^2)}{3d(b+d)(d+2b))}$
3. $\frac{2A{ϵ}_0(d^2+2bd+b^2)}{3d(d+b)(d+2b))}$
4. $\frac{A{ϵ}_0(d^2+2bd+b^2)}{3d(d+b)(d+2b))}$

Q.

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :

1. E, C
2. 6E, 6C
3. $\frac{E}{6},6C$
4. E, 6C