MCQ Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 3

The charge on an electron was calculated by:

0%
Faraday
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J.J Thompson
0%
Millikan
0%
Einstein

Torque acting on an electric dipole in a uniform electric field is maximum if the angle between

$\overrightarrow{P}$ ; and

$\overrightarrow{E}$ is:

0%
$180^o$
0%
$0^o$
0%
$90^o$
0%
$45^o$

Explanation

The torque acting on an electric dipole

$\tau =\overrightarrow{p}\times\overrightarrow{E}$

$=pE\, sin \, \theta$
Torque will maximum when value of

$sin \, \theta$ will be maximum or

$\theta=90^o$

Electric lines of force about a positive point charge are:

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radially outwards
0%
circular clockwise
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radially inwards
0%
parallel straight lines
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circular anti-clockwise

The charge is negative, then the electric lines of forces are

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straight lines converging towards the charge
0%
concentric circle with charge at the centre
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Straight lines radiating away from the charge
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Non of these

The unit of electric dipole moment is:

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newton
0%
coulomb
0%
farad
0%
debye

Explanation

The unit of electric dipole moment is -

$2ql=$

$C-m$ =Debye

Coulomb's law relates two charges and distance between them describing the electric force as being:

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proportional to the sum of the charges
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inversely proportional to the distance between charges
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proportional to the product of the charges and inversely proportional to the distance
0%
proportional to the product of the charges and inversely proportional to the square of the distance

Explanation

Coulomb's law can be given by:

$F = k\cfrac{q_{1} q_{2} }{r^{2}}$

$\therefore F \propto \cfrac{1}{r^{2}}$

$\oint _{ s }^{ }{ \overrightarrow { E } .\overrightarrow { d } s } =0$ over surface, then

0%
the electric field inside the surface and on it is zero
0%
the electric field inside the surface is necessarily uniform
0%
all charges must necessarily be outside the surface
0%
all of these

Explanation

$\oint { \vec { E } .d\vec { s } } =\cfrac { q }{ { \varepsilon }_{ 0 } }$
where

$q$ is charge enclosed by the surface
when

$\quad \oint { \vec { E } .d\vec { s } } =0,q=0$
i.e., net charge enclosed by the surface must be zero. Therefore, all other charges must be outside the surface. This is because charges outside the surface do not contribute to the electric flux.

Electric lines of force about negative point charge are

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Circular, anticlockwise
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Circular, clockwise
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Radial, outward
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Radial, inward

An electric dipole consists of two opposite charges each of magnitude

$1.6 \times 10^-19$ coulomb at separation

$1\ Armstrong$ . The diploe moment is:

0%
$1.6 \times 10^{-19} c-m$
0%
$1.6 \times 10^{-29} c-m$
0%
$3.2 \times 10^{-29} c-m$
0%
$3.2 \times 10^{-19} C-m$

Explanation

Dipole moment

$P = qd$

So,

$P = 1.6\times 10^{-19}\times 10^{-10} = 1.6\times 10^{-29} \ C-m$

When a glass rod is rubbed with a piece of silk cloth the rod ?

0%
And the cloth both acquire positive charge.
0%
Becomes positively charged while the cloth has a negative charge.
0%
And the cloth both acquire negative charge.
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Becomes negatively charged while the cloth has a positive charge.

A dipole is placed in a non-uniform field. It will

0%
Experience net Force
0%
Experience net Moment
0%
Experiences both Force and Moment
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None of the above

Explanation

REF.Image.

In a non-uniform field, the two change at end of dipole may

experience different electric force (given by

$= q\vec{E},\vec{E}$ at that point)

or at different directions, resulting in some net force.

Now due to the force, it may experience a couple, an changes are

some distance a parts.

