MCQ Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 13

The potential at a point

$(x, 0, 0)$ is given by

$V = \left(\dfrac{1000}{x} + \dfrac{1500}{x^2} + \dfrac{500}{x^3}\right)$ . The intensity of the electric field at

$x = 1$ will be

0%
$550 V/m$
0%
$55 V/m$
0%
$55000 V/m$
0%
$5500 V/m$

Explanation

$V=(\cfrac{1000}{x}+\cfrac{1500}{x^{2}}+\cfrac{500}{x^{3}})$

We know that,

$E=\cfrac{-dv}{dr}$

So,

$E=\cfrac{-d(\cfrac{1000}{x}+\cfrac{1500}{x^{2}}+\cfrac{500}{x^{3}})}{dx}$ at

$x=1$

$=-[\cfrac{(-1000)}{x^{2}}-\cfrac{2\times 1500}{x^{3}}-\cfrac{3\times 500}{x^{4}}]_{x=1}$

$=1000+3000+1500=5500\,V/m$

Four changed particles, each having a charge

$q$ , are placed at the four vertices of a regular pentagon. The sides of the Pentagon have equal length '

$a$ '. The electric field at the centre of the Pentagon is :

0%
$\dfrac{q}{4\pi \varepsilon_0 a}$
0%
$\dfrac{2q}{4\pi \varepsilon_0 a^2}$
0%
$\dfrac{q}{4\pi \varepsilon_0 a^2}$
0%
None of these

Explanation

Suppose if charge

$'q'$ is placed at E as well as

$\overset{\rightarrow}E(at \,point\, O)=0$ because of symmetry.

$\overset{\rightarrow}E$ (center because of charge,A,B,C,D)

$=\overset{\rightarrow}E$ (center because of charge E alone)$$

$\overset{\rightarrow}E$ (at center due to charge

$q$ )

$=\cfrac{q}{4\pi\epsilon_{0}a^{2}}$

$\therefore$ Electric field at O will be same i.e.

$\cfrac{q}{4\pi\epsilon_{0}a^{2}}$ along OE due to given system of charge.

1102124_1016438_ans_3c2bb9d535d545058149b62d8457860b.png

An electron is projected from a distance

$d$ and with initial velocity

$u$ parallel to a uniformly charged flat conducting plate as shown. It strikes the plate after travelling a distance

$\ell$ along the direction of projection. The surface charge density of the conducting plate is equal to

0%
$\dfrac{2d\varepsilon_0mu^2}{e\ell^2}$
0%
$\dfrac{2d\varepsilon_0mu}{e\ell^2}$
0%
$\dfrac{d\varepsilon_0mu^2}{e\ell}$
0%
$\dfrac{d\varepsilon_0mu}{e\ell}$

Explanation

For the motion of the electron along

$Y$ axis

$d = \dfrac{1 \,eE}{2 \,m}.t^2$ ...(1)

Where

$E$ is the electric field due to the charged plate given by

$E = \dfrac{\sigma}{\varepsilon_0}$ ...(2)

$\sigma$ is the surface density of charge of the plate.
That time taken by the electron is

$t = \dfrac{\ell}{u}$ ...(3)

From (1), (2) and (3) we obtain,

$d = \dfrac{1}{2} \dfrac{e\sigma}{\varepsilon_0 m}\left(\dfrac{\ell}{u}\right)^2 \Longrightarrow \sigma = \dfrac{2d\varepsilon_0 mu^2}{e\ell^2}$ Y

An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the z-direction, then.

0%
Positive ions deflects towards $+$ y-direction and negative ions towards $-$ y direction
0%
All ions deflect towards $+$ y -direction
0%
All ions deflect towards $-$ y direction
0%
Positive ions deflect towards $-$ y direction and negative ions towards y-direction

Explanation

As the electric field is switched on positive ion will start to move along

$(+ve)$

$x-$ direction and negative ion along

$x-$ direction. Current associated with the motion of both types of ions is along

$(+ve)$ direction.

