MCQ Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 9

A charge

$+q$ having mass

$m$ is released from rest in a uniform electric field

$E$ momentum acquired by the charge after time

$t$ is:

0%
$\frac{{{q^2}{E^2}{t^2}}}{{2m}}$
0%
$\dfrac{q Et}{m}$
0%
$q Et$
0%
None of these.

Explanation

initial velocity

$u=0,$

acceleration

$a=qE/m,$

time

$t=t,$

final velocity

$v=?$

$v=u+at,$ we get

$v = 0 + \left( {\frac{{qE}}{m}} \right)t$

$or\,v = \frac{{qEt}}{m}$

so the final kinetic energy is

$K\left( t \right) = \frac{1}{2}m{\left( {\frac{{qEt}}{m}} \right)^2}$

$\Rightarrow K\left( t \right) = \frac{{{q^2}{E^2}{t^2}}}{{2m}}$

Hence,

option

$(A)$ is correct answer.

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

0%
positive charge
0%
negative charge
0%
both positive and negative charge
0%
no charge

For a dipole, the value of each charge is

$10^{-10}\ stat\ coulomb$ and separation is

$1A^0$ , then its dipole moment is :

0%
$1 debye$
0%
$2 debye$
0%
$10^{-3} debye$
0%
$3\ \times\ 10^{-20} debye$

Explanation

Given,

$q=10^{-10} stat\, coulomb$

$d=1\times 10^{-10}=10^{-8}cm$

Dipole moment,

$P=qd=10^{-10}\times 10^{-8}=10^{-18}esu\, cm$

$P=1debye$

The correct option is A.

An extremely long wire is uniformly charged. An electron is revolving around the wire and making

${ 10 }^{ 8 }$ revolutions per second in an orbit of radius

$2cm$ . Linear charge density of the wire is nearly ?

0%
$50 nC/m$
0%
$25 nC/m$
0%
$40 nC/m$
0%
$12.5 nC/m$

Explanation

Given,

$\omega=2\pi\times 10^8 rad/s$

$r=2\times 10^{-2}m$

We know,

Electric field=

$E=\dfrac{2k\lambda}{r}$

$F=qE=m\omega^2r$

$\dfrac{2ke\lambda}{r}=m\omega^2r$

$\lambda=\dfrac{m\omega^2r^2}{2ke}$

$\lambda=\dfrac{9.1\times 10^{-31}\times 4\pi^2 \times 10^{16}\times 4\times 10^{-4}}{2\times 9\times 10^{9}\times 1.6\times 10^{-19}}$

$\lambda=50nC/m$

Option

$\textbf A$ is the correct answer

Charges

$\theta_1$ and

$\theta_2$ lie inside and out side respectively for a closed surface s. Let E be the field at a point on S and

$\phi$ be flux of E over S

0%
If only $\theta_1$ changes, both E & $\phi$ charges
0%
If only $\theta_2$ changes, E will change but $\phi$ will not change
0%
If $\theta_1=0, \theta_2\neq 0$ then $E \neq 0$ but $\phi = 0$
0%
If $\theta_1=0, \theta_2\neq 0$ then $E = 0$ but $\phi \neq 0$

Explanation

Charges

${ \theta }_{ 1 }$ and

${ \theta }_{ 2 }$ lie inside and out side respectively.

For a closed surface

$=S$

Let

$E=$ the field at a point on

$S$ and

$\varphi$ be flux of

$E$ over

$S$ .

In this context if only

${ \theta }_{ 1 }$ changes, both

$E$ and

$\varphi$ charges and if any

${ \theta }_{ 2 }=$ changes,

$E$ will change but

$\varphi$ will not change.

${ \theta }_{ 1 }=0$ and

${ \theta }_{ 2 }\neq 0$

$E\neq 0$ but

$\varphi =0$

This point is to be noted.

A point charge

$q$ is placed at a distance

$\dfrac {a}{2}$ above the center of a square surface of side

$a$ . The electric flux through the square is:-

0%
$\dfrac {q}{\epsilon_{0}}$
0%
$\dfrac {q}{\pi\ \epsilon_{0}}$
0%
$\dfrac {q}{4\ \epsilon_{0}}$
0%
$\dfrac {q}{6\ \epsilon_{0}}$

Explanation

$\textbf{Step 1: Gauss Law}$

We are given that a charge is placed at the centre of a square of side

$a$ at distance

$a/2$ :

$\bullet$ Let us consider a cube that is enclosed and the square having a point charge at the centre.