1104345_1179363_ans_ebca571912074233a9c352fd26aaa401.JPG

A large flat metal surface has uniform charge density $+\sigma$ . An electron of mass $m$ and charge $e$ leaves the surface at point A with speed $v$ , and return to it at point B. The maximum value of AB is:

0%
$\dfrac{vm\epsilon _{0}}{\sigma e}$
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$\dfrac{v^{2}m\epsilon _{0}}{\sigma e}$
0%
$\dfrac{v^{2}e}{\epsilon _{0}\sigma m}$
0%
$\dfrac{v^{2}\sigma e}{\epsilon _{0}m}$

Explanation

Field near metal surface

$E = \displaystyle\frac{\sigma}{\epsilon_o}$

Force on electron

$= eE = \displaystyle\frac{e \sigma}{\epsilon_o}$

Acceleration of electron

$= a = \displaystyle\frac{e \sigma}{m \epsilon_o}$

Max Range

$= \displaystyle\frac{v^2}{a} = \displaystyle\frac{v^2 m \epsilon_o}{e \sigma}$

A cube is arranged such that its length, breadth, height are along X,Y and Z directions. One of its corners is situated at the origin. Length of each side of the cube is $25\ cm$ . The components of electric field are $E_{x}=400\sqrt{2}\ N/C,\ E_{y}=0$ and $E_{Z}=0$ respectively. The flux coming out of the cube at one end, whose plane is perpendicular to X axis, will be:

0%
$25\sqrt{2}\ Nm^{2}/C$
0%
$5\sqrt{2}\ Nm^{2}/C$
0%
$250\sqrt{2}\ Nm^{2}/C$
0%
$25\ Nm^{2}/C$

Explanation

Flux coming out on one side is

$\vec E.d \vec a$

So flux coming out of the cube through the surface normal to X-axis is given by:

$\phi =E_x \times (25\times 25\times 10^{-4})$

$=\dfrac{400 \sqrt{2}\times 25\times 25}{10000}$

$=25\sqrt{2}$

$Nm^2/C$

An electron moves with a velocity

$\vec{v}$ in an electric field

$\vec{E}$ . If the angle between

$\vec{v}$ and

$\vec{E}$ is neither 0 nor

$\pi$ , then the path followed by the electron is a(an):

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straight line
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circle
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ellipse
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parabola

Explanation

The perpendicular component of velocity is given by:

$v_y=v \sin\theta$
Now in the X direction, the electron experiences a force of

$E(-e)$
So,

$F= -eE$
or,

$a =\dfrac {-Ee}{m}$

$\therefore$ velocity in X direction:

$v_x= u + at$

$= v cos\theta-\dfrac {Ee}{m}\times t$

We know,

$y=\int v_ydt$ and

$x=\int (v_x)dt$

$\Rightarrow$

$y=v \sin\theta \cdot t, x=v cos\theta \cdot t-\dfrac {Ee}{m}\dfrac {t^2}{2}$

Substituting

$t=\dfrac {y}{v sin\theta}$ , we get

$x=v \cos\theta \dfrac { y}{v sin\theta}-\dfrac {Ee}{m_2}\cdot \dfrac {y^2}{v^2 sin^2\theta}$

$\Rightarrow$

$x=y \cot\theta -\dfrac {Ee}{2m v^2sin^2\theta}y^2$ , where

$\theta$ is the initial angle of

$v$ with

$E$ .

$\therefore$ path followed is a parabola

The electric field in a region of space is given by $\vec{E}=(\hat{5i}+\hat{2j} )Nc^{-1}$ . The electric flux due to this field through an area $2m^{2}$ lying in the Y-Z plane in S.I. units is

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$10$
0%
$20$
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$10\sqrt{2}$
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$2\sqrt{29}$

Explanation

$2m^{2}$ area is in y-z plane its area vector which is perpendicular to y-z plane is in x axis.

