Case study (4 Marks)

Question 14 Marks

The mean free path of electrons in a gas in a discharge tube is inversely proportional to the pressure inside it. The Crookes dark space occupies half the length of the discharge tube when the pressure is 0.02mm of mercury. Estimate the pressure at which the dark space will fill the whole tube.

Answer

Let the mean free path be ‘L’ and pressure be ‘P’
$\text{L}\propto\frac{1}{\text{p}}$ for L = half of the tube length, P = 0.02mm of Hg

As ‘P’ becomes half, ‘L’ doubles, that is the whole tube is filled with Crook’s dark space.
Hence the required pressure $=\frac{0.02}{2}=0.01\text{m of Hg}$

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Question 24 Marks

Consider the situation shown in figure. The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf $\in.$ All surfaces are frictionless. Calculate the value of M for which the dielectric slab will stay in equilibrium.

Answer

We knows
In this particular case the electricfield attracts the dielectric into the capacitor with a force $\frac{\in_0\text{bV}^2(\text{k}-1)}{2\text{d}}$
Where, b = Width of plates
k = Dielectric constant
d = Separation between plates
V = E = Potential difference.
Hence in this case the surfaces are frictionless, this force is counteracted by the weight.
So, $\frac{\in_0\text{bE}^2(\text{k}-1)}{2\text{d}}=\text{Mg}$
$\Rightarrow\text{M}=\frac{\in_0\text{bE}^2(\text{k}-1)}{2\text{dg}}$

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Question 34 Marks

Potential difference $(\triangle\text{V})$ between two points A and B separated by a distance x, in a uniform electric field E is given by $\triangle\text{V}=-\text{Ex},$ where xis measured parallel to the field lines. If a charge $q_0$ moves from P to Q, the change in potential energy $(\triangle\text{U})$ is given as $\triangle\text{U}=-\text{q}_0\triangle\text{V}.$ A proton is released from rest in uniform electric field of magnitude $4.0 \times 10^8Vm^{-1}$ directed along the positive X-axis. The proton undergoes a displacement of 0.25m in the direction of E. Mass of a proton $= 1.66 \times 10^{-27}kg$ and charge of proton $= 1.6 \times 10^{-19}C.$

The change in electric potential of the proton between the points A and B is:

$-1 \times 10^8V$
$1 \times 10^8V$
$6.4 \times 10^{-19}V$
$-6.4 \times 10^{-19}V$

The change in electric potential energy of the proton for displacement from A to B is:

$1.6 \times 10^{11}J$
$0.5 \times 10^{23}J$
$-1.6 \times 10^{-11}J$
$3.2 \times 10^{22}J$

The mutual electrostatic potential energy between two protons which are at a distance of $9 \times 10^{-15}$m, in $_{92}U^{235}$ nucleus is:

$1.56 \times 10^{-14}J$
$5.5 \times 10^{-14}J$
$2.56 \times 10^{-14}J$
$4.56 \times 10^{-14}J$

If a system consistsoftwocharges 4mC and -3mC with no external field placed at (-5cm, 0, 0) and (5cm, 0, 0) respectively. The amount of work required to separate the two charges infinitely away from each other is:

-1.1J
2J
2.5J
3J

As the proton moves from P to Q, then:

The potential energy of proton decreases.
The potential energy of proton increases.
The proton loses kinetic energy.
Total energy of the proton increases.

Answer

(a) $ -1 \times 10^8V$

Explanation:
As $\triangle\text{V}=-\text{E}\triangle\text{x}=-(4.0\times10^8\frac{\text{V}}{\text{m}})(0.25\text{m})$
$=-10^8\text{V}.$

Explanation:
As $\triangle\text{U}=\text{q}_0\triangle\text{V}$
$=-(1.6\times10^{-19})\times(-1.0\times10^8\text{V})$
$=-1.6\times10^{-11}\text{V}.$

Explanation:
Here, $q_1 = q_2 = 1.6 \times 10^{-19}C, r = 9 \times 10^{-15}m$
$\text{U}=\frac{9\times10^9\times1.6\times10^{-19}\times1.6\times10^{-19}}{9\times10^{-15}}$
$=2.56\times10^{-14}\text{J}.$

(a) -1.1J

Explanation:
Here, $\text{q}_1=4\mu\text{C},\text{ q}_2=3\mu\text{C}$
r = 10cm = 0.1m
Electrostatic potential energy,
$\text{U}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}}$
$=9\times10^9\times\frac{4\times10^{-6}\times(-3)\times10^{-6}}{0.1}$
= -1.1J

(a) The potential energy of proton decreases.

Explanation:
As proton moves in the direction of the electric field, then its potential energy decreases.