$\bullet$ A cube has six surfaces. Therefore flux is passing through six surface.

By using Gauss's law

$\Phi=\dfrac{Q{in}}{\epsilon_0}$

$\{Q_{in}=Q_{inclosed}\}$

$Q_{in}=q$

$\Phi=\dfrac{q}{\epsilon_0}$

$\Phi$ is the flux electric flux passing through cube.

$\textbf{Step 2: Electric flux through each side of the cube:}$

$\Phi'=\dfrac{\Phi}{6}$

$\Phi'=\dfrac{q}{6\in_0}$

Hence option

$D$ is correct

An electron enters an electric field with its velocity in the direction of the electric lines of field then:

0%
the path of the electron will be a circle
0%
the path of the electron will be a parabola
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the velocity of the electron will decrease just after entry
0%
the velocity of the electron will increase just after entry

In a region of space, the electric field is given by

$\vec {E} = 8\hat {j} + 4\hat {j} + 3\hat {k}$ . The electric flux through a surface of area of

$100$ units in

$x - y$ plane is?

0%
$800\ units$
0%
$300\ units$
0%
$400\ units$
0%
$1500\ units$

Explanation

$\vec {E} = 8\hat {i} + 4\hat {j} + 3\hat {k}, \vec {A} = 100\hat {k}$ .

Electric flux

$\phi=\vec E\cdot A=(8\hat I+4 \hat j+3\hat k)\cdot 100\hat k=300$

Force of interaction between two co-axial short electric dipoles whose centres are R distance apart varies as :

0%
$\frac{1}{R}$
0%
$\frac{1}{R^{2}}$
0%
$\frac{1}{R^{3}}$
0%
$\frac{1}{R^{4}}$

The electric potential is directly proportional to from the region. The intensity of electric field

0%
Is proportional to $r^{2}$
0%
Is changing with uniform rate
0%
Is uniform
0%
Is non-uniform

A small ball of mass m carrying a charge of Q is dropped in a uniform horizontal magnetic field B . Depth of deepest point of its path from point release is h. choose the correct option:

0%
Speed at deepest point is $\sqrt { 2gh }$
0%
Speed at deepest point is $\frac { QBh }{ m }$
0%
Speed at deepest point is $\frac { 2mg }{ QB}$
0%
Speed at deepest point is $\frac { 2QBh }{ m}$

Explanation

In this contest mass

$=M$ , charge

$=Q$ and momentum filed

$=B$ height

$=h$

Now, speed is independent of mass, charge and magnetic field

now,

$\dfrac{1}{2} mV^2 = mgh$

$V=\sqrt { 2gh }$

A particle of mass

$m$ and charge

$+q$ approaches from a very large distance towards a uniformly charged ring of radius

$R$ and charge, mass same as that of particle, with initial velocity

$v_0$ along the axis of the as shown in the figure. What is the closet distance of approach between the ring and the particle? Assume the space to be gravity free and frictionless.

0%
$\sqrt{\dfrac{q^4}{\pi^2 \varepsilon_{0}^{2} m^2 v_{0}^{2}} - R^2}$
0%
$\sqrt{\dfrac{3q^4}{2\pi^2 \varepsilon_{2}^{0} m^2 v_{4}^{0}} - R^2}$
0%
$\sqrt{\dfrac{m^2 v_{4}^{0}}{2\pi^2 q^4 \varepsilon_{2}^{0}} - R^2}$
0%
$\sqrt{\dfrac{q^4}{4\pi^2 \varepsilon_{2}^{0} m^2 v_{4}^{0}} - R^2}$

Explanation

A mass of particles

$=m$

Charge

$=+q$

radius

$=R$

Velocity

$={ v }_{ 0 }$

$d=?$

Now, Using Coulomb's law

$F=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { a }^{ 2 } }{ { r }^{ 2 } } -{ R }^{ 2 }$

or,

${ F }^{ 2 }=\dfrac { { q }^{ n } }{ { \pi }^{ 2 }{ \epsilon }_{ 0 }^{ 2 }{ v }^{ 2 } } -{ R }^{ 2 }$

or,

$\left( ma \right) =\sqrt { \dfrac { { q }^{ 4 } }{ { \pi }^{ 2 }{ \epsilon }_{ 0 }^{ 2 }{ v }_{ 0 }^{ 2 }{ m }^{ 2 } } -{ R }^{ 2 } }$