$\therefore$ area vector

$=2(\hat{i})$
given

$E=5i+2j$
We know flux

$\phi=E.A$

$=(2i).(5i+2j)$

$\phi=10$

Drawings I and II show two samples of electric field lines. Then :

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the electric fields in both I and II are produced by negative charge located somewhere on the left and positive charges located somewhere on the right
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in both I and II the electric field is the same everywhere
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in both cases the field becomes stronger on moving from left to right
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the electric field in I is the same everywhere, but in II the electric field becomes stronger on moving from left to right

Two thin infinite parallel sheets have uniform surface densities of charge

$+\sigma$ and

$-\sigma$ . Electric field in the space between the two sheets is:

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$\sigma/\epsilon _{0}$
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$\sigma/2\epsilon _{0}$
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$2\sigma/\epsilon _{0}$
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zero

Explanation

Electric field due to positive charged plate

$E'= \dfrac{\sigma}{{2 \epsilon }_{ 0 }}$
Similarly for negative charged plate

$E''= \dfrac{\sigma}{{2 \epsilon }_{ 0 }}$

$\therefore$ total electric field between the fixed plates

$= \dfrac{\sigma}{2{ \epsilon }_{ 0 }} + \dfrac{\sigma}{2{ \epsilon }_{ 0 }}$

$= \dfrac{\sigma}{{ \epsilon }_{ 0 }}$

A cylinder of radius R and length L is placed in the uniform electric field E parallel to the cylinder axis. The total flux from the curved surface of the cylinder is given by :

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$2\pi R^{2}E$
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$\dfrac{\pi R^{2}}{E}$
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$\dfrac{\pi R^{2}-\pi R}{E}$
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$zero$

Explanation

As shown in the diagram angle between the curved surface and the field

$E$ is

$90^0$

We know than flux

$=E.da$

$flux=E.da cos 90^0$

$flux=0$

$\therefore flux\ from \ curved\ surface =0$

A hemisphere of radius r is placed in a uniform electric field of strength E. The electric flux through the hemisphere is :

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$2E\pi r^{2}$
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$-E\pi r^{2}$
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$-2E\pi r^{2}$
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zero

A particle that carries a charge $-q$ is placed at rest in uniform electric field $10\ N/C$ . It experiences a force and moves in a certain time t, it is observed to acquire a velocity $10\vec{i}-10\vec{j}$ m/s. The given electric field intersects a surface of area $A$ $m^{2}$ in the X -Z plane. Electric flux through surface is:

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$5\sqrt{2}A\ Nm^{2}/C$
0%
$5A\ Nm^{2}/C$
0%
$\sqrt{2}A\ Nm^{2}/C$
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$2\sqrt{5}A\ Nm^{2}/C$

Explanation

$v=10\hat{i}-10\hat{j}$

As velocity is in x and y components
Electric field also has two components

Let

$E=Ex\hat{i}+Ey\hat{j}$

Now force on

$(-q) =E(-q)$

Force

$F=-Exq\hat{i}+-Eyq\hat{j}$

Acceleration

$a =(-\dfrac{Exq}{m})\hat{i} +(-\dfrac{EyV}{m})\hat{i}$

Now

$v_x=10$ and

$v_y=-10$

we know

$v =u+at \ \ (given \ u=0 )$

$\therefore V=at$

$\Rightarrow v_x=(-\dfrac{Exq}{m})t --(1) \ \ and\ \ v_y=(\dfrac{-Eyq}{m})t --(2)$

$Given |E|=10\ N/C$

Now from 1 and 2 we get,

$-1=\dfrac{-Exq}{m}\dfrac{m}{Eyq}$

$\Rightarrow Ex=Ey$

$10=\sqrt{Ex^{2}+Ey^{2}}$

$\Rightarrow Ex=5\sqrt{2}=Ey$

Now than

$\sigma = \int E.dA$

As the area is in X-Z plane

Area =

$A$

$\hat{i}$

$\therefore Ea=Ey.A$

$\phi=5\sqrt{2}A$

The electric dipole moment is:

0%
$15.36 \times 10^{-29}C-m$
0%
$15.36 \times 10^{-19}C-m$
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$7.68 \times 10^{-29}C-m$
0%
$7.68 \times 10^{-19}C-m$