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Question 44 Marks

A dielectric slab is a substance which does not allow the flow of charges through it but permits them to exert electrostatic forces on one another. When a dielectric slab is placed between the plates, the field $E_0$ polarises the dielectric. This induces charge $-Q_P$ on the upper surface and $+ Q_P$ on the lower surface of the dielectric. These induced charges set up a field $E_P$ inside the dielectric in the opposite direction of $\vec{\text{E}_0}$ as shown.

ln a parallel plate capacitor, the capacitance increases from $4\mu\text{F}$ to $80\mu\text{F},$ on introducing a dielectric medium between the plates. What is the dielectric constant of the medium?

A parallel plate capacitor with air between the plates has a capacitance of 8pE The separation between the plates is now reduced half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in second case.

8pF
10pF
80pF
100pF

A dielectric introduced between the plates of a parallel plate condenser:

Decreases the electric field between the plates.
Increases the capacity of the condenser.
Increases the charge stored in the condenser.
Increases the capacity of the condenser.

A parallel plate capacitor of capacitance 1 pF has separation between the plates is d. When the distance of separation becomes 2d and wax of dielectric constant x is inserted in it the capacitance becomes 2pE What is the value of x?

A parallel plate capacitor having area A and separated by distanced is filled by copper plate of thickness b. The new capacity is:

$\frac{\in_0\text{A}}{\text{d}+\frac{\text{b}}{2}}$
$\frac{\in_0\text{A}}{\text{2d}}$
$\frac{\in_0\text{A}}{\text{d-b}}$
$\frac{2\in_0\text{A}}{\text{d}+\frac{\text{b}}{2}}$

Answer

(b) 20

Explanation:
$\text{k}=\frac{\text{Capacitance with dielectric }}{\text{Capacitance without dielectric }}$
$=\frac{80\mu\text{F}}{4\mu\text{F}}=20$

Explanation:
Capacitance of the capacitor with air between plates
$\text{C'}=\frac{\in_0\text{A}}{\text{d}}=8\text{pF}$
With the capacitor is filled with dielectric (k = 5) between its plates and the distance between the plates is reduced by half, capacitance become,
$\text{C'}=\frac{\in_0\text{A}}{\frac{\text{d}}{2}}=\frac{\in_0\times5\times\text{A}}{\frac{\text{d}}{2}}$
$=\text{10C'}=10\times8=80\text{pF}.$

(d) Increases the capacity of the condenser.

Explanation:
If a dielectric medium of dielectric constant K is filled completely between the plates then capacitance increases by K times.

(b) 4

Explanation:
$\text{C'}=\frac{\in_0\text{A}}{\text{d}}=1\text{pF}$ (i)
$\text{C'}=\frac{\text{x}\in_0\text{A}}{\text{(2d})}=2\text{pF}$ (ii)
Divide (ii) by (i), $\frac{\text{x}}{2}=\frac{2}{1}$
⇒ x = 4

Explanation:
As capacitance, $\text{C}_0=\frac{\in_0\text{A}}{\text{d}}$
$\therefore$ After inserting copper plate, $\text{C}=\frac{\in_0\text{A}}{\text{d-b}}$

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Question 54 Marks

A capacitor is a device to store energy. The process of charging up a capacitor involves the transferring of electric charges from its one place to another. This work done in charging the capacitor is stored as its electrical potential energy.

If q is the charge and V is the potential difference across a capacitor at any instant during its charging, then small work done in storing an additional small charge dq against the repulsion of charge q already stored on it is $\text{dW}=\text{V.dq}=(\frac{\text{q}}{\text{C}})\text{dq}.$

A system of 2 capacitors of capacitance $2\mu\text{F}$ and $4\mu\text{F}$ is connected in series across a potential difference of 6 V. The energy stored in the system is:

$3\mu\text{J}$
$24\mu\text{J}$
$30\mu\text{J}$
$108\mu\text{J}$

A capacitor of capacitance of $10\mu\text{F}$ is charged to 10V. The energy stored in it is:

$100\mu\text{J}$
$500\mu\text{J}$
$1000\mu\text{J}$
$1\mu\text{J}$

A parallel plate air capacitor has capacity C farad, potential V volt and energy E joule. When the gap between the plates is completely filled with dielectric:

Both V and E increase.
Both V and E decrease.
V decreases, E increases.
V increases, E decreases.

A capacitor with capacitance $5\mu\text{F}$ s charged to $5\mu\text{C}.$ If the plates are pulled apart to reduce the capacitance to $2\mu\text{F},$ how much work is done?