1238482_1130904_ans_4d43e713cf8247af8146afe1bed746dc.png

A particle of mass

$m$ and positive charge

$q$ is projected towards an infinitely long line of a charge (having linear density of charge

$+\lambda$ ) from a distance

$t_0$ . The direction of initial velocity

$v_0$ makes an angle

$30^\circ$ with the normal to the line of charge as shown in figure. The minimum distance of approach of the charge particle with the line of charge will be (neglect gravity). Take

$\lambda = \dfrac{\pi \varepsilon_0 mv_0^2}{4q}$ .

0%
$\dfrac{r_0}{e}$
0%
$\dfrac{r_0}{e^2}$
0%
$\dfrac{r_0}{e^3}$
0%
$\dfrac{r_0}{2}$

Explanation

Particle of mass

$=m$ positive charge

$=\phi$

$\lambda =\dfrac { \pi { \epsilon }_{ 0 }m{ V }_{ 0 }^{ 2 } }{ 4q } \quad ;\quad \theta ={ 30 }^{ 0 }$

distance

$={ t }_{ 0 }$

initial velocity

$={ V }_{ 0 }$

the

$d=\dfrac { { r }_{ 0 } }{ { e }^{ 3 } } \cos{ 30 }^{ 0 }$

$=\dfrac { { r }_{ 0 } }{ { e }^{ 3 } } .\dfrac { 1 }{ 2 }$

$\bullet \dfrac { 1 }{ 2 }$

$2d=\dfrac { { r }_{ 0 } }{ { e }^{ 3 } }$

1238395_1130874_ans_c77cab38e44e4ea0a7d4c29d150441be.png

The electric field at a point is:

0%
Always continuous
0%
Continuous if there is no charge at that point.
0%
Discontinuous only if there is a negative charge at that point.
0%
Discontinuous if there is a charge at that point.

Explanation

The electric field in a region is always continuous provided that there is no charge at the point of consideration. If a charge is present at that point, then the field is discontinuous there,

Option

$(B, D)$ are correct

A point charge + Q is positioned at the centre of the base of square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is

0%
$Q/16\varepsilon_0$
0%
$Q/4\varepsilon_0$
0%
$Q/8\varepsilon_0$
0%
none

The electric dipole moment of an electron and a proton

$4.3\ nm$ apart is

0%
$6.88\times 10^{-28}Cm$
0%
$2.56\times 10^{-29} C^{2}/ m$
0%
$3.72\times 10^{-14} C/m$
0%
$13.76\times 10^{-29} C^{2}/m$

A charged body has an electric flux F associated with it . Now if the body is place inside a conducting shell then the electric flux outside the shell is:

0%
zero
0%
Greater than F
0%
Less than F
0%
Equal to F

Two coaxial coils are very close to each other and their mutual inductance is

$5$ mH. If a current

$50\sin 500r$ is passed in one of the coils then the peak value of induced e.m.f. in the secondary coil will be?

0%
$5000$ V
0%
$500$ V
0%
$150$ V
0%
$125$ V

Explanation

by the principle of mutual inductance we know that flux induced in coil 2 will depends on the current flowing in coil 1

$\phi _{ 2 }=Mi_{ 1 }$

M = mutual inductance

$i1$ = current in first coil

now by Faraday's law induced EMF in 2nd coil will be

$\dfrac { d\phi }{ dt } =EMF$

$EMF=M\dfrac { di_{ 1 } }{ dt }$

now plug in value of current and mutual inductance

$EMF=5\times 10^{ -3 }\dfrac { d }{ dt } (50sin500t$

$EMF=5\times 10^{ -3 }50\times 500\times cos500t$

$EMF=125cos500t$

so above is the equation of induced EMF

so maximum value of induced EMF will be

$EMF_{ max }=125volts$

Two conducting spheres of radii

$r_!$ and

$r_2$ have same electric field near their surfaces. The ratio of their electric potentials is :-