Explanation

We know electric dipole moment

$P= q .d$

where q is the magnitude of each charge

d is the distance between them

On putting the values we get

$P=3.2\times 10^{-19}\times2.44$

$P=7.68\times 10^{-29}\ C-m$

Two charges when kept at a distance of $1m$ apart in vacuum have some force of repulsion. If the force of repulsion between these two charges be same, when placed in an oil of dielectric constant $4$ , the distance of separation is :

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$0.25m$
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$0.4m$
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$0.5m$
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$0.6m$

Explanation

In vacuum force of repulsion

$= \dfrac{1}{4\pi \varepsilon _0}\dfrac{q _1q _2}{(1)^2}$

given the force of repulsion is same in oil so let the distance in oil be X.

$\Rightarrow \dfrac{1}{4\pi \varepsilon_0}\dfrac{q _1q _2}{(1)^2}= \dfrac{1}{4\pi K {\varepsilon}_{0} } \dfrac{q _1q _2}{X^2}$

$\Rightarrow \dfrac{K}{1}= \dfrac{1}{X^2}$

$\Rightarrow 4= \dfrac{1}{X^2}$

$\Rightarrow 2= \dfrac{1}{X}$

$\Rightarrow X= 0.5m$

An electric dipole is placed in a non uniform electric field increasing along the $+ve$ direction of X - axis. The dipole moves along $\underline{\hspace{0.5in}}$ and rotates $\underline{\hspace{0.5in}}$

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$+ ve$ direction of X - axis, clockwise
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$- ve$ direction of X - axis, clockwise
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$+ ve$ direction of X - axis, anti clockwise
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$- ve$ direction of X - axis, anti clockwise

Explanation

As electric field is more towards positive side of x axis force acting on +q greater than force acting on -q.
total force

$F_{net}=q(E_1(x)-E_2(x)) \ where\ E_1> E_2 \ are\ different$

$\therefore$ the dipole will move along the +ve x axis.
Torque is given as

$P\times E$
As shown in the diagram the dipole will rotate in clockwise direction.

A small electric dipole is placed at origin with its dipole moment directed along positive x -axis. The direction of electric field at point

$\left ( 2,2\sqrt{2},0 \right )$ is:

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along z-axis
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along y-axis
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along negative y-axis
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along negative z-axis

Explanation

The electric field due to a dipole at any point in space is given by,

$E = 1/(4\pi\epsilon_0 r^3) (3 (\overrightarrow p . \hat{r} )\hat{r} - \overrightarrow p )$

$r = 2\hat{i} +2\sqrt{2}\hat{j} \Rightarrow \hat{r} = 1/\sqrt{3} \hat{i} + \sqrt{\dfrac{2}{3}}\hat{j}$

$E = 1/(4\pi\epsilon_0 r^3) ( \sqrt{3}p \hat{r} - p\hat{i} )$
The i component cancels out leaving only the j component.

Two electric dipoles each of dipole moment $p=6.2\times 10^{^{-30}}C-m$ are placed with their axis along the same line and their centres at a distance d $=10^{-8}cm$ . The force of attraction between dipoles is:

0%
$2.1\times 10^{-16}N$
0%
$2.1\times 10^{-12}N$
0%
$2.1\times 10^{-10}N$
0%
$2.1\times 10^{-8}N$

Explanation

Electric field due to a dipole on its axis is given by

$2\overrightarrow{p}/(4\pi\epsilon_0 r^3)$

Therefore using necessary approximation and

$F = qE$ we get,

$F = 2qp/(4\pi\epsilon_0) (1/(d-l)^3 - 1/(d+l)^3 ) = 2qp/(4\pi\epsilon_0) 6d^2l/d^6 = 6p^2 / 4\pi\epsilon_0 d^4$
putting in the values,

$F = 6\times 9\times 10^9\times 6.2^2 \times 10^{-60} \times 10^{40} = 2.1 \times 10^{-8} N$

A metallic shell has a point charge q kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces?

Explanation

Electric field inside the conductor is zero, thus there should be no lines of forces in the shaded region. Also positive charge

$q$ inside the cavity will induce

$-q$ charge on the interior surface of the conductor and a

$+q$ charge on the outermost surface.