According to Fleming left hand rule force on both types of ions will be along negative

$y-$ directions.

$W=K\left( { Q }_{ 1 }-{ Q }_{ 2 } \right) \left[ \dfrac { 1 }{ \sqrt { { R }^{ 2 }+{ r }^{ 2 } } } -\dfrac { R }{ r } \right]$

Six point charges are kept at the vertices of a regular hexagon of side

$L$ and centre

$O$ , as shown in the figure. Given that

$\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { q }{ { L }^{ 2 } }$ , which of the following statement(s) is(are) correct?

0%
The electric field at $O$ is $6K$ along $OD$ .
0%
The potential at $O$ is zero.
0%
The potential at all points on the line $PR$ is zero.
0%
The potential at all points on the line $ST$ is same.

Explanation

Given

$k=\dfrac{1}{4\pi { \varepsilon }_{ 0 }}\dfrac{q}{L^2}$

(B)

$v_0=\dfrac{1}{4\pi { \varepsilon }_{ 0 }}\left[ \dfrac { { q }_{ A } }{ L } +\dfrac { { q }_{ B } }{ L } +......\dfrac { { q }_{ F } }{ L } \right] =0$

$[As\;q_A+q_B+.......+q_F=0]$

At line

$PR,$ potential is same and is equal to zero.

Hence, solved.

Fill in the blanks.
A field normal to the plane of a circular wire n turns and radius r which carries a current I is measured on the axis of the coil at small h distance h from the centre of the coil. This is smaller than the field at the centre by a friction ____

0%
$3\dfrac{3h^2}{2r^2}$
0%
$\dfrac{h^2}{2r^2}$
0%
$\dfrac{4h^2}{2r^2}$
0%
$\dfrac{3h^2}{2r^2}$

Explanation

Turns

$=n$ friction

$=\dfrac { { 3h }^{ 2 } }{ { 2r }^{ 2 } }$

radius

$=r$

current

$=I$

where field at the centre,

${ B }_{ 1 }=\dfrac { { \mu }_{ 0 } }{ 4\pi } \times \dfrac { 2\pi in }{ r }$

$=\dfrac { { \mu }_{ 0 } }{ 2 } \times \dfrac { ni }{ r }$

field at a distance

$h$ from the centre

${ B }_{ 2 }=\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { 2\pi ni{ r }^{ 2 } }{ { \left( { r }^{ 2 }+{ h }^{ 2 } \right) }^{ 3/2 } }$

$=\dfrac { { \mu }_{ 0 } }{ 2 } \dfrac { { nir }^{ 2 } }{ { r }^{ 3 }{ \left( 1+\dfrac { { h }^{ 2 } }{ { r }^{ 2 } } \right) }^{ 3/2 } }$

$={ B }_{ 1 }{ \left( 1+\dfrac { { h }^{ 2 } }{ { r }^{ 2 } } \right) }^{ 3/2 }$

$={ B }_{ 1 }\left( 1-\dfrac { 3 }{ 2 } \dfrac { { h }^{ 2 } }{ { r }^{ 2 } } \right)$ by Binomial expansion.

Hence

${ B }_{ 2 }<{ B }_{ 1 }$ by a friction

$=\dfrac { 3 }{ 2 } \dfrac { { h }^{ 2 } }{ { r }^{ 2 } }$

A "dipole" is formed from a rod of length 2 a and two charges +q and -q. Two such dipoles are oriented as shown in figure, their centers being separated by the distance R. The force exerted on the left dipole is .