0%
$r_1^2 / r_2^2$
0%
$r_2^2 / r_1^2$
0%
$r_1 / r_2$
0%
$r_2 / r_1$

Explanation

Two conducting spheres of radii

${ r }_{ 1 }$ and

${ r }_{ 2 }$

$E$

$\alpha$

$\dfrac { 1 }{ { r }^{ 3 } } \widehat { r }$

$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { { r }_{ 2 }^{ 2 } }{ { r }_{ 1 }^{ 2 } } \times \dfrac { { \widehat { r } }_{ 1 } }{ { \widehat { r } }_{ 2 } }$

or,

$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { { r }_{ 2 } }{ { r }_{ 1 } }$

Eight point charges (can be assumed as small spheres uniformly charged and their centres at the corner of the cube) having values q each are fixed at vertices of a cube. The electric flux through square surface ABCD of the cube is

0%
$\frac {q}{24 \in_0}$
0%
$\frac {q}{12 \in_0}$
0%
$\frac {q}{6 \in_0}$
0%
$\frac {q}{8 \in_0}$

Explanation

Using Gauss law we said that,

$\phi =\dfrac { q }{ { \epsilon }_{ 0 } }$

Since, all the c/s are in the

$8$ corners, Then,

$\phi =\dfrac { 1 }{ 8 } \dfrac { q }{ { \epsilon }_{ 0 } }$

1238281_1185561_ans_affb7e0de75d45b49b60cc31c0f7a712.jpg

Following figure shows four situations in which positive and negative charges moves horizontally through a region and gives the rate at which each charge moves. Rank the situations according to the effective current through the region greatest first

0%
$i = i i = i i i = i v$
0%
$i > i i > i i i > i v$
0%
$i = i i = i i i > i v$
0%
$i = i i = i i i < i v$

Explanation

For figure (i)

$i_1=7A$

For figure (ii)

$i_2=4+3=7A$

For figure (iii)

$i_3=5+2=7A$

For figure (iv)

$i_4=6-1=5A$

A molecule with a dipole moment p is placed in electric field of strength E. Initially the dipole is aligned parallel to the field. If the dipole is to be rotated to be anti parallel to the field, the work required to be done by an external agency is

0%
-2pE
0%
-pE
0%
pE
0%
2pE

Explanation

$\textbf {Hint :}$ When a dipole moment placed in an external electric field then potential energy

$U=-pEcos\theta$

$\textbf{Step 1: Calculating initial and final potential energy}$

Initially the direction of dipole is parallel to the electric field

$U_i=-pEcos\theta = -pEcos0^o= -pE$

Finally the direction of dipole is anti parallel to the electric field

$U_i=-pEcos\theta = -pEcos180^o= pE$

$\textbf{Step 1: Calculating work done}$

As we know, workdone = change in potential energy

$W = \Delta U$

$\Rightarrow$

$W = U_f - U_i = pE-(-pE)=2pE$

${\textbf{Correct option: D}}$

Electric ield on the axis of small electric dipole at a distance r is

$\vec{E_1}$ and

$\vec{E_2}$ at a distance

$2$ r on a line of perpendicular dissector. Then

0%
$\vec{E_2}=-\dfrac{\vec{E_1}}{8}$
0%
$\vec{E_2}=-\dfrac{\vec{E_1}}{16}$
0%
$\vec{E_2}=-\dfrac{\vec{E_1}}{4}$
0%
$\vec{E_2}=\dfrac{\vec{E_1}}{8}$

Explanation

Electric field on the axis of small electric diapole at a distance r

$\overrightarrow{E_1}$ and

$\overrightarrow{E_2}$ at a distance

$=2r$

On a line of perpendicular dissector

Then,

Solution

$E_{axis}=\cfrac{2KP}{r^2}(along\quad \overrightarrow{P})\\E=\cfrac{1}{4\pi E_0}\cfrac{2Pr}{(r^2-l^2)^2}\\ \quad=\cfrac{1}{4\pi E_0}\cfrac{2Pr}{r^4}\\ \quad=-\cfrac{E_{axis}}{8}$

An electron at rest has a charge of

$1.6\ \times 10^{-19}\ C$ . It starts moving with a velocity

$v=c/2$ , where

$c$ is the speed of light, then the value the new charge on it is-

0%
$1.6\ \times 10^{-19}$ Coulomb
0%
$1.6\ \times 10^{-19}\sqrt {1-\left(\dfrac {1}{2}\right)^{2}}$ Coulomb
0%
$1.6\ \times 10^{-19}\sqrt {\left(\dfrac {1}{2}\right)^{2}-1}$ Coulomb
0%
$\dfrac {1.6\ \times 10^{-19}}{\sqrt {1-\left(\dfrac {1}{2}\right)^{2}}}$ Coulomb

Explanation

$\textbf{From the property of invariance of electric charge :}$ The magnitude of the charge on a body does not vary with speed or frame of reference.

i.e Magnitude of Charge at rest = Charge in motion

$q_{rest}=1.6\times 10^{-19}C$ (given)

Therefore electron charge in motion will also be same

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

Hence, Option(A) is correct.

