MCQ Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 14

A dipole is kept at origin along y-axis. As one moves from

$A$ to

$B$ along the curve, the direction of the electric field changes from negative y-direction to positive y-direction. The angle

$\theta$ (with the dipole moment) at which y-component of electric field is zero is

0%
$45^{o}$
0%
$\tan^{-1}{\dfrac{1}{2}}$
0%
$\tan^{-1}{\sqrt{2}}$
0%
$\tan^{-1}{{B}}$

Work done in turning dipole through an angle

$60^{o}$ is

0%
$zero$
0%
$pE/4$
0%
$pE$
0%
$pE/2$

An isosceles right angle triangle of side d is placed in a horizontal plane. A point charge q is placed at a distance d vertically above from one of the corner as shown in the figure. Flux of electric field passing through the triangle is

0%
$\dfrac{q }{36\varepsilon_0}$
0%
$\dfrac{q }{18\varepsilon_0}$
0%
$\dfrac{q }{24\varepsilon_0}$
0%
$\dfrac{q }{48\varepsilon_0}$

Explanation

Let there is a cube of edge d with

$ABCD$ as on face and charge a one corner

$\therefore \phi = \left( {\frac{q}{{6{ \in _0}}}} \right).\frac{1}{3}.\frac{1}{2}$

$\phi = \frac{q}{{36{ \in _0}}}$

Hence,

option

$(A)$ is correct answer.

1201275_1235294_ans_011955c81aaa4f4d8173de6993e6c1e3.PNG

Three charges

${q_1} = 1 \times {10^{ - 8}},{q_2} = 2 \times {10^{ - 6}},{q_3} = - 3 \times {10^{ - 6}}$ have been placed, as shown in figure, in four surfaces

${S_1},{S_2},{S_3}$ and

${S_4}$ electrical flux emitted from the surface

${S_2}$ in

$N - {m^2}/C$ wil be

0%
$36\pi \times {10^3}$
0%
$-36\pi \times {10^3}$
0%
$36\pi \times {10^9}$
0%
$-36\pi \times {10^9}$

Below figure shows a closed surface which intersects a conducting sphere. if a positive charge is placed at

$P,$ the flux of the electric field through the closed surface

0%
Will remain zero
0%
will be positive
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will be negative
0%
will be undefined

In an electroscope, if aluminium strips are replaced by plastic strips, and a charged object is touched with the metal top of the electroscope:

0%
the plastic strips will diverge
0%
the plastic strips will not diverge
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one strip will move away while other will remain
same.
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the plastic strips will come more closer.

Electric charges

$q,q, -2q$ are placed at the corners of an equilateral triangle

$ABC$ of side

$1$ .The magnitude of electric dipole moment of the system is

0%
$ql$
0%
$2ql$
0%
$\sqrt{3} ql$
0%
$4ql$

Explanation

$\therefore {P_{net}} = \sqrt 3 P$

$= \sqrt 3 q \times a$

$= \sqrt 3 q \times 1$

Hence

$,$ option

$(C)$ is correct

$.$

1195533_1218035_ans_c398c922b70743d3ab02ace1665d9fba.PNG

If the field are parallel to surface then electric flux, linked linked, with the surface will be

0%
Greater than zero
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Equal to zero
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May be less than zero
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Can't determine

A flat square surface with sides of length L is described by the equations

$x = L,0 \le y \le L,0 \le z \le L$ . the electric flux through the square due to a positive point charge

$q$ located at the origin

$\left( {x = 0,y = 0,z = 0} \right)$

0%
$\dfrac{q}{{4{\varepsilon _0}}}$
0%
$\dfrac{q}{{6{\varepsilon _0}}}$
0%
$\dfrac{q}{{24{\varepsilon _0}}}$
0%
$\dfrac{q}{{48{\varepsilon _0}}}$

The electric field just outside a conducting sphere of charge density

$\sigma$ is given by

0%
$\dfrac { \sigma }{ { \varepsilon }_{ 0 } }$
0%
$\dfrac { \sigma }{ { 2\varepsilon }_{ 0 } }$
0%
$\dfrac { 2\sigma }{ { \varepsilon }_{ 0 } }$
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$zero$

A cylinder of length

$l$ , radius

$R$ is kept in the uniform electric field as shown in the figure. If electric field strength is