0%
$\dfrac{6kP_1P_2}{r^4}$
0%
$\dfrac{3kP_1P_2}{r^4}$
0%
$\dfrac{2kP_1P_2}{r^4}$
0%
$\dfrac{4kP_1P_2}{r^4}$

Two identical conducting spheres $M$ and $N$ has charges ${q_m}$ and ${q_n}$ respectively. A third identical neutral sphere $P$ is brought in contact with $M$ and then separated. Now sphere $P$ is brought in contact with $N$ and then separated. Final charge on sphere $P$ will be:

0%
$\dfrac{q_m + 2q_n}{6}$
0%
$\dfrac{q_m + q_n}{4}$
0%
${q_m}$ + $\dfrac{ q_n}{4}$
0%
$\dfrac{q_m + 2q_n}{4}$

An elliptical cavity is charge within perfect conductors. A positive charge

$q$ is placed at the centre of the cavity. The points

$A$ and

$B$ are on the cavity surface as shown, Then:

0%
Electric field near $A$ in the cavity $=$ electric field near $B$ in the cavity.
0%
Charge density at $A=$ charge density at $B$ .
0%
Potential at $A=$ potential at $B$ .
0%
Total electric field flux through the surface of the cavity is $q/\ \varepsilon_{0}$

A charge of

$1 \mu C$ is divided into two parts such that their charges are in the ratio of

$1:3$ . These two charges are kept at distance

$1m$ apart in vacuum. Then, the electric force between them (in newton) is :

0%
$1.7 \times 10^{-3}$
0%
$1.7 \times 10^{-4}$
0%
$3.4\times 10^{-3}$
0%
$3.4\times 10^{-5}$

Explanation

The two parts of charge according to ratio are given as,

${q_1} = \dfrac{1}{4}\;{\rm{\mu C}}$

${q_2} = \dfrac{3}{4}\;{\rm{\mu C}}$

The electric force is given as,

$F = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$

$F = 9 \times {10^9} \times \dfrac{{\dfrac{1}{4} \times {{10}^{ - 6}} \times \dfrac{3}{4} \times {{10}^{ - 6}}}}{1}$

$F = 1.6875 \times {10^{ - 3}}\;{\rm{N}}$

Consider a hemispherical surface of radius

$r$ , a positive point charge

$q$ is kept at the centre of hemisphere. The electric flux through this hemisphere is :

0%
zero
0%
$\dfrac{q}{\varepsilon_0}$
0%
$\dfrac{q}{2\varepsilon_0}$
0%
$\dfrac{2q}{\varepsilon_0}$

Explanation

if this charge is put at the centre of a sphere then net flux

$=\phi =q/\varepsilon_0$ ( hauss s law)

$\Rightarrow$ But this is hemisphere and it contain only half of the field lines and halve flux will become halved

Hence net flux through hemisphere

$=\dfrac{q}{2\varepsilon_0}$

1456855_1075541_ans_f1972c60126c432290a0d151206b6b21.png

A thin non-conducting ring of radius

$R$ has a linear charge

$\lambda = \lambda_0 \cos \theta$ where

$\lambda_0$ is the value of

$\lambda$ at

$\theta = 0$ . The net electric dipole moment for this charge distribution is :

0%
$\pi R^2 / \lambda_0$
0%
$\lambda_0 / \pi R^2$
0%
$\pi R^2 \lambda_0$
0%
$\dfrac {\lambda_0 \pi}{ R^2}$

Explanation

The small length is given as,

$dl = Rd\theta$

The small charge is given as,

$dq = \lambda dl$

$dq = {\lambda _0}\cos {\rm{\theta }}Rd\theta$

The position of charge is given as,

$\vec r = \left( {R\cos {\rm{\theta }},R\sin {\rm{\theta }},0} \right)$

The electric dipole moment is given as,

$\vec p = \int {\vec rdq}$

$= \int_0^{2\pi } {{\lambda _0}\cos {\rm{\theta }}Rd\theta \left( {R\cos {\rm{\theta }},R\sin {\rm{\theta }},0} \right)}$

${\vec p_x} = \int_0^{2\pi } {{\lambda _0}{R^2}{{\cos }^2}{\rm{\theta }}d\theta }$

$= \frac{{{\lambda _0}{R^2}}}{2}\int_0^{2\pi } {\left( {1 + \cos 2{\rm{\theta }}} \right)d\theta }$

$= \frac{{{\lambda _0}{R^2}}}{2}\left[ {\theta + \frac{1}{2}\sin 2{\rm{\theta }}} \right]_0^{2\pi }$