An infinite cylinder of radius 'R' carrying charge density .

$p = ar +br^{2}$ where 'r' distance of point from the axis and a b are non-zero constant. Find the ratio of a/b if field out of the cylinder is zero

0%
$\frac{R}{4}$
0%
- $\frac{R}{4}$
0%
- $\frac{3R}{4}$
0%
$\frac{3R}{4}$

Three charger

$q. Q$ and

$4q$ are placed in a straight, line of length

$L$ at points distant

$0, L/2$ and

$L$ respectively from one end in other to make the net force on

$q$ zero the charge

$Q$ must be eqyal to:

0%
$-q$
0%
$2q$
0%
$-\dfrac{q}{2}$
0%
$q$

Explanation

Three charges

$q,Q$ and

$4q$ .

Length

$=L$

Points distant.

$0,\dfrac { L }{ 2 } ,L$ respectively.

force on

${ q }_{ 0 }$

$Q=-q$ (opposite and cancel to each other).

Two charged plates with charges

$Q$ and

$3Q$ are placed

$d$ distance apart. Another plate with charge

$Q$ is placed between two plates.
Final charges

$q_{3}$ and

$q_{5}$ on the surface of plates are

0%
$\dfrac{-Q}{2},\dfrac{-3Q}{2}$
0%
$\dfrac{-Q}{2},\dfrac{-5Q}{2}$
0%
$\dfrac{-3Q}{2},\dfrac{-5Q}{2}$
0%
$\dfrac{-3Q}{2},\dfrac{-3Q}{2}$

Explanation

The charge

$Q$ and

$3Q$

distance

$=d$

another plate with charge

$=Q$

Final charges

${ q }_{ 3 }$ and

${ q }_{ 5 }$ on the surface of plates are

$\rightarrow$

${ q }_{ 3 }=\dfrac { -Q }{ 2 }$

${ q }_{ 5 }=\dfrac { -3Q }{ 2 }$

1238287_1191093_ans_f9824fdacaa24f38b09532bb0df9a6c9.jpg

8 four point charge of +

$10^{-7}C,-10^{-7},C,-2\times 10^{-7}$ C and +

$2\times 10^{-7}$ are placed respectively at the corner A, B,C, of a 0.05 m square. Find the magnitude of the resultant force on the charge at D.

0%
0.2 dyne
0%
0.2 newton
0%
2 dyne
0%
0.02 newton

Explanation

$8$ four point charges of

Four charges are placed at corners of square at side length

$O$ ,

$0.5m$ and a charge is placed at centre

$(O)$ of square as shown in the fig.

Forces of repulsion on

$1\mu C$ charges at

$O$ due to

$2\times { 10 }^{ -7 }C$ charge at

$A$ and

$C$ are equal and opposite directions.

${ F }_{ C0 }={ F }_{ A0 }=\dfrac { K\left( { 10 }^{ -7 }\times 2\times { 10 }^{ -7 } \right) }{ { \left( \sqrt { { 0.05 }^{ 2 }+{ 0.05 }^{ 2 } } \right) }^{ 2 } } =0.02N$

Therefore force are cancel to each other.

1238686_1192146_ans_419604980da2451a82abcaed07504409.png

A charge +Q is located in space at point

$(x=1 m,y=10m, z=5m)$ . What is the total electric flux that passes through the

$y-z$ plane?

0%
$\cfrac Q {\varepsilon_0}$
0%
$\cfrac Q {3\varepsilon_0}$
0%
$\cfrac Q {6\varepsilon_0}$
0%
$\cfrac Q {2\varepsilon_0}$

Explanation

A charge

$+Q$

Point (

$x=1m$ ,

$y=10m$ ,

$z=5m$ )

Using the Gaussian theorem,

total electric flux

$=\phi =\dfrac { Q }{ { \epsilon }_{ 0 } }$

Flux through the rectangular piece of area 10 cm by 20 cm when placed ( as shown in an uniform electric field of 200 N/C, ) is

0%
4.0 $Nm^{2}/C$
0%
3.5 $Nm^{2}/C$
0%
2.5 $Nm^{2}/C$
0%
zero

Explanation

$E=3\times { 10 }^{ 3 }N/C$

and square is parallel to

$y-z$ plane, So, flux will be simply one multiplication of area of electric field which is as follows.