Thus option C is correct.

A dipole of moment p is placed in uniform electric field E then torque acting on it is given by : -

0%
$\vec\tau=\vec P\cdot \vec E$
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$\vec\tau=\vec P\times \vec E$
0%
$\vec\tau=\vec P+ \vec E$
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$\vec\tau=\vec P-\vec E$

Explanation

Force on charge

$+ q = {F_A} = q E$ along direction of vector

$\vec E$

Force on charge

$- q = {F_B} = qE$ opposite to direction of

$\vec E$

As there two forces are equal, they from a couple

$\begin{array}{l} Torque\left( \tau \right) =Force\times Arm\, \, of\, \, couple \\ \tau =F\times AB\sin \theta \\ \tau =qE \times AB\sin \theta \\ \tau =qE \times 2a\sin \theta \\ \tau =\left( { q\times 2a } \right) E \sin \theta \\ \tau =PE \sin \theta ,where\, \, P\, \, in\, \, dipole\, \, movement \\ \tau =P\times E\end{array}$

A hollow cylinder has a charge q coulomb distributed uniformly within it. If

$\phi$ is the electric flux in units of voltmeter associated with the curved surface

$B$ , the flux linked with the plane surface

$A$ in units of voltmeter will be

0%
$\dfrac{q}{\epsilon _{0}}-\phi$
0%
$\dfrac{1}{2}\left ( \frac{q}{\epsilon _{0}}-\phi \right )$
0%
$\dfrac{q}{2\epsilon _{0}}$
0%
$\dfrac{\phi }{3}$

The dipolemoment of the given system is:

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$\sqrt{3}$ $ql$ along perpendicular bisector of $q - q$ line
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$2\ ql$ along perpendicular bisector of $q - q$ line
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$ql$ $\sqrt{2}$ along perpendicular bisector of $q - q$ line
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$0$

Explanation

See the diagram.
Dipole moment of a system of charges is given by,

$\overrightarrow p = \Sigma q_i \overrightarrow{r_i}$

$\vec P = -ql\hat{i} + ql\hat{i} + (-2q)(\sqrt{3}/2\times l )\hat{j} = -\sqrt{3} ql \hat{j}$

A square surface of side L metres is in the plane of the paper. A uniform electric field

$\underset{E}{\rightarrow}$ (volt/m),also in the plane of the paper, is limited only to the lower half of the square surface, (see figure).The electric flux in SI units associated with the surface is:-

0%
$EL^{2}$ / $\left (2\varepsilon_{0} \right )$
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$EL^{2}$ /2
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Zero
0%
$EL^{2}$

Fill in the blank :

The mutual electrostatic force between two point charges

$q_{1}$ and

$q_{2}$ is _____ to the product of charges

$q_1$ and

$q_{2}$ .

0%
inversely proportional
0%
directly proportional
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square proportional
0%
inversely square proportional

The force of interaction between any two point charges is proportional to :

0%
$q_{1}q_{2}$
0%
$r^{2}$
0%
$r^{-2}$
0%
$q_1/q_2$

Explanation

This is the formula of Coulomb's Law which ca be written as

$F= \dfrac{kq_1q_2}{r^2}$

The formation of a dipole is due to two equal and unlike point charges placed at a :

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short distance
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long distance
0%
both (a) and (b)
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None of these

The ratio of electric force between two electrons to that between two protons separated by the same distance in air is :

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$10^{0}$
0%
$10^{6}$
0%
$10^{4}$
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None of the above

Electric lines of force are:

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Exist everywhere
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Exist only in the immediate vicinity of electric charges
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Exist only when both positive and negative charges are near one another
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Are imaginary

A charge

$q$ is enclosed in a cube. What is the electric flux associated with one of the faces of cube ?