$= {\lambda _0}{R^2}\pi$

${\vec p_y} = \int_0^{2\pi } {{\lambda _0}{R^2}\cos {\rm{\theta }}\sin {\rm{\theta }}d\theta }$

$= \frac{{{\lambda _0}{R^2}}}{2}\int_0^{2\pi } {\sin 2{\rm{\theta }}d\theta }$

$= \frac{{{\lambda _0}{R^2}}}{2}\left[ {\cos 2{\rm{\theta }}} \right]_0^{2\pi }$

$= 0$

${\vec p_z} = 0$

Thus, net electric dipole moment is $\pi {R^2}{\lambda _0}$ .

A particle of mass

$m$ and charge

$q$ at rest is released in a uniform electric field between parallel planes of charge

$+q$ and

$-q$ respectively. The particle accelerates towards the other place a distance

$'d'$ away. The speed at which it strikes the opposite plane is:

0%
$\sqrt{qEd/m}$ and $j$
0%
$\sqrt{2qEd/m}$ along $-j$
0%
$\sqrt{qm/Ed}$ along $j$
0%
$\sqrt{qm/Ed}$ along $-j$

Explanation

We have

$\dfrac{1}{2}m{v^2} = qEd$

${v^2} = \dfrac{{2qEd}}{m}$

$v = \sqrt {\dfrac{{2qEd}}{m}}$

Thus the particle strikes the opposite plane with speed $\sqrt {\dfrac{{2qEd}}{m}}$ along $- {\rm{j}}$ .

The diode 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%
$\frac{2\sqrt{2} qR}{\pi}$
0%
$\frac{\sqrt{2} qR}{\pi}$
0%
$\frac{qR}{\pi}$
0%
$\frac{2qR}{\pi}$

Electric field at a distance

$4R$ from the surface of a charge cylinder is

$E$ . If

$R$ is the radius of the cylinder and

$L$ is the length of the cylinder then surface charge density of the cylinder is:

0%
$\dfrac { 4 \epsilon E }{L}$
0%
$\dfrac { 5 \epsilon E }{L}$
0%
$5 \epsilon E$
0%
$4 \epsilon E$

A cubical region of side

$'a'$ has its centre at the origin. It encloses three fixed point charges,

$-q$ at

$(0, -a/4, 0), +3q$ at

$(0, 0, 0)$ and

$-q$ at

$(0, +a/4, 0)$ . Choose the incorrect option

0%
The net electric flux crossing the plane $x = + \dfrac {a}{2}$ is equal to the net electric flux crossing the plane $x = -\dfrac {a}{2}$
0%
The net electric flux crossing the plane $y = +\dfrac {a}{2}$ is more than the net electric flux crossing the plane $y = -\dfrac {a}{2}$
0%
The net electric flux crossing the entire region is $\dfrac {q}{\epsilon_{0}}$
0%
The net electric flux crossing the plane $z = \dfrac {a}{2}$ is equal to the net electric flux crossing the plane $x = + \dfrac {a}{2}$

An electric dipole is kept on the axis of a uniformly charged ring at distance

$\sqrt 2 R$ from the centre of the ring. The direction of the dipole moment is along the axis. the dipole moment is

$P$ , charge of the ring is

$Q$ and radius of the ring is

$R$ . The force on the dipole is:

0%
$\dfrac{4 k p Q}{3 \sqrt 3 R^S}$
0%
$\dfrac{k p Q}{3 \sqrt 3 R^S}$
0%
$\dfrac{2 k p Q}{3 \sqrt 3 R^S}$
0%
Zero

A square surface of side

$L$ metre in the plane of the paper is placed in a uniform electric field

$E(V/m)$ acting along the same place at an angle

$\theta$ with the horizontal side of the square as shown in figure. The electric flux linked to the surface in unit of

$V-m$ , is

0%
$EL^{2}$
0%
$EL^{2}\cos \theta$
0%
$EL^{2}\sin \theta$
0%
Zero

A negative charged object reples another charged object kept close to it. What is the nature of the charge on the other object?