$6.25 \times 10^{-6}J$
$3.75 \times 10^{-6}J$
$2.16 \times 10^{-6}J$
$2.55 \times 10^{-6}J$

A metallic sphere ofradius 18cm has been given a charge of $5 \times 10^{-6}C$. The energy of the charged conductor is:

0.2J
0.6J
1.2J
2.4J

Answer

(b) $24\mu\text{J}$

Explanation:
As, $\text{C}_1=2\mu\text{F},\text{C}_2=4\mu\text{F}$
ln series combination, the equivalent capacitance will be,
$\text{C}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\Big(\frac{2\times4}{2+4}\Big)\mu\text{F}$
$=\frac{4}{3}\mu\text{F}$
Potential difference applied, V = 6V
Energy stored in the system, $\text{U}=\frac{1}{2}\text{CV}^2$
$=\frac{1}{2}\times\frac{4}{3}\times10^{-6}\times(6)^2$
$\text{J}=24\mu\text{J}.$

(b) $500\mu\text{J}$

Explanation:
The energy stored in a capacitor is
$\text{U}=\frac{1}{2}\text{CV}^2$
$=\frac{1}{2}\times(10\times10^{-6})(10)^2=500\mu\text{J}.$

(b) Both V and E decrease.

Explanation:
When the gap between the plates is completely filled with dielectric of dielectric constant K, then potential is:
$\text{V}=\frac{\text{Qd}}{\text{A}\in_0\text{K}}$ (i)
and electric field is
$\text{E}=\frac{\text{Q}}{\text{A}\in_0\text{K}}$ (ii)
From equations (i) and (ii), both electric field and potential decrease.

(b) $3.75 \times 10^{-6}J$

Explanation:
Work done $=\text{U}_\text{f}-\text{U}_\text{i}=\frac{1}{2}\frac{\text{q}^2}{\text{C}_\text{f}}-\frac{1}{2}\frac{\text{q}^2}{\text{C}_\text{i}}$
$=\frac{\text{q}^2}{2}\Big[\frac{1}{\text{C}_\text{f}}-\frac{1}{\text{C}_\text{i}}\Big]$
$=\frac{(5\times10^{-6})^2}{2}\Big[\frac{1}{2\times10^{-6}}-\frac{1}{5\times10^{-6}}\Big]$
$= 3.75 \times 10^{-6}J$

(b) 0.6J

Explanation:
Here $r = 18cm = 18 \times 10^{-2}m, q = 5 \times 10^{-6}C$
As, $\text{C}=4\pi\in_0\text{r}=\frac{18\times10^{-2}}{9\times10^9}=2\times10^{-11}\text{F}$
Energy of charged conductor is
$\text{U}=\frac{\text{q}^2}{\text{2C}}=\frac{(5\times10^{-6})^2\text{C}}{2\times2\times10^{-11}\text{F}}=0.625\text{J}.$

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Question 64 Marks

When an insulator is placed in an external field, the dipoles become aligned. Induced surface charges on the insulator establish a polarization field $\vec{\text{E}}_\text{i}$ in its interior. The net field $\vec{\text{E}}$ in the insulator is the vector sum of $\vec{\text{E}}_0$ and $\vec{\text{E}}_\text{i}$ as shown in the figure.

On the application of external electric field, the effect of aligning the electric dipoles in the insulator is called polarisation, and the field $\vec{\text{E}}_\text{i}$ is known as the polarisation field. The dipole moment per unit volume of the dielectric is known as polarisation $(\vec{\text{P}}).$ For linear isotropic dielectrics, $\vec{\text{P}}=\chi\vec{\text{E}},$ where $\chi=$ electrical susceptibility of the dielectric medium.

Which among the following is an example of polar molecule?

$O_2$
$H_2$
$N_2$
$HCI$

When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance:

Increases K times.
Remains unchanged.
Decreases K times.
Increases 2K times.

Which of the following is a dielectric?

Copper.
Glass.
Antimony (Sb).
None of these.

For a polar molecule, which of the following statements is true?

The centre of gravity of electrons and protons coincide.
The centre of gravity of electrons and protons do not coincide.
The charge distribution is always symmetrical.
The dipole moment is always zero.

When a comb rubbed with dry hair attracts pieces of paper. This is because the?

Comb polarizes the piece of paper.
Comb induces a net dipole moment opposite to the direction of field.
Electric field due to the comb is uniform.
Comb induces a net dipole moment perpendicular to the direction of field.

Answer

(d) HCI

Explanation:
A polar molecule is one in which the centre of gravity for positive and negative charges are separated.
(a) Comb polarizes the piece of paper.
Explanation:
As $\text{F}_\text{m}=\frac{\text{F}_0}{\text{K}}$
$\therefore$ The maximum force decreases by Klimes.
(b) Glass.
(b) The centre of gravity of electrons and protons do not coincide.
Explanation:
ln polar molecule the centres of positive and negative charges are separated even when there is no external field. Such molecule have a permanent dipole moment. Ionic molecule like HCI is an example of polar molecule.
(c) Decreases K times.