0%
$q/ \varepsilon _{0}$
0%
$\varepsilon /q$
0%
$6q/ \varepsilon _{0}$
0%
$q/ 6\varepsilon _{0}$

Explanation

According to Gauss theorem - the total electric flux

$\phi = \dfrac{1}{\varepsilon _{0}} \times total \ enclosed \ charge = \dfrac{1}{\varepsilon _{0}} \times q.$

Since cube has six faces. Hence electric flux linked with each face =

$= (1/6 \times \phi )= q/6\varepsilon _{0}$

The value of

$\dfrac {1}{4\pi \epsilon_0}$ is _________

$m^{-3} \ kg^{-1} \ s^4 \ A^2$

0%
$8.854\times 10^{-12}$
0%
$4\pi \times 10^{-7}$
0%
$6.25\times 10^{12}$
0%
$9\times 10^9$

Explanation

$\displaystyle \frac {1}{4\pi \epsilon_0}=\frac {1}{4\times \frac {22}{7}\times 8.854\times 10^{-12}}$

$=9\times 10^9$

Two charges are placed a certain distance apart in air. If a glass slab is introduced between them, the force between them will:

0%
increase
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decrease
0%
remains the same
0%
be zero

Explanation

$F_{glass} = \dfrac{q_{1}q_{2}}{4\pi K \epsilon _{0}r^{2}}= \dfrac{F}{K}$

As

$K$ (for glass)

$>1 \therefore F_{glass}< F$

Dielectric constant of glass is greater than 1 hence, the force will decrease.

Electric flux at a point in an electric field is

0%
positive
0%
negative
0%
zero
0%
None of these

SI unit of permittivity is:

0%
$F m^{-1}$
0%
$N m^2C^{-2}$
0%
$N m^{-2}C^{-1}$
0%
$Am^{-1}$

Explanation

SI unit of permittivity is farad

$metre^{-1}(Fm^{-1}).$

Choose the correct statement from the following:

0%
Electric lines of force never cross each other
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Electric lines of force end at a positive charge
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The electric field lines inside a conductor are infinity
0%
Lines of electric field point towards regions of lower potential.

A charge Q is placed at the corner of a cube. The electric flux through all the faces of the cube is

0%
$Q/ \varepsilon _{0}$
0%
$Q/6 \varepsilon _{0}$
0%
$Q/8 \varepsilon _{0}$
0%
$Q/ 3\varepsilon _{0}$

Explanation

We require eight cubes to surround completely this corner charge. So, the flux through one cube is

$\frac{1}{8} \times \frac{Q}{\varepsilon _{0}}= \frac{Q}{8\varepsilon _{0}}$

A point charge

$+10\mu C$ is at a distance

$5\ cm$ directly above the centre of a square of side

$10\ cm$ , as shown in the figure. What is the magnitude of the electric flux through the square?

0%
$1.2\times 10^5 Nm^2/C$
0%
$1.9\times 10^5 Nm^2/C$
0%
$3.2\times 10^5 Nm^2/C$
0%
$4.2\times 10^5 Nm^2/C$

Explanation

Imagine a square cube around the charge with the charge at the centre of the cube and the square being one surface of the cube..
Now apply Gauss law for finding flux through the cube. Divide by 6 to get flux through one face, this is the answer.

$6\phi =\frac {\phi}{\epsilon_0}$ where

$\phi =$ flux through one of the faces of square cube.

$\phi = 1.9\times 10^5 Nm^2/C$

A positively charged body has deficiency of ______.

0%
Electrons
0%
Protons
0%
Neutrons
0%
None of the above

Explanation

A body is normally neutral.

When the body has deficiency of electron, the positive charge inside body increases the negative charge inside and hence the body becomes overall

$positive$ charge.

Answer-(A)

$4$ charges are placed each at a distance 'a' from origin. The dipole moment of configuration is :

0%
$2qa\widehat{j}$
0%
$3qa\widehat{j}$
0%
$2aq[\widehat{i}+\widehat{j}]$
0%
none

Explanation

As the total charge of the configuration is zero so the origin is independent of choice. Now we can find the dipole moment about given origin.