0%
positive
0%
negative
0%
both
0%
none.

electric intensity in axis of dipole at dist.

$2m$ from its centre is

$9\times 10^{3}\ N/C$ . Electric dipole moment of dipole is

0%
$4\times 10^{-6}\ cm$
0%
$8\times 10^{-6}\ cm$
0%
$2\times 10^{-6}\ cm$
0%
$16\times 10^{-6}\ cm$

A square loop of side b is rotating with angular speed

$\omega$ about one of the diagonals as axis. At

$t=0$ , the plane of the loop is perpendicular to the magnetic field B. If the number of turns of the loop be N then what is the instantaneous flux through the coil

0%
BAN
0%
BAN $sin \omega t$
0%
BAN $cos \omega t$
0%
BAN $\omega t$

A negative charged object attracts another charged object kept close to it. What is the nature of the charge on other object?

0%
positive
0%
negative
0%
both
0%
none

Two charges, each equal q, are kept at

$x=-a$ and

$x=a$ on the x-axis. A particle of mass

$m$ and charge

${q_0} = \frac{q}{2}$ is placed at the origin. If charge

${q_0}$ is given a small displacement

$(y<<a)$ along the y-axis, the net force acting on the particle is proportional to:

0%
$y$
0%
$-y$
0%
$\frac{1}{y}$
0%
$-\frac{1}{y}$

A negatively charged rod is held close to one side of a metal ball and the other side is earthed. Which of the following diagrams is correctly shows the charge distribution?

A positively charged glass rod is brought near the disc of uncharged gold leaf electroscope. The leaves diverge. Which of the following statements is correct?

0%
No charge is present on the leaves
0%
Positive charge is induced on the leaves
0%
Negative charge is induced on the leaves
0%
Positive charge is induced on the leave and negative charge on the other.

When alternating current flows through a conductor, the rate of flux change

0%
Is higher in the inner part of the conductor
0%
is lower in the inner part of the conductor
0%
is uniform throughout the conductor
0%
Depends on the resistivity of the conductor

A point charge

$q=10^{-11}\ C$ , is placed at

$4\ cm$ above a square plate

$(8\ cm\ \times 8\ cm)$ , having charge density

$0.5\ \times 10^{-8}\ C/m^{2}$ . Find the flux related with it.

0%
$0.188\ V-m$
0%
$0.12\ V-m$
0%
$0.288\ V-m$
0%
$0.388\ V-m$

A charge q is distributed uniformly on a ring of radius R. A sphere of equal radius R is constructed with its centre at the periphery of the ring, Find the flux of the electric field through the surface of the sphere

0%
$\dfrac{q}{\epsilon_0}$
0%
$\dfrac{q}{2\epsilon_0}$
0%
$\dfrac{q}{3\epsilon_0}$
0%
Zero.

Explanation

Let point of intersection of ring & sphere are

$P.Q.........$ let center of ring is

$O$ & center of sphere is

$O'$ ........

$O O'$ bisects

$P Q$ in two equal ,if

$S$ is the point of intersection of

$O O'$ &

$P Q$ then

In triangle right angles

$POS$

$PO=R, OS=R/2$ so angle

$POS=\theta$

Using trigonometry,

$cos \theta =OS/PS=1/2$

$\theta =\pi/3$

total angle subtended by arc is

$2 \pi/3.....$ this is the arc of ring which lies inside sphere & substends

$2 \pi /3$ angle at the center of ring.....

for

$2 \pi$ radian charge =q

for unit radian

$= \dfrac{q}{{2\pi }}$

for

$\dfrac{q}{{2\pi }}$ radian

${q_1} = \left( {\dfrac{q}{{2\pi }}} \right)\left( {\dfrac{{2\pi }}{3}} \right) = \dfrac{q}{3}$

total flux through sphere= total charge enclosed

$\dfrac{q}{{3{\varepsilon _0}}}$

$\therefore$ Option

$C$ is correct.

What is the electric flux linked with closed surface?