$E$ , then the outgoing electric flux through the cylinder is

0%
$Zero$
0%
$\pi R^{2}E(l)$
0%
$R/ \pi$
0%
$2R/E$

What will be the total flux through faces of the cube as given in the figure with side of length

$'a'$ if charge

$q$ is placed at a corner of the cube

0%
$\frac { q } { 8 \varepsilon _ { 0 } }$
0%
$\frac { q } { 4 \varepsilon _ { 0 } }$
0%
$\frac { q } { 2 \varepsilon _ { 0 } }$
0%
$\frac { q } { 3 \varepsilon _ { 0 } }$

Consider a hollow sphere of radius

$R$ , Its quarter part is cut and kept that geometrical centre lies at origin as shown in figure.

$x-y$ axis lies in plane of paper and

$z$ -axis is

$\bot$ to the plane of the plane the paper. Sphere is kept in a manner such that plane portions of sphere lies in

$x-z$ and

$y-z$ plane. Uniform electric field

$\vec{E}=E_{0}\hat{i}+2E_{0}\hat{j}+2E_{0}\hat{k}$

$(N/c)$ lies in the complete space Electric flux coming out of the curved surface of sphere is:

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$\dfrac{3}{2}\pi R^{2}E_{0}$
0%
$\dfrac{5}{2}\pi R^{2}E_{0}$
0%
$2\pi R^{2}E_{0}$
0%
$Zero$

Explanation

Magnitude of electric flux entering the XZ plane surface and YZ plane will be equal to electric flux coming out of curved surface.

Flux through XZ plane,

Area vector for XZ plane,

$A = -\dfrac{\pi R^2}{2} \hat{j}$

$\phi_1 = E.A = (E_0\hat{i} + 2E_0\hat{j} +2E_0\hat{k}).(-\dfrac{\pi R^2}{2}\hat{j})$

$\phi_1 = -\pi R^2E_0$

Flux through YZ plane,

Area vector for YZ plane,

$A = -\dfrac{\pi R^2}{2} \hat{i}$

$\phi_2 = E.A = (E_0\hat{i} + 2E_0\hat{j} +2E_0\hat{k}).(-\dfrac{\pi R^2}{2}\hat{i})$

$\phi_2 = -\dfrac{\pi R^2E_0}{2}$

Total flux,

$\phi = \phi_1+\phi_2 = \dfrac{3\pi R^2E_0}{2}$

A charge Q is placed at the centre of the open end of a cylindrical vessel of radius R and height 2R as shown in figure. the flux of the electric field through the surface ( curved surface + base ) of the vessel is

0%
$\dfrac { Q }{ \varepsilon_0 }$
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$\dfrac { Q }{ 2 \varepsilon_0 }$
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$\dfrac { Q }{ 4 \varepsilon_0 }$
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$\dfrac { Q }{ \sqrt {5} \varepsilon_0 }$

Explanation

In the figure, we have shown in the figure part a charge

$q$ at the centre of one open end of a cylinder. Note that the other end is closed.

Now, take an another identical cylinder and joint's open end with the open of end of the first end of the first cylinder. Now, we have closed surface of two cylinders.

This electric flux through this closed surface, according to Gauss's according to Gauss's theorem is

$q/{ \epsilon }_{ 0 }$ where

${ \epsilon }_{ 0 }$ is electric permittivity of space.

Then, from the symmetry, we can say that flux through surface of any one of the two joined surface will be

$\dfrac { q }{ 2{ \epsilon }_{ 0 } }$ .

1240358_1252069_ans_30c088078dea48be85645d5b15901ef6.jpg

There exist uniform electric field E= -E in space, four point charges are placed on vertices of a square as shown in the figure. What in minimum work required to exchange positions of point charges placed at(a, 0)and(0, 0)

0%
-4qEa
0%
0
0%
2qEa
0%
4qEa

Explanation

The distance

$=\sqrt { { a }^{ 2 }+0 } =a$ unit

Work done

$=F.d$

$=-4qEa$

$=-4qF$

$F=-4qE$

for four charges

$=q+q+q+q=4q$

$(-Ve)$ sign because electric field in space is

$(-ve)$ in sign

$E=$ Electric filed.