Thus, dipole moment about the origin is

$\vec p=(-2q)(a\hat i)+3q(a\hat j)+(-2q)(-a\hat i)+q(-a\hat j)=2qa\hat j$

$96,000\ C$ deposit on 1 gm atom of mono-valent silver. The positive charge on a silver ion is equal in magnitude to that of an electron and Avogadro's number is

$6 \times { 10 }^{ 23 }$ . Electronic charge will be:

0%
$-1.6\times { 10 }^{ -19 }\ C$
0%
$5.76\times { 10 }^{ 28 }\ C$
0%
$-6.2\times { 10 }^{ 18 }\ C$
0%
$- 160\ C$

Explanation

One gm atom of silver will include

$N_A$ particles or electron

We know that:

$Q={N}_{A}\times e$
Given,

$Q=96000\ C$ and

${N}_{A}=6\times{10}^{23}$
So,

$e=\dfrac{Q}{N_A}$

$=\dfrac{96000}{6\times{10}^{23}}$

$=16000\times{10}^{-23}$

$=1.6\times{10}^{-19}\ C$

In Coulomb's law, the constant of proportionality K has the units

0%
$N$
0%
$Nm^2$
0%
$NC^2/m^2$
0%
$Nm^2/C^2$

Explanation

force between two charged particles is given by

$[F= \dfrac{(K\times q_{1}\times q_{2})}{r^{2}}]$

$[K=\dfrac{(F\times r^{2})}{(q_{1} \times q_{2})}]$

Therefore unit of K will be

$[(N \times m^{2})/(coulomb)^{2}]$

The electrostatic force between two point charges

$q_{1}$ and

$q_{2}$ at separation 'r' is given by

$F = \dfrac {Kq_{1}q_{2}}{r^{2}}$ . The constant K:

0%
Depends on the system of units only
0%
Depends on the medium between the charges only
0%
Depends on both the system of units and the medium between the charges
0%
Is independent of both the system of units and the medium between the charges

Explanation

$K=\dfrac{1}{4\pi\epsilon_ok}$ and its units is

$N-m^2/C^2$

It depend on system of unit and dielectric constant (k), that's medium between charges

The dipole moment of a system of charge

$+q$ distributed uniformly on an arc of radius

$R$ subtending an angle

$\pi{/2}$ at its centre where another charge

$-q$ is placed is :

0%
$\displaystyle \frac{2\sqrt{2}qR}{\pi}$
0%
$\displaystyle \frac{\sqrt{2}qR}{\pi}$
0%
$\displaystyle \frac{qR}{\pi}$
0%
$\displaystyle \frac{2qR}{\pi}$

Explanation

The definition of dipole moment is

$\vec{p}=\int \vec{r}dq$
If

$\lambda$ is the line charge density, charge on element

$dl$ is

$dq=\lambda Rd\theta$
In polar coordinate,

$\vec{r}=R\cos\theta \hat i+R\sin\theta \hat j$
thus,

$\displaystyle \vec{p}=\int (R\cos\theta \hat i+R\sin\theta \hat j)\lambda Rd\theta$

$\displaystyle p_x=\lambda R^2\int_0^{\pi/2}\cos\theta d\theta=-\lambda R^2$

$\displaystyle p_y=\lambda R^2\int_0^{\pi/2}\sin\theta d\theta=\lambda R^2$
thus,

$\displaystyle p=\sqrt{p_x^2+p_y^2}=\sqrt{2}\lambda R^2=\sqrt{2}R^2 \frac{2q}{\pi R}$

$\displaystyle (q=\int_0^{\pi/2}\lambda Rd\theta=\frac{\lambda R\pi}{2})$

$\displaystyle \therefore p=\frac{2\sqrt{2}qR}{\pi}$

Select the correct statement : (Only force on a particle is due to electric field)

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A charged particle always moves along the electric line of force
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A charged particle may move along the line of force
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A charge particle never along the line of force
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A charged particle moves along the line of force only if released from rest

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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