$=3\times \left( { 10 }^{ 3 } \right) \times 20\times \left( { 10 }^{ -9 } \right)$

$=3Vm$

1240039_1226258_ans_799be431c5d649c3ad5a7cbe50aebb8d.png

A dipole of dipole moment P is placed at a distance r form a point change Q(as shown in figure).Choose the incorrect statement.

0%
Torque acting on the dipole is zero
0%
Force acting on the dipole due to the electrie field produce by Q is zero
0%
potential energy of the dipole due to the point charge Q is $\frac { Qp }{ { 4\pi }{ \varepsilon }_{ 0 }{ r }^{ 2 } }$
0%
Force acting on the dipole due to the point charge Q is $\frac { Qp }{ { 4\pi }{ \varepsilon }_{ 0 }{ r }^{ 3 } }$

Figures below show regular hexagons, with charges at the vertices. In which case is the electric filed at the center zero?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Four charges

$+Q,-Q,+Q$ and

$-Q$ are situated at the corners of a square; in a sequence then at the centre of the square:

0%
$E=0,V=0$
0%
$E=0,V\neq 0$
0%
$E\neq 0,V=0$
0%
$E\neq 0,V\neq 0$

Explanation

Four charges

$+Q$ ,

$-Q$ ,

$+Q$ ,

$-Q$ .

In this context,

$E\neq 0$ but

$V=0$ because.

$E$ is not cancel out to each other but

$V$ is cancel out each other.

To construct an air filled capacitor which can store

$12 \mu C$ of charge, What can be the minimum plate area of the capacitor? (Dielectric strength of air is

$3\times 10^6 V/m$ )

0%
$0.25 m^2$
0%
$0.45 m^2$
0%
$4.5 m^2$
0%
$0.35 m^2$

Two equally charged identical metal spheres A and B repel other with a force of

$2 \times {10^{ - 5}}$ .Another identical uncharged sphere C is touched to B and then placed at the mid point between A and B.The net electric force on C is.

0%
$2 \times {10^{ - 5}}\;N\;along\;\overrightarrow {AB}$
0%
$2 \times {10^{ - 5}}\;N\;along\;\overrightarrow {BA}$
0%
$4 \times {10^{ - 5}}\;N\;along\;\overrightarrow {AB}$
0%
$4 \times {10^{ - 5}}\;N\;along\;\overrightarrow {BA}$

Explanation

$F_{AB}=\cfrac{KQ^2}{r^2}=2\times10^{-5}$

Now,

'C' is touched to 'B' and both are identical so the charge on 'B' will be equally distributed on both 'B' and 'C' i.e,

So,

$F_{AC}=\cfrac{K(Q)(Q/2)}{(r/2)^2}\\ \quad=\cfrac{2KQ^2}{r^2}\\ F_{BC}=\cfrac{K(Q/2)(Q/2)}{(r/2)^2}\\ \quad=\cfrac{KQ^2}{r^2}$

So, Net force on 'C' is given by

$\Rightarrow F_{AC}-F_{BC}\\ \Rightarrow \cfrac{2KQ^2}{r^2}-\cfrac{KQ^2}{r^2}\\ \Rightarrow \cfrac{KQ^2}{r^2}=F_{AB}\\ \Rightarrow 2\times 10^{-5}N\quad along \quad\overrightarrow{AB}$

1198073_1249820_ans_26db9d70bea44761ad710561ab713b67.png

A light beam travelling in the x-direction is described by the electric field

${ E }_{ y }=(300V{ m }^{ -1 })sin\quad \omega (t-x/c).$ An electron is constrained to move along the y-direction with a speed of

$2.0\times {10}^7 { m }^{ -1 }$ Find the maximum electric force and the maximum magnetic force on the electron

0%
$4.8\times {10}^{-17} N, zero$
0%
$4.2\times {10}^{-18} N, 1.8\times 10^{-8}N$
0%
$4.8\times {10}^{-17} N, 3.2\times 10^{-18}N$
0%
Zero, Zero

Explanation

A light beam travelling

${ E }_{ y }=\left( 300V/m \right) sinw\left( t-x/c \right)$ .