1240354_1258871_ans_8d2ff1e43f0e4381a9781a2122f065c3.png

Charge

$q$ is kept at centre of cylindrical body, flux of circle square of body is

$\phi_{0}$ then flux of curve surface of body is

0%
$\dfrac { q }{ { \varepsilon }_{ 0 } }$
0%
$\dfrac { q }{ { \varepsilon }_{ 0 } } +\pi$
0%
$\dfrac { q }{ { 2\varepsilon }_{ 0 } }+\phi$
0%
$\dfrac { q }{ { \varepsilon }_{ 0 } }-\phi$

Total electric force on an electric dipole placed in an electric field of a point charge is:

0%
Always zero
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never zero
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zero when midpoint of dipole coincides with the point charge
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zero when dipole axis is along any electric line of force.

If the given two identical charged rings lie in

$xy$ plane both having linear charge density

$\lambda$ varies as per

$\lambda = \lambda_0 \cos \theta$ (

$\lambda_0$ = constant) where

$\theta$ is measured from +x-axis. Radius for both the rings is

$R$ . Electric force between the two rings is

$\dfrac{x K \lambda_{0}^{2} \pi^{2} R^{4}}{d^4}$ then

$x$ is.

0%
$6$
0%
$3$
0%
$2$
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$1$

The ratio of the energy required to set up in cube of side 10 cm uniform magnetic field of 4

${ Wb/m }^{ 2 }$ and a uniform electric field of

$10^{ 6 }V/m$ is:

0%
$1.4x{ 10 }^{ 7 }$
0%
$1.4x{ 10 }^{ 5 }$
0%
$1.4x{ 10 }^{ 6 }$
0%
$1.4x{ 10 }^{ 3 }$

Explanation

Side

$=10cm$

magnetic field

$=4wb/{ m }^{ 2 }$

electric field

$={ 10 }^{ 6 }V/m$

$B=E\times r$

or,

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

or,

$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { 5.6\times { 10 }^{ 6 }\times 10 }{ 4 }$

$=\dfrac { 5.6\times { 10 }^{ 7 } }{ 4 }$

$=1.4x{ 10 }^{ 7 }$

At some point in space the electric field is

$5\ NC^{-1}$ . The electric lines of force crossing a unit area placed at right angles to electric field at this point is:-

0%
$\varepsilon_0$
0%
$\dfrac{\varepsilon_0}{4\pi}$
0%
$4 \pi \varepsilon_0$
0%
$5$

Three point charges 2q,-q and -q are located respectively at (0,a,a),(0,a,-a) and (0,0,-a) as shown. The dipole moment of this distribution is:-

0%
2qa in the y-z plane at ${ tan }^{ -1 }\left( \dfrac { 1 }{ 4 } \right)$ with z-axis
0%
$\sqrt { 17 }$ qa in the y-z plane at ${ tan }^{ -1 }\left( \dfrac { 1 }{ 4 } \right)$ with z-axis
0%
$\sqrt { 5 }$ qa in the x-y plane at ${ tan }^{ -1 }(4)$ with y-axis
0%
4qa in the x-y plane at ${ tan }^{ -1 }(4)$ with y-axis

The electric field of a plane polarized electromagnetic Wave in free space at time t = 0 i given by an expression

$\vec{E} (x,y) = 10\hat{j} cos [(6x + 8z)]$
The magenetic field

$\vec{B}$ (x,z,t) is given by ; (c is the velocity of light will be:

0%
$\dfrac{1}{c}(4 \hat{k} + 8 \hat{i}) cos [(2x - 8z + 20ct)]$
0%
$\dfrac{1}{c}(6 \hat{k} - 8 \hat{i}) cos [(6x + 8z - 10ct)]$
0%
$\dfrac{1}{c}(5 \hat{k} + 8 \hat{i}) cos [(6x + 8z - 80ct)]$
0%
$\dfrac{1}{c}(4 \hat{k} - 8 \hat{i}) cos [(6x + 8z + 70ct)]$

Explanation

$\overset{\rightarrow} E = 10\hat{j} cos [(6 \hat{i} + 8 \hat{k}). (x\hat{i} + z\hat{k})]$