0%
$10^{11} Nm^2/C$
0%
$10^{12} Nm^2/C$
0%
$10^{10} Nm^2/C$
0%
$8.85\times10^1{3} Nm^2/C$

The electric field in a region of space is given by

$E=(5\ \hat {i}+2\hat {j})N/C$ . The electric flux through an area of

$2\ m^{2}$ lying in the

$YZ$ plane, in

$S.I.$ . units is ?

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

The volume charge density as a function of distance

$x$ from one face inside a unit cube is varying as shown in the figure. Then the total flux

$(in S.I units)$ through the cube if

$(p-{0}=8085\times10^{-12}Cm^{3})$ is

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{4}$
0%
$1$

A spherical shell of radius

$R$ has two holes

$A$ and

$B$ through which a ring having charge positive through without touching the shell. The flux through spherical shell is

$\dfrac{q}{3 \varepsilon_0}$ . Find the radius of the ring:-

0%
$R\sqrt{2}$
0%
$2 R$
0%
$0.5 R$
0%
$R$

If the electronic charge is

$1.6\times 10^{-19}$ C, then the number of electrons passing through a section of wire per second, when the wire carries a current of

$2$ ampere is?

0%
$1.25\times 10^{17}$
0%
$1.6\times 10^{17}$
0%
$1.25\times 10^{19}$
0%
$1.6\times 10^{19}$

Explanation

Current is the amount of charge passing through a conductor in a particular interval of time. So according to the definition,the formula for the current is:

$I=\dfrac { q }{ tI } =2 A$ and

$t=1$ sec Therefore,

$Q=I\times tQ=2\times1=2C$

One electon consists of a charge of

$1.6\times10^{-19}C$

Then the number of electrons passing would be:

$Q=n\times en=\dfrac{q}{en}$

$=2/(1.6\times10^{-19})$

$=1.25\times10^{19}electrons$

Four charge particle of charges -q,-q,-q and 3q are place at the vertices pf a square of side as shown.The magnitude of dipole moment of the arrangement is

0%
$2\sqrt 2 \;qa$
0%
qa
0%
3qa
0%
$\left( {2 + \sqrt 2 } \right)qa$

Three point charges

$+q,\ -2q$ and

$+q$ , are passed at points

$(x=0,\ y=a,\ z=0),\ (x=0,\ y=0,\ z=0)$ and

$(x=a,\ y=0,\ z=0 )$ respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are ?

0%
$\sqrt{2}\ qa$ along + $x$ direction
0%
$\sqrt{2}\ qa$ along + $y$ direction
0%
$\sqrt{2}\ qa$ along the line joining points $(x=0,\ y=0,\ z=0)$ and $(x=a,\ y=a,\ z=0)$
0%
$qa$ along the line joining points $(x=0,\ y=0,\ z=0)$ and $(x=a,\ y=z,\ z=0)$

Twelve infinite long wire of uniform linear charge density

$\left( \lambda \right)$ are passing along the twelve edges of a cube. Find electric flux through any face of cube

0%
$\left( \frac { \lambda l }{2 \varepsilon _{ { 0 } } } \right)$
0%
$\left( \frac { \lambda l }{ \varepsilon _{ { 0 } } } \right)$
0%
$\left( \frac { \lambda l }{3 \varepsilon _{ { 0 } } } \right)$
0%
$\left( \frac { 3\lambda l }{ \varepsilon _{ { 0 } } } \right)$

A conducting sphere of radius a has charge Q on it. It is enclosed by a neutral conducting concentric concentric spherical shell having inner radius 2a and outer radius 5a. Find electrostatic energy of system.

0%
$\dfrac{5}{12} \dfrac{kQ^2}{a}$
0%
$\dfrac{11}{12} \dfrac{kQ^2}{a}$
0%
$]dfrac{kQ^2}{2a}$
0%
$none$

Charges

$q$ and

$q$ are given to the conducting shells in figure (i) and

$q$ and

$-q$ are given to the identical pair in figure (ii).