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Question 74 Marks

This energy possessed by a system of charges by virtue of their positions. When two like charges lie infinite distance apart, their potential energy is zero because no work has to be done in moving one charge at infinite distance from the other.
In carrying a charge q from point A to point B, work done $W = q(V_A - V_B)$. This work may appear as change in $\frac{\text{KE}}{\text{PE}}$ of the charge. The potential energy of two charges $q_1$ and $q_2$ at a distance r in air is $\frac{\text{q}_1\text{q}_2}{1\pi\in_0\text{r}}.$ It is measured in joule. It may be positive, negative or zero depending on the signs of $q_1$ and $q_2.$

Calculate work done in separating two electrons form a distance of 1m to 2m in air, where e is electric charge and k is electrostatic force constant.

$ke^2$
$\frac{\text{e}^2}{2}$
$-\frac{\text{ke}^2}{2}$
Zero

Four equal charges q each are placed at four corners of a square of side a each. Work done in carrying a charge -q from its centre to infinity is:

$\text{Zero}$
$\frac{\sqrt{2}\text{q}^2}{\pi\in_0\text{a}}$
$\frac{\sqrt{2}\text{q}}{\pi\in_0\text{a}}$
$\frac{\text{q}^2}{\pi\in_0\text{a}}$

Two points A and Bare located in diametrically opposite directions of a point charge of $+2\mu\text{C}$ at distances 2m and 1m respectively from it. The potential difference between A and B is:

$3 \times 10^3V$
$6 \times 10^4V$
$-9 \times 10^3V$
$-3 \times 10^3V$

Two point charges A = + 3nC and B = + 1nC are placed 5cm apart in air. The work done to move charge B towards A by 1cm is:

$2.0 \times 10^{-7}J$
$1.35 \times 10^{-7}J$
$2.7 \times 10^{-7}J$
$12.1 \times 10^{-7}J$

A charge Q is placed at the origin. The electric potential due to this charge at a given point in space is V. The work done by an external force in bringing another charge q from infinity up to the point is:

$\frac{\text{V}}{\text{q}}$
Vq
V + q
V

Answer

Explanation:

$W = (P.E.)_{final} - (P.E.)_{initial}$

$=\frac{\text{ke}^2}{2}-\frac{\text{ke}^2}{1}=\frac{\text{-ke}^2}{2}$

(b) $\frac{\sqrt{2}\text{q}^2}{\pi\in_0\text{a}}$

Explanation:

Potential at the centre of the square due to four equal charges q at four corners,

$\text{V}=\frac{\text{4q}}{\frac{4\pi\in_0\text{(a}\sqrt{2})}{2}}=\frac{\sqrt{2}\text{q}}{\pi\in_0\text{a}}$

$\text{W}_{0\rightarrow\infty}=-\text{W}_{0\rightarrow\infty}=-(\text{-q})\text{V}$

$=\frac{\sqrt{2}\text{q}^2}{\pi\in_0\text{a}}$

Explanation:

Here, $\text{q}=2\mu\text{C}=2\times10^{-6}\text{C},\text{r}_{\text{A}}=2\text{m},\text{r}_{\text{B}}=1\text{m}.$

$\therefore\text{V}_\text{A}-\text{V}_\text{B}=\frac{\text{q}}{4\pi\in_0}\Big[\frac{1}{\text{r}_\text{A}}-\frac{1}{\text{r}_\text{B}}\Big]$

$=2\times10^{-6}\times9\times10^9\Big[\frac{1}{2}-\frac{1}{1}\Big]$

$\text{V}=-9\times10^3\text{V}.$

(b) $1.35 \times 10^{-7}J$

Explanation:

Required work done = Change in potential energy of the system,

$\text{W}=\text{U}_\text{f}-\text{U}_\text{i}=\text{k}\frac{\text{q}_1\text{q}_2}{\text{r}_\text{f}}-\text{k}\frac{\text{q}_1\text{q}_2}{\text{r}_\text{i}}$

$=\text{k }\text{q}_1\text{q}_2\Big[\frac{1}{\text{r}_\text{f}}-\frac{1}{\text{r}_\text{i}}\Big]$

$\therefore\text{W}=(9\times10^9)(3\times10^{-9}\times1\times10^{-9})$

$\times\Big[\frac{1}{4\times10^{-2}}-\frac{1}{5\times10^{-2}}\Big]$

$= 27 \times 10^{-7} \times (0.05) = 1.35 \times 10^{-7}J.$

(b) Vq

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Question 84 Marks

Calculate $\frac{\text{n(T)}}{\text{n}(1000\text{K})}$ for tungsten emitter at T = 300K, 2000K and 3000K, where n(T) represents the number of thermions emitted per second by the surface at temperature T. Work function of tungsten is 4.52eV.