$V=2.0\times { 10 }^{ 7 }{ m }^{ -1 }$

${ B }_{ 0 }=\dfrac { { F }_{ 0 } }{ C } =\dfrac { 300 }{ 2\times { 10 }^{ 7 } } =150\times { 10 }^{ -7 }T$

${ F }_{ m }={ B }_{ 0 }qV$

$=9.1\times { 10 }^{ -31 }\times 150\times { 10 }^{ -7 }\times 2\times { 10 }^{ 7 }$

$=2.8\times { 10 }^{ -28 }\times 2$

$=4.8\times { 10 }^{ -17 }N$

magnetic force

$=0$

because they are perpendicular to each other.

Two charges Q and -2Q are placed at some distance. the locus of points in the plane of the charges where the potential is zero will be

0%
Straight line
0%
Circle
0%
Parabola
0%
ellipse

Explanation

Two charges

$Q$ ,

$-2Q$

some distance

Potential

$=0$

hence,

$\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \times \dfrac { Q\times \left( -2Q \right) }{ R } =0$

When we use it the parabolic condition then,

${ y }^{ 2 }=4ax\quad \longrightarrow \left( 1 \right)$

Now,

$\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \times \dfrac { Q\times 2Q }{ R } =0$

$\dfrac { { 2Q }^{ 2 } }{ R } =0\quad \longrightarrow \left( 2 \right)$

Hence, equating

$(1)$ and

$(2)$ and we get, the system is getting parabola.

Two or more cells joined together form a...................

0%
Dry cell
0%
Battery
0%
Transformer
0%
MCB

The charges 1,2,3 are moving in uniform transverse magnetic field then :-

0%
Particle 1 positive and particle 3 negative
0%
Particle 1 negative and particle 3 positive
0%
Particle 1 negative and particle 2 neutral
0%
Particle 1 and 3 are positive and particle 2 neutral

A ring of radius R is charged uniformly with a charge +Q. The electric field at a point on its axis at a distance r from any point on the ring will be:-

0%
$\dfrac{{KQ}}{{\left( {{r^2} - {R^2}} \right)}}$
0%
$\dfrac{{KQ}}{{{r^2}}}$
0%
$\dfrac{{KQr}}{{{r^3}}}{\left( {{r^2} - {R^2}} \right)^{\dfrac{1}{2}}}$
0%
$\dfrac{{KQr}}{{{R^3}}}$

Explanation

form Charge Q, at distance x ,electric field:

$E = \dfrac{kQ}{x^2}$ ---eq.1

$r^2 = {R^2 + x^2}$

from eq.1

$E = \dfrac{kQ}{r^2-R^2}$

Hence option 'A' is correct.

The electrical force is a:

0%
contact force
0%
force at distance
0%
consequential force
0%
none of these

Explanation

The electrical force is

$\rightarrow$

Consequential force.

Consequential force is defined that consequential forces are the forces which occur as a result of some action on a body force of friction is a consequential force because its acts as a result of an applied contact force. This force always oppose the motion of the bodies and ultimately stops them.

The electric field intensity due to a dipole of 10 cm and having a charge of

$500 \mu C$ , on the axis at a distance

$20 cm$ .

0%
$6.25 \times 10^7 N/C$
0%
$6.28 \times 10^7 N/C$
0%
$1.31 \times 11^{11} N/C$
0%
$20.5 \times 10^7 N/C$

Explanation

Lenght of dipole

$=10\ cm=0.1\ m$

Change on dipole

$q=500\mu C$

$=500\times 10^{-6}\ C$

Distance of the point on the axis of the midpoint of the dipoler

$=20+5=25\ cm$

$=0.25\ m$

$E=\dfrac{1}{4\pi \varepsilon_0} \dfrac{2q.2lr}{(r^2-l^2)^2}=$

$E=9\times 10^9\times \dfrac{2(500\times 10^{-6}\times 0.1)\times 0.25}{[(0.25)^2-(0.05)^2]^2}$

$E=\dfrac{225\times 10^3}{3.6\times 10^{-3}}=6.25\times 10^2\ N/C$

1346267_1239670_ans_f20955ce8a3c4360a184f6d5d7e77eb3.PNG

The total probability of finding a particles in space under normalized condition according to quantum mechanics is

0%
Zero
0%
Infinity
0%
One
0%
Uncertain

Explanation

The total probability of finding a particle in space under normalized condition is

$'1'.$

Hence,

option

$(C)$ is correct answer.