0%
force of interaction between the shells in the figure (i) will be greater than that in figure (ii)
0%
force of interaction between the shells in the figure (i) will be smaller than that in figure (ii)
0%
force of interaction will be same in both the figures
0%
nothing can be said

Figure shows electric field lines in which an electric dipole

$\rho$ is placed as shown. Which of the following statements is correct?

0%
The dipole will not experience any force
0%
The dipole will not experience a force towards right
0%
The dipole will not experience a force towards left
0%
The dipole will not experience a force upwards

A non conducting sphere of radius

$'a'$ has a type charge

$'+q'$ uniformly distributed throughout in volume. A hollow spherical conductor having inner and outer, radii

$'b'$ and

$'c'$ and net charge

$'-q'$ is concentric with the sphere (see the figure)
Read the following statements
(i) The electric field at a distance

$r$ from the center of the sphere

$= \frac{1}{{4\pi {\varepsilon _0}}}\frac{{qr}}{{{a^3}}}$ for

$r<a$ .
(ii) The electric field at distance

$r$ from the center of the sphere for

$a<r<b=0$
(iii) The electric field at distance

$r$ from the center of the sphere for

$b<r<c<=0$
(iv) The charge on the inner surface of the hollow sphere

$=-q$
(v) The charge on the outer surface of the hollow sphere

$=+q$ .

0%
(i),(ii) and (v)
0%
(i),(iii) and (iv)
0%
(ii),(iii) and (v)
0%
(ii),(iii) and (iv)

Explanation

Solution

Let

$q^{1}$ is the inner surface charge.

then,

$q_{outer}= -q-q^{1}$ such that

$q_{inner}+q_{outer}= -q$

Electric field inside

the conductor should

be zero ( r > b and r < e)

so in the arbitrary

Gaussian surface

Electric flux =

$\dfrac{q_{inside}}{\varepsilon _{0}}= 0$

$\Rightarrow q_{inside}( b < r < c)= q^{1}+q=0$

$q^{1}= -q$

So,

$q_{inner}= -q$ and

$q_{outer}=0$

(i) Electric field for r < a ia,

$E.4\pi r^{2}= \dfrac{q_{inside}}{\varepsilon _{0}}$

As a charge is uniformly distributed

q inside =

$(\dfrac{r^{3}}{a^{3}})q$

So,

$E. 4\pi r^{2}= \dfrac{r^{3}q}{a^{3}\varepsilon _{0}}$

$\Rightarrow E= \dfrac{1}{4\pi }\times \dfrac{q}{a^{3}\varepsilon _{0}}$

(ii) Electric field for a < r < b,

$E= \dfrac{kq}{r^{2}}$

(iii) Electric field for b < r < c

As told earlier, E = 0

(iv) Electric field for c < r

$E= \dfrac{k_{qouter}}{r^{2}}= k\dfrac{(0)^{2}}{r^{2}}=0$

Option - B is correct.

1121123_1137796_ans_b62c44c4400b49f793d413b0f18ec7f9.jpg

An inductor having

$400$ turns, carrying of

$8 m A$ and inductances

$5 mH$ then flux is:

0%
$\mu_0$
0%
$4 \pi/\mu_0$
0%
$\mu_0/4 \pi$
0%
$4 \pi$

A point charge

$50 \: \mu C$ is located in the

$XY$ plane at the point of position vector

$r_1 = 2i + 3j$ . What is the electric field at the point of position vector

$r_2 = 8i-5j$ .

0%
$1200 v/m$
0%
$0.04 V/m$
0%
$900 V/m$
0%
$4500 V/m$

The ratio of the forces between two charges placed at a certain distance apart in air and at half of the distance apart in medium of dielectric '2' is

0%
1 : 4
0%
1 : 2
0%
4 : 1
0%
2 : 1

A point charge $+ 20\mu C$ is at a distance 4 cm above the center of a square of side 8 cm. What is the magnitude of electrostatic flux through the square?