Answer

$\text{i}=\text{ne or n}=\frac{\text{i}}{\text{e}}$
‘e’ is same in all cases.
We know,
$\text{i}=\text{AST}^2\text{e}^{\frac{-\phi}{\text{RT}}}\ \phi=4.52\text{eV},\ \text{K}=1.38\times10^{-23}\text{J/k}$
$\text{n(1000)}=\text{As}\times(1000)^2\times\text{e}^{-4.52\times1.6\times10^{-19}/1.38\times10^{-23}\times1000}$
$1.7396\times10^{-17}$

$\text{T}=300\text{K}$

$\frac{\text{n(T)}}{\text{n(1000K)}}=\frac{\text{AS}\times(300)^2\times\text{e}^{-4.52\times1.6\times10^{-19}/1.38\times10^{-23}\times300}}{\text{AS}\times1.7396\times10^{-17}}=7.05\times10^{-55}$

$\text{T}=2000\text{K}$

$$$\frac{\text{n(T)}}{\text{n(1000K)}}=\frac{\text{AS}(2000)^2\times\text{e}^{-4.52\times1.6\times10^{-19}/1.38\times10^{-23\times2000}}}{\text{AS}\times1.7396\times10^{-17}}=9.59\times10^{11}$

$\text{T}=3000\text{K}$

$\frac{\text{n(T)}}{\text{n(1000K)}}=\frac{\text{AS}\times(3000)^2\times\text{e}^{-4.52\times1.6\times10^{-19}/1.38\times10^{-23}\times3000}}{\text{AS}\times1.7396\times10^{-17}}=1.340\times10^{16}$

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Question 94 Marks

For the various charge systems, we represent equipotential surfaces by curves and line of force by full line curves. Between any two adjacent equipotential surfaces, we assume a constant potential difference the equipotential surfaces of a single point charge are concentric spherical shells with their centres at the point charge. As the tines of force point radially outwards, so they are perpendicular to the equipotential surfaces at all points.

Identify the wrong statement.

Equipotential surface due to a single point charge is spherical.
Equipotential surface can be constructed for dipoles too.
The electric field is normal to the equipotential surface through the point.
The work done to move a test charge on the equipotential surface is positive.

Nature of equipotential surface for a point charge is:

Ellipsoid with charge at foci.
Sphere with charge at the centre of the sphere.
Sphere with charge on the surface of the sphere.
Plane with charge on the surface.

A spherical equipotential surface is not possible:

Inside a uniformly charged sphere.
For a dipole.
Inside a spherical condenser.
For a point charge.

The work done in carrying a charge q once round a circle of radius a with a charge Q at its centre is:

$\frac{\text{qQ}}{4\pi\in_0\text{a}}$
$\frac{\text{qQ}}{4\pi\in_0\text{a}^2}$
$\frac{\text{q}}{4\pi\in_0\text{a}}$
$\text{Zero}$

The work done to move a unit charge along an equipotential surface from P to Q:

Must be defined as $-\int\limits_{\text{P}}^{\text{Q}}\vec{\text{E}}\cdot\text{d}\vec{\text{l}}.$
Is zero.
Can have a non-zero value.
Both (a) and (b) are correct.

Answer

(d) The work done to move a test charge on the equipotential surface is positive.
(b) Sphere with charge at the centre of the sphere.
(b) For a dipole.
(d) $\text{Zero}$

Explanation:

The electrical potential at any point on circle of radius a due to charge Q at its centre is $\text{V}=\frac{1}{4\pi\in_0}\frac{\text{Q}}{\text{a}}$

It is an equipotential surface.

Hence, work done in carrying a charge q round the circle is zero.

(d) Both (a) and (b) are correct.

Explanation:

Work done to move a unit charge along an equipotential surface from P to Q,

$\text{W}=-\int\limits_{\text{P}}^{\text{Q}}\vec{\text{E}}\cdot\text{d}\vec{\text{l}}$

On equipotential surface $\vec{\text{E}}\perp\text{d}\vec{\text{l}}$

$\text{W}=-\int\limits_{\text{P}}^{\text{Q}}\text{E}(\text{dl})\cos90^\circ=0$

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Question 104 Marks

A tungsten cathode and a thoriated-tungsten cathode have the same geometric dimensions and are operated at the same temperature. The thoriated-tungsten cathode gives 5000 times more current than the other cathode. Find the operating temperature. Take relevant data from the previous problem.