$= 10 \hat{j} cos [\overrightarrow{K}.\overrightarrow{r}]$

$\overrightarrow{K} = 6\hat{i} + 8\hat{K}$ ; direction of waves travel. i.e direction of 'c'
Direction of

$\hat{B}$ will be along

$\hat{C} \times \hat{E} = \dfrac{-4\hat{i} + 3\hat{k}}{5}$
Mag. of

$\vec{B} = \dfrac{E}{C} = \dfrac{10}{C}$

$\therefore \overrightarrow{B} = \dfrac{10}{C}\left(\dfrac{-4\hat{i} + 3\hat{k}}{5}\right) = \dfrac{(-8\hat{i} + 6\hat{k})}{c}$

A hollow conducting sphere has inner radii

$5\ cm$ and outer radius

$8\ cm$ then find maximum possible potential of conductor if breakdown electric field of air is

$3 \times 10^6 V/m$

0%
$240\ kV$
0%
$480\ kV$
0%
$960\ kV$
0%
$720\ kV$

A uniform electric field exists in the region as shown. The total electric flux through the cube of side

$l$ , is

0%
$l^{ 2 }E$
0%
$\sqrt { 2l } ^{ 2 }E$
0%
Zero
0%
$6{ l }^{ 2 }E$

A metallic ring is inserted through a sot iron cylinder which is wound with a copper wire and fixed to the wooden base as shown in the diagram when we flow A.C current through the copper wire then metal ring is leviated on the coil because,
Choose the correct option from above and give explanation

0%
The metal ring availed sufficient lifting force equal to its weight by getting repelled by the coil continuously.
0%
the metal ring behaved like a magnet by varying its polarities opposites to that of coil in the same time intervals.
0%
the metal ring is induced with an alternating current by which it opposed the variation of polarities produced in the coil
0%
the variation of flux produced by the coil has induced some alternating current in the metal ring.

A charge

$q$ is placed inside a cube of side

$a$ as shown in the figure.

$O$ is the centre of the face

$ABCD$ and

$q$ is just above the point

$O$ at a distance

$\cfrac{a}{4}$ . The electric flux passing through

$ABCD$ is

0%
$\cfrac{q}{6{\epsilon}_{0}}$
0%
$< \cfrac{q}{6{\epsilon}_{0}}$
0%
$> \cfrac{q}{6{\epsilon}_{0}}$
0%
$\cfrac{q}{{\epsilon}_{0}}$

$E_{a}$ be the electric field strength of a short dipole at a point on its axial line and

$E_{e}$ that on the equatorial line at the same distance, then

0%
$E_{e} = 2E_{a}$
0%
$E_{a} = 2E_{e}$
0%
$E_{a} = E_{e}$
0%
None of the above

A small sphere of mass

$m$ and having charge

$q$ is suspended by a silk thread of length

$l$ in a uniform horizontal electric filed. If it stands at a distance x from the vertical line from point of suspension, then magnitude of electric field is :

0%
$\dfrac{mg}{q}$
0%
$\dfrac{mg}{q} \dfrac{x}{l}$
0%
$\dfrac{mg x}{q\sqrt{l^2 - x^2}}$
0%
$\dfrac{mg l}{q \sqrt{x^2 - l^2}}$

Explanation

$E=?$

the distance

$=\sqrt { { l }^{ 2 }-{ x }^{ 2 } }$

${ E }_{ q }\sqrt { { l }^{ 2 }-{ x }^{ 2 } } =mgx$

where

$mg$ is equivalent to

$E$

now,

or,

$E=\dfrac { mgx }{ q\sqrt { { l }^{ 2 }-{ x }^{ 2 } } }$

A loop of area

$4 m ^ { 2 }$ is placed flat in the x -y plane. There is a constant magnetic field

$4 \hat { j }$ in the region. Find the flux through the loop

0%
2 units
0%
4 units
0%
6 units
0%
0 units

An observer moves past an electron at rest. His instrument measures :

0%
an electric field only
0%
a magnetic field only
0%
both fields
0%
any of the two fields, depending upon his speed.