0%
$3.76 \times {15^5}{\text{N}}{{\text{m}}^2}{{\text{C}}^{ - 1}}$
0%
$4.2 \times {15^5}{\text{N}}{{\text{m}}^2}{{\text{C}}^{ - 1}}$
0%
$2.9 \times {15^5}{\text{N}}{{\text{m}}^2}{{\text{C}}^{ - 1}}$
0%
$5.1 \times {15^5}{\text{N}}{{\text{m}}^2}{{\text{C}}^{ - 1}}$

Two charges

$q$ each attract each by a force of

$100 N$ . Now, the entire space is filled with a medium of relative permittivity

$10$ . The force exerted on any one of the charges by the medium is

0%
$90 N$
0%
$10 N$
0%
$110 N$
0%
$Zero$

If the electric lines of force are as shown in figure. Electric field intensities at

$A$ and

$B$ are

$E_{A}$ and

$E_{B}$ respectively.

0%
$E_{A} < E_{B}$
0%
$E_{A} > E_{B}$
0%
$E_{A} = E_{B}$
0%
$E_{A} = E_{B} = 0$

A capacitor is a perfect insulator for:

0%
constant direct current
0%
alternating current
0%
direct as well as alternating current
0%
variable direct current.

What is the value of electric flux in SI unit in the Y-Z plane of area $2{m^2}$ , if intensity of electric field is $\overrightarrow {\text{E}} {\text{ = }}\left( {{\text{5}}\widehat {\text{i}}{\text{ + 2}}\widehat {\text{j}}} \right){\text{N/C}}$

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

An electric dipole ( dipole moment p) is placed at a radial distance r>>a ( Where a is dipole length) from an infinite line of charge having linear charge density $+ \lambda$ . dipole moment vector is aligned along the radial vector $\overrightarrow r$ force experienced by the dipole is:-

0%
$\dfrac{{\lambda p}}{{2\pi {\varepsilon _o}{r^2}}},{\text{attractive}}$
0%
$\dfrac{{\lambda p}}{{2\pi {\varepsilon _o}{r^3}}},{\text{attractive}}$
0%
$\dfrac{{\lambda p}}{{2\pi {\varepsilon _o}{r^2}}},{\text{repulsive}}$
0%
$\dfrac{{\lambda p}}{{2\pi {\varepsilon _o}{r^3}}},{\text{repulsive}}$

Explanation

We know,

$E=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { q }{ { \left( r-a \right) }^{ 2 } } +\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { -q }{ { \left( r+a \right) }^{ 2 } }$

$=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \left[ \dfrac { q }{ { \left( r-a \right) }^{ 2 } } -\dfrac { q }{ { \left( r+a \right) }^{ 2 } } \right]$

$=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } 4aq\dfrac { r }{ { \left( { r }^{ 2 }-{ a }^{ 2 } \right) }^{ 2 } }$

The quality

$2qa$ is defined as the dipole moment

$(P)$ and so we can substitute

$p=2qa$ with the substitution we now get

$E=\dfrac { 1 }{ 2\pi { \epsilon }_{ 0 } } \dfrac { Pr }{ { \left( { r }^{ 2 }-{ a }^{ 2 } \right) }^{ 2 } }$

Consider the far field case (i.e the case where

$r>>a$ )

Then one can see that

${ r }^{ 2 }>>{ a }^{ 2 }$ and that

${ r }^{ 2 }-{ a }^{ 2 }$ approaches

${ r }^{ 2 }$ , which allows us to simply simplify the equation in the far field.

$E=\dfrac { 1 }{ 2\pi { \epsilon }_{ 0 } } \dfrac { P }{ { r }^{ 3 } }$

$=\dfrac { 1 }{ 2\pi { \epsilon }_{ 0 } } \dfrac { \lambda P }{ { r }^{ 2 } }$

A cone of base radius R and height

$h$ is located in a uniform electric field

${\vec E}$ parallel to its base. The electric flux entering the cone is

0%
$EhR$
0%
$\frac{1}{2}EhR$
0%
$4EhR$
0%
$2EhR$

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 13 - 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 13 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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