Answer

Pure tungsten:
$\phi=4.5\text{eV}$
$\text{A}=60\times10^{4}\text{A/m}^2-\text{k}^2$
$\text{i}=\text{AST}^2\text{e}^{\frac{\phi}{\text{KT}}}$
Thoriated tungsten:
$\phi=2.6\text{eV}$
$\text{A}=3\times10^{4}\text{A/m}^2-\text{k}^2$
$\text{i}_{\text{Thoriated Tungsten}}=5000\text{i}_{\text{Tungsten}}$
So, $5000\times\text{S}\times60\times10^{4}\times\text{T}_2\times\text{e}^{\frac{-4.5\times1.6\times10^{-19}}{1.38\times\text{t}\times10^{-23}}}$
$\Rightarrow\text{S}\times3\times10^{4}\times\text{T}^2\times\text{e}^{\frac{-2.56\times1.6\times10^{-19}}{1.38\times\text{T}\times10^{-23}}}$
$\Rightarrow3\times10^{8}\times\text{e}^{\frac{-4.5\times1.6\times10^{-19}}{1.38\times\text{t}\times10^{-23}}}=\text{e}^{\frac{-2.56\times1.6\times10^{-19}}{1.38\times\text{T}\times10^{-23}}}\times3\times10^4$
Taking 'in'
$\Rightarrow9.21\text{T}=220.29$
$\Rightarrow\text{T}=\frac{22029}{9.21}=2391.856\text{K}$

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Question 114 Marks

Electrostatic potential energy of a system of point charges is defined as the total amount of work done in bringing the different charges to their respective positions from infinitely charge mutual separations. The work is stored in the system of two point charges in the form of electrostatic potential energy U of the system. Electric potential difference between any points A and B in an electric field is the amount of work done in moving a unit positive test charge from A to B along any path agents the electrostatic force $\text{V}_\text{B}-\text{V}_\text{A}=\frac{\text{W}_\text{AB}}{\text{q}_\text{o}}=\int\vec{\text{E}}.\text{dl}.$

A test charge is moved from lower potential point to a higher potential point. The potential energy of test charge wiII.

Remain the same.
Increase.
Decrease.
Become zero.

Which of the following statement is not true?

Electrostatic force is a conservative force.
Potential energy of charge q at a point is the work done per unit charge in bringing a charge from any point to infinity.
Spring force and gravitational force are conservative force.
Both (a) and (c).

Work done in moving a charge from one point to another inside a uniformly charged conducting sphere is:

Always zero.
Non-zero.
May be zero.
None of these.

The work done in bringing a unit positive charge from infinite distance to a point at distance x from a positive charge Q is W. Then the potential $\phi$ at that point is:

$\frac{\text{WQ}}{\text{x}}$
W
$\frac{\text{W}}{\text{x}}$
WQ

If $1\mu\text{C}$ charge is shifted from A to B and it is found that work done by an external force is $40\mu\text{J}.$ ln doing so against electrostatics force, the potential difference $V_A - V_B$ is:

40V
-40V
20V
-60V

Answer

(c) Decrease.
(b) Potential energy of charge q at a point is the work done per unit charge in bringing a charge from any point to infinity.
(a) Always zero.

Explanation:

Since, E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface.

(b) W

Explanation:

The work done in bringing unit positive charge from infinity to a point which is at a distance x from the positive charge Q is defined as the potential at the given point due to the charge Q. Therefore $\phi=\text{W}.$

(b) -40V

Explanation:

$\text{W}_\text{ext}=\text{q}_0\triangle\text{V}$

$(\text{W}_{\text{AB}})_{\text{ext}}=\text{q}(\text{V}_\text{B}-\text{V}_\text{A})$

$40\mu\text{J}=1\mu\text{C}(\text{V}_\text{B}-\text{V}_\text{A})$

$\text{V}_\text{A}-\text{V}_\text{B}=-40\text{V}.$

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Question 124 Marks

The potential at any observation point P of a static electric field is defined as the work done by the external agent (or negative of work done by electrostatic field) in slowly bringing a unit positive point charge from infinity to the observation point. Figure shows the potential variation along the line of charges. Two point charges $Q_1$ and $Q_2$ lie along a line at a distance from each other.

At which of the points 1, 2, and 3 is the electric field is zero?

1
2
3
Both (a) and (b)

The signs of charges $Q_1$ and $Q_2$ respectively are:

Positive and negative.
Negative and positive.
Positive and positive.
Negative and negative.

Which of the two charges $Q_1$ and $Q_2$ is greater in magnitude?

$Q_2$
$Q_1$
Same.
Can't determined.

Which of the following statement is not true?

Electrostatic force is a conservative force.
Potential energy of charge q at a point is the work done per unit charge in bringing a charge from any point to infinity.
When two like charges lie infinite distance apart, their potential energy is zero.
Both (a) and (c).