A charged water drop whose radius is

$0.1\mu m$ is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be

$(g = 10 { m }^{ -1 }$ )

0%
$1.61 N/C$
0%
$26.2 N/C$
0%
$262 N/C$
0%
$1610 N/C$

The charge density of an insulating infinite surface is

$(e/\pi)$

$\mathrm { C/m } ^ { 2 }$ then the field intensity at a nearby point in volt/meter will be-

0%
$2.88 \times 10 ^ { - 12 }$
0%
$2.88 \times 10 ^ { - 10 }$
0%
$2.88 \times 10 ^ { - 9 }$
0%
$2.88 \times 10 ^ { - 19 }$

There is an electric field of

$100N/C$ X-direction . the flux passing through a square of 10 cm sides placed on XY plane inside the electric field is _____

$Nm^2/C.$

0%
$0$
0%
$10$
0%
$1$
0%
$2$

Two infinitely long parallel wires having linear charge densities

${ \lambda }_{ 1 }$ and

${ \lambda }_{ 2 }$ respectively are placed at a distance R. The force per unit length on either wire will be :-

0%
$\dfrac { 2k{ \lambda }_{ 1 }{ \lambda }_{ 2 } }{ { R }^{ 2 } }$
0%
$\dfrac { 2k{ \lambda }_{ 1 }{ \lambda }_{ 2 } }{ { R } }$
0%
$\dfrac { k{ \lambda }_{ 1 }{ \lambda }_{ 2 } }{ { R }^{ 2 } }$
0%
$\dfrac { k{ \lambda }_{ 1 }{ \lambda }_{ 2 } }{ { R } }$

Explanation

${\text{Force on }}l{\text{ length of the wire }}2{\text{ is }}$

${F_2} = Q{E_1} = \left( {{\lambda _2}l} \right)\dfrac{{2k{\lambda _1}}}{R}$

$\Rightarrow \dfrac{{{F_2}}}{l} = \dfrac{{2k{\lambda _1}{\lambda _2}}}{R}$

${\text{ Also }}\dfrac{{{F_1}}}{l} = \dfrac{{{F_2}}}{l} = \dfrac{F}{l} = \dfrac{{2k{\lambda _1}{\lambda _2}}}{R}$

Therefore, option b is correct.

Choose the correct option that shows the rod and fur after charging

Four charges equal to -Q are placed at the four corners of a square and a charge q is at its center. If the system is in equilibrium the value of q is

0%
$-\frac{Q}{4}(1+2\sqrt{2})$
0%
$\frac{Q}{4}(1+2\sqrt{2})$
0%
$-\frac{Q}{2}(1+2\sqrt{2})$
0%
$\frac{Q}{2}(1+2\sqrt{2})$

Electric charges

$q , q , - 2 q$ are placed at the corners of an equilateral triangle

$A B C$ of side

$l.$ The magnitude of electric dipole moment of the system is

0%
$q l$
0%
$2 q l$
0%
$\sqrt { 3 } q l$
0%
$4 q l$

Two point charges

$q _ { 1 } \text { and } q _ { 2 }$ terminates at are kept as shown.
One of the electric field line coming out from

$q _ { 1 }$ makes an angle

$30 ^ { \circ }$ with the line joining

$q _ { 1 } q _ { 2 }$ terminates at

$q _ { 2 }$ makes an angle

$60 ^ { \circ }$ .The ratio

$\frac { q _ { 1 } } { q _ { 2 } }$ is

0%
$\frac { 1 } { 2 - \sqrt { 3 } }$
0%
$-\frac { 1 } { 2 - \sqrt { 3 } }$
0%
$\frac { - 1 } { 2 }$
0%
$2$

Three charges are placed at three corners of an equilateral triangle as shone in figure. The electric potential at the centroid of the triangle will be

0%
$\frac{q}{4\pi \varepsilon _{0}r}$
0%
$\frac{3\sqrt{3}q}{4\pi \varepsilon _{0}r}$
0%
Zero
0%
$\frac{3q}{4\pi \varepsilon _{0}r}$

A gold leaf electroscope is positively charged. When an earthed plate is brought above the electroscope then the changes in charge, potential and capacity will be following -
1.CHARGE POTENTIAL 3.CAPACITY

0%
Increases , Decreases , Remains constant
0%
Remains constant , Increases , Decrease
0%
Remains constant , Decreases , Increases
0%
Decreases , Decreases , Remains constant