Figure shows an electric line of force which curves along a circular arc. The magnitude of electric field intensity is same at all points on this curve and is equal to E. If the potential at A is V, then the potential at B is:

0%
$V-ER\theta$
0%
$V-2ER sin \dfrac {\theta} {2}$
0%
$V+ER\theta$
0%
$V+2ER sin \dfrac {\theta} {2}$

Explanation

In the given figure.

Distance between

$AB$

$d=2R\sin \dfrac{\theta }{2}$

change in potential

$=V - 2Ed$

$=V - 2ER\sin \dfrac{\theta }{2}$

Hence,

option

$(B)$ is correct answer.

Which of the following is correct for the case of an uniform electric field?

0%
All points are at the same potential
0%
No two points can have the same potential
0%
Pairs of points separated by the same distance must have the same difference in potential
0%
None of the above

A hemispherical surface of radius R is kept in a uniform electric field E as shown in figure. The flux through the curved surface is:

0%
$E2\pi R^2$
0%
$E\pi R^2$
0%
$E4\pi R^2$
0%
$Zero$

Explanation

Flux passing through the curved surface is equal to the flux passing through the flat surface.

Since flux through the flat surface

$\phi_{flat} = E\times Area = E\times \pi R^2$

Flux through the curved surface

$\phi_{curved} = \phi_{flat} = E\times \pi R^2$

Correct answer is option B.

A small sphere of radius

$r_1$ and charge

$q_1$ is enclosed by a spherical shell of radius

$r_2$ and

$q_2$ . charged sphere and shell are connnected by a wire. if

$q_1$ is positive then

0%
charge will flow from sphere to shell
0%
charge will flow from shell to sphere
0%
change flow will depend on magnitude of $q_2$
0%
change flow will depend on magnitude of $q_1$

Explanation

We know that potential difference doesn't depends on

$q_2$ we can thus conclude that the charge will always flow from the sphere to shell, no matter whatsoever charge is on the shell.

Electric charges q, q, -2q are placed at the corners of an equilateral triangle ABC of sideThe magnitude of electric dipole moment of the system is

0%
${ q }^{ 1 }$
0%
${ 2q }^{ 1 }$
0%
${ \sqrt { 3 } q }^{ 1 }$
0%
${ 4q }^{ 1 }$

Explanation

Electric field

$q,q,-2q$ are placed at the corners of an equilibrium.

Total charge

$=\sqrt { 2+1 } =\sqrt { 3 } q$

The magnitude of electric dipole momentum of this system

$=\sqrt { 3 } q\times l=\sqrt { 3 } ql$ .

Electric field lines in a certain region are shown.

0%
${E}_{x}={E}_{y}={E}_{z}$
0%
${E}_{x}<{E}_{y}<{E}_{z}$
0%
${E}_{x}>{E}_{y}>{E}_{z}$
0%
${E}_{x}={E}_{y}<{E}_{z}$

Explanation

Here,

${E_Z} = {E_y} = {E_z}$

Option

$A$ is correct answer to given figure satisfy.

$\begin{array} { l } { \text { Three infinitely long charges sheets are placed as shown in figure. The electric field at } } \\ { \text { point } P \text { is } } \end{array}$

0%
$\frac { 2 \sigma } { \varepsilon _ { 0 } } \hat { k }$
0%
$- \frac { 2 \sigma } { \varepsilon _ { 0 } } \hat { k }$
0%
$\frac { 4 \sigma } { \varepsilon _ { 0 } } \hat { k }$
0%
$- \frac { 4 \sigma } { \varepsilon _ { 0 } } \hat { k }$

Explanation

$\begin{array}{l} E=\frac { \sigma }{ { 2{ \in _{ 0 } } } } \left( { -\widehat { k } } \right) +\frac { { 2\sigma } }{ { 2{ \in _{ 0 } } } } \left( { -\widehat { k } } \right) +\left( { \frac { { -\sigma } }{ { 2{ \in _{ 0 } } } } } \right) \widehat { k } \\ =\frac { { 2\sigma } }{ { { \in _{ 0 } } } } \left( { -\widehat { k } } \right) \end{array}$

Hence,

option

$(B)$ is correct answer.

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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