Positive and negative point charges of equal magnitude are kept at $\Big(0,0,\frac{\text{a}}{2}\Big)$ and $\Big(0,0,\frac{\text{-a}}{2}\Big)$ respectively.

The work done by the electric field when another positive point charge is moved from (-a, 0, 0) to (0, a, 0) is to:

Positive.
Negative.
Zero.
Depends on the path connecting the initial and final positions.

Answer

Explanation:

As $\frac{\text{-dV}}{\text{dr}}=\text{E}_\text{r},$ the negative of the slope of V versus r curve represents the component of electric field along r. Slope of curve is zero only at point 3. Therefore, the electric field vector is zero at point 3.

(a) Positive and negative.

Explanation:

Near positive charge, net potential is positive and near a negative charge, net potential is negative. Thus, charge $Q_1$ is positive and $Q_2$ is negative.

(b) $Q_1$

Explanation:

From the figure, it can be seen that net potential due to two charges is positive everywhere in the region left to charge $Q_1$. Therefore the magnitude of potential due to charge $Q_1$ is greater than due to $Q_2.$

(b) Potential energy of charge q at a point is the work done per unit charge in bringing a charge from any point to infinity.
(c) Zero.

Explanation:

It can be seen that potential at the points both A and B are zero. When the charge is moved from A to B, work done by the electric field on the charge will be zero.

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Question 134 Marks

The electrical capacitance of a conductor is the measure of its ability to hold electric charge. An isolated spherical conductor of radius R. The charge Q is uniformly distributed over its entire surface. It can be assumed to be concentrated at the centre of the sphere. The potential at any point on the surface of the spherical conductor will be $\text{V}=\frac{1}{4\pi\in_0}\frac{\text{Q}}{\text{R}}.$

Capacitance of the spherical conductor situated in vacuum is $\text{C}=\frac{\text{Q}}{\text{V}}=\frac{\text{Q}}{\frac{1}{4\pi\in_0}.\frac{\text{Q}}{\text{R}}}$ or $\text{C}=4\pi\in_0\text{R}$ Clearly, the capacitance of a spherical conductor is proportional to its radius. The radius of the spherical conductor of 1F capacitance is $\text{R}=\frac{1}{4\pi\in_0}.$ C and this radius is about 1500 times the radius of the earth $(\sim6\times10^3\text{km}).$

If an isolated sphere has a capacitance 50pE Then radius is:

90cm
45cm
45m
90m

How much charge should be placed on a capacitance of 25 pF to raise its potential to 105V?

$1\mu\text{C}$
$1.5\mu\text{C}$
$2\mu\text{C}$
$2.5\mu\text{C}$

Dimensions of capacitance is:

$[M L^{-2} T^4 A^2]$
$[M^{-1} L^{-1} T^3 A^1]$
$[M^-L^{-2} T^4 A^2]$
$[M^0 L^{-2} T^4 A^1]$

Metallic sphere of radius R is charged to potential V. Then charge q is proportional to:

V
R
Both V and R
None of these

If 64 identical spheres of charge q and capacitance C each are combined to form a large sphere. The charge and capacitance of the large sphere is:

64q, C
16q, 4C
64q, 4C
16q, 64C

Answer

(b) 45cm

Explanation:
Here $C = 50pF = 50 \times 10^{-12}F, V = 10^4V$
$\text{R}=\frac{1}{4\pi\in_0}\cdot$
$C = 9 \times 109 mF^{-1} \times 50 \times 10^{-12}F$
$= 45 \times 10-2m = 45cm$

(d) $2.5\mu\text{C}$

Explanation:
As $q = CV = 25 \times 10^{-12}\times 10^5$
$=2.5\mu\text{C}$

Explanation:
As charge, $\text{q}=\text{CV}=(4\pi\in_0\text{R})\text{V}$
$\therefore$ q depends on both V and R.

Explanation:
64 drops have formed a single drop of radius R. Volume of large sphere= 64 × Volume of small sphere,
$\therefore\frac{4}{3}\pi\text{R}^3=64\frac{4}{3}\pi\text{r}^3$
$\Rightarrow R = 4r$ and $Q_{total}= 64q$
$\text{C'}=4\pi\in_0\text{R}\Rightarrow\text{C'}=(4\pi\in_0)\text{4r}$
$\Rightarrow\text{C'}=\text{4C}.$

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Question 144 Marks

The simplest and the most widely used capacitor is the parallel plate capacitor. It consists of two large plane parallel conducting plates, separated by a small distance. In the outer regions above the upper plate and below the lower plate, the electric fields due to the two charged plates cancel out. The net field is zero. In the inner region between the two capacitor plates, the electric fields due to the two charged plates add up. The net field is $\frac{\sigma}{\in_0}.$

For a uniform electric field, potential difference between the plates = Electric field x distance between the plates. Capacitance of the parallel plate capacitor is the charge required to supplied to either of the conductors of the capacitor so as to increase the potential difference between then by unit amount.