The total flux associated with the given cube will be where

$'a'$ is side of the cube:-

$\left(\dfrac{1}{\epsilon_0} = 4\pi \times 9 \times 10^{9}\ SI\ units\right)$

0%
$162\pi \times 10^{-3} Nm^{2}/C$
0%
$162\pi \times 10^{3} Nm^{2}/C$
0%
$162\pi \times 10^{-6} Nm^{2}/C$
0%
$162\pi \times 10^{6} Nm^{2}/C$

Identify the wrong statement in the following:
Coulomb's law correctly describes the electric force that

0%
Binds the electrons of an atom to its nucleus
0%
Binds the protons and neutrons in the nucleus of an atom
0%
Binds atoms together to form molecules
0%
Binds atoms and molecules together to form solids

The shaft of an electric motor starts from rest and on the application of torque, it gains an angular acceleration given by

$\alpha =3t-t^{2}$ during the first

$2$ seconds after it starts after which

$\alpha=0$ . The angular velocity after

$6$ sec will be -

0%
$\dfrac{10}{3}$ rad/sec
0%
$\dfrac{3}{10}$ rad/sec
0%
$\dfrac{30}{4}$ rad/sec
0%
$\dfrac{4}{30}$ rad/sec

Figure shows the electric field lines emerging from any charge configuration. If the electric field at A and B are

$E_A$ and

$E_B$ respectively and if the displacement between A and B is r then:

0%
$E_A$ > $E_B$
0%
$E_A$ < $E_B$
0%
$E_A$ = $\frac{E_B}{r}$
0%
$E_A$ = $\frac{E_B}{r^2}$

${ \varepsilon }_{ 0 }, \mu_0$ and

$c$ represent the relative permittivity of free space, the magnetic permeability of free space and the velocity of light respectively, which of the following combinations of correct?

0%
$c=\cfrac{1}{{ \varepsilon }_{ 0 }, \mu_0}$
0%
$c=\cfrac{1}{{\sqrt{ \varepsilon }_{ 0 }, \mu_0}}$
0%
$c={ \varepsilon }_{ 0 }, \mu_0$
0%
$c=\sqrt{{ \varepsilon }_{ 0 }, \mu_0}$

If the force between two charged objects is to be left unchanged, even though the charge on one of the objects is halved, keeping the other the same, the original distance of separation

$d$ should be changed to:

0%
$d / \sqrt{2}$
0%
$d / 2$
0%
$\sqrt{2} d$
0%
$2d$

An electric dipole is formed due to two particles fixed at the ends of a light rigid rod od length I. The mass of each particle is m and charges are -q and +q The system is suspended by a torsionless thread in an electric filed of intensity E such that the dipole axis is parallel to the filed if it is slightly displaced, the period of angular motion is

0%
$\cfrac {1}{2 \pi} \sqrt {\cfrac {2qe}{ml}}$
0%
$2 \pi \sqrt {\cfrac {ml}{qE}}$
0%
$2 \pi \sqrt {\cfrac {ml}{2qE}}$
0%
$\cfrac {1}{2 \pi} \sqrt {\cfrac {ml}{4qE}}$

A charge q is located at the center of a cube . The electric flux through any face is

0%
$\dfrac { \pi q }{ 6(4\pi \epsilon _{ 0 }) }$
0%
$\dfrac { q }{ 6(4\pi \epsilon _{ 0 }) }$
0%
$\dfrac { 2\pi q }{ 6(4\pi \epsilon _{ 0 }) }$
0%
$\dfrac { 4\pi q }{ 6(4\pi \epsilon _{ 0 }) }$

Electric and magnetic field are directed as

${ E }_{ 0 }\hat { i }$ and

${ B }_{ 0 }\hat { k }$ , a particle of mass m and charge +q is released from position

$(0, 2, 0)$ from rest. The velocity of that particle at

$(x, 5, 0)$ is

$\left( 5\hat { i } +12\hat { j } \right)$ the value of x will be

0%
$\dfrac { 169m }{ 2q{ E }_{ 0 } }$
0%
$\dfrac { 25m }{ 2q{ E }_{ 0 } }$
0%
$\dfrac { 25m }{ 12q{ E }_{ 0 } }$
0%
$\dfrac { 144m }{ 12q{ E }_{ 0 } }$

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

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 14 - 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 14 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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