A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are:

Increases, decreases, decreases.
Constant, increases, decreases.
Constant, decreases, decreases.
Constant, decreases, increases.

In a parallel plate capacitor, the capacity increases if:

Area of the plate is decreases.
Distance between the plates increases.
Area of the plate is increases.
Dielectric constant decreases.

A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plates are $+\sigma$ and $-\sigma$ respectively. In the region between the plates the magnitude of the electric field is:

$\frac{\sigma}{2\in_0}$
$\frac{\sigma}{\in_0}$
0
None of these.

If a parallel plate air capacitor consists of two circular plates of diameter 8cm. At what distance should the plates be held so as to have the same capacitance as that of sphere of diameter 20cm?

If a charge of $+ 2.0 \times 10^{-8}C$ is placed on the positive plate and a charge of $- 1.0 \times 10^{-8}C$ on the negative plate of a parallel plate capacitor of capacitance $1.2\times10^{-3}\mu\text{F},$ then the potential difference developed between the plates is:

6.25V
3.0V
12.5V
25V

Answer

(b) Constant, increases, decreases.

Explanation:

As the capacitor is isolated after charging, charge Q on it remains constant. Plate separation d increases, capacitance decreases as $\text{C}=\frac{\in_0\text{A}}{\text{d}}$ and hence, potential increases as $\text{V}=\frac{\text{Q}}{\text{C}}.$

Explanation:

ln a parallel plate capacitor, the capacity of capacitor $\text{C}=\frac{\text{K}\in_0\text{A}}{\text{d}}$ i.e., $\text{C}\propto\text{A}$

The capacity of capacitor increases if area of the plate increases.

(b) $\frac{\sigma}{\in_0}$

Explanation:

The magnitude of the electric field between the plates is

$\text{E}=\frac{\sigma}{2\in_0}-\Big(-\frac{\sigma}{2\in_0}\Big)=\frac{\sigma}{\in_0}$

(b) 4mm

Explanation:

As, $\frac{\in_0\text{A}}{\text{d}}=4\pi\in_0\text{R}$ or $\frac{\in_0\text{D}^2}{\text{4d}}=4\pi\in_0\text{R}$

Or $\text{d}=\frac{\text{D}^2}{\text{16R}}=\frac{(0.8)^2}{16\times0.10}$

$=4\times10^{-3}\text{m}=\text{4mm}$

Explanation:

Here, $\text{V}=\frac{\text{q}_1-\text{q}_2}{\text{2C}}$

$=\frac{2.0\times10^{-8}+1.0\times10^{-8}}{2\times1.2\times10^{-9}}=12.5\text{V}$

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Question 154 Marks

The saturation current from a thoriated-tungsten cathode at 2000K is 100mA. What will be the saturation current for a pure-tungsten cathode of the same surface area operating at the same temperature? The constant A in the Richardson-Dushman equation is $60 \times 10^4Am^{-2}K^{-2}$ for pure tungsten and $3.0 \times 10^4Am^{-2}k^{-2}$ for thoriated tungsten. The work function of pure tungsten is 4.5eV and that of thoriated tungsten is 2.6eV.

Answer

$\text{i}=\text{AST}^2\text{ e}^{\frac{-\phi}{\text{KT}}}$
$\text{i}_1=\text{i},\ \text{i}_2=100\text{mA}$
$\text{A}_1=60\times10^4,\ \text{A}_2=3\times10^4$
$\text{S}_1=\text{S},\ \text{S}_2=\text{S}$
$\text{T}_1=2000,\ \text{T}_2=2000$
$\phi_1=4.5\text{eV},\ \phi_2 =2.6\text{eV}$
$\text{i}=(60\times10^{4})(\text{S})\times(2000)^2\frac{-45\times1.6\times10^{-19}}{\text{e}^{1.38\times10^{-23}\times2000}}$
$100=(3\times10^{4})(\text{S})\times(2000)^2\frac{-2.6\times1.6\times10^{-19}}{\text{e}^{1.38\times10^{-23}\times2000}}$
Dividing the equation,
$\Rightarrow\frac{\text{i}}{100}=\text{e}^{\big[\frac{-4.5\times1.6\times10}{1.38\times2}\big(\frac{-2.6\times1.6\times10}{1.38\times20}\big)\big]}$
$\Rightarrow\frac{\text{i}}{100}=20\times\text{e}^{-11.014}$
$\Rightarrow\frac{\text{i}}{100}=20\times0.000016$
$\Rightarrow\text{i}=20\times0.0016=0.0329\text{mA}=33\mu\text{A}$

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