MCQ Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 10

Which charge is not possible

0%
$4.8\times 10^{-16}C$
0%
$8.0\times 10^{-17}C$
0%
$11.2\times 10^{-17}C$
0%
$12.0\times 10^{-20}C$

A point charge

$Q$ is placed at the centre of a circular wire of radius

$R$ having charge

$q$ . The force of electrostatic interaction between point charge and the wire is:

0%
$\cfrac{qQ}{4\pi\varepsilon_oR^2}$
0%
$\cfrac{q}{4\pi\varepsilon_oR}$
0%
Zero
0%
none of these

Explanation

A point charge

$=Q$

radius

$=R$

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

There is only one point charge. But according to Coulomb's law interaction force create with two point charges hence none of this is correct.

The tangent at any point of field line gives the direction of

0%
Electric field at that point
0%
Electric force on positive charge at that point
0%
Both $(1)$ & $(2)$
0%
Rotation of charge

Under the influence of the Coulomb field of charge

$+Q$ , a charge

$-q$ is moving around it in an elliptical orbit. Find out the correct statements.

0%
The angular momentum of the charge $-q$ is constant.
0%
The linear momentum of the charge $-q$ is constant.
0%
The angular velocity of the charge $-q$ is constant.
0%
The linear speed of the charge $-q$ is constant.

Explanation

Coulomb field of charge

$=+Q$

a charge

$=-q$

The angular momentum of the charge

$-q$ is constant.

angular momentum of the charge

$=(-q)$

$K$ .

$F=\dfrac { kq\left( -Q \right) }{ { r }^{ 2 } }$

angular momentum

$=K(mV)$

Five positively charged particles are kept fixed on the x-axis at point

$(1\, m ,0) (2 \,m, 0), (3 \,m 0) (4\, m 0)$ and

$(5 \, m 0)$ . The magnitude of charges kept at these points are

$1\, \mu C, 4\, \mu C, 9\, \mu C, 16\, \mu C,$ and

$25 \, \mu C$ respectively. The magnitude of net electric force on charge

$1 \,\mu C$ placed at origin is

0%
$4.5 \times 10^3 N$
0%
$4.5 \times 10^2 N$
0%
$4.5 \times 10^{-2} N$
0%
$4.5 \times 10^{-4} N$

Explanation

According to Coulomb's law, the net electric force on charge

$q$ due to five charges is

$F_{net}=\dfrac{1}{4\pi\epsilon_0}\left[\dfrac{qq_1}{r_1^2}+\dfrac{qq_2}{r_2^2}+\dfrac{qq_3}{r_3^2}+\dfrac{qq_4}{r_4^2}+\dfrac{qq_5}{r_5^2}\right]$

$=(9\times 10^9)\left[\dfrac{(1)(1)}{(1-0)^2}+\dfrac{(1)(4)}{(2-0)^2}+\dfrac{(1)(9)}{(3-0)^2}+\dfrac{(1)(16)}{(4-0)^2}+\dfrac{(1)(25)}{(5-0)^2}\right]\times 10^{-12}$

$=45\times 10^{-3}=4.5\times 10^{-2} N$

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

0%
$\frac{Q}{{{\varepsilon _0}}}$
0%
$\frac{Q}{{{6\varepsilon _0}}}$
0%
$\frac{Q}{{{8\varepsilon _0}}}$
0%
$\frac{Q}{{{3\varepsilon _0}}}$

Explanation

The charge is kept at a corner of a cube, so it can be visualized as being at the center of a cube of a double its side i. e., comprising of 8 such cubes kept besides and top of each other.

The flux to each small cube is

$\dfrac{Q}{8 \varepsilon_0}$ .

The correct option is C.

The SI unit of electric flux is

0%
Voltmetre
0%
Joule/Metre
0%
Newton
0%
None

Explanation

$SI$ unit of electric flux

$E=V=dE$

$E=\dfrac { dv }{ dx } =V/m$

A hollow cylinder has a charge q coulomb within it. If f the electric flux in units of volmeter associated with the curved surface B , the flx linked with the plane surface A in units of V-m will be :

0%
$\dfrac { q }{ 2{ \varepsilon }_{ 0 } }$
0%
$\dfrac { \Phi }{ 3 }$
0%
$\dfrac {q}{\varepsilon _0}$
0%
$\dfrac {4q}{\varepsilon_0}$

Explanation

Charge

$=qC$

associated with the curved surface

$=B$

Now, using Gauss law,

$\phi =$ the flux linked with the plane surface

$A$ in units of

$V-m$

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

1244371_1340015_ans_fd5af2062ee14a6885ad91b711b81dc4.png

In the hydrogen atom the electron moves around the proton with a speed of

$2.0 \times 10 ^ { 6 } \mathrm { ms } ^ { - 1 }$ in a circular orbit of radius

$5.0 \times 10 ^ { - 11 }$ What is the equivalent dipole moment ?

0%
$2 \times 10 ^ { - 24 } \mathrm { Am } ^ { 2 }$
0%
$4 \times 10 ^ { - 24 } \mathrm { Am } ^ { 2 }$
0%
$8 \times 10 ^ { - 24 } \mathrm { Am } ^ { 2 }$
0%
$16 \times 10 ^ { - 24 } \mathrm { Am } ^ { 2 }$

Explanation

$\begin{array}{l} v=2\times { 10^{ 6 } } \\ t=\dfrac { { 2\pi r } }{ v } =\dfrac { { 2\pi \times 5\times { { 10 }^{ -11 } } } }{ { 2\times { { 10 }^{ -6 } } } } \\ =5\pi \times { 10^{ -17 } } \\ i=\dfrac { q }{ t } =\dfrac { { 1.6\times { { 10 }^{ -19 } } } }{ { 5\pi \times { { 10 }^{ -17 } } } } \\ =\dfrac { { 1.6\times { { 10 }^{ -2 } } } }{ { 5\pi } } \\ =\dfrac { { 16\times { { 10 }^{ -3 } } } }{ { 5\pi } } \end{array}$

$\begin{array}{l} \mu =iA \\ =\dfrac { { 16\times { { 10 }^{ -3 } } } }{ { 5\pi } } \times \pi \, 25\times { 10^{ -22 } } \\ =80\times { 10^{ -25 } } \\ =8\times { 10^{ -24 } } \end{array}$

Option

$C$ is correct .

A and B are two points on the axis and the perpendicular bisector respectively of an electric dipole. A and B are far away from the dipole and at equal distances from its centre. The fields at A and B i.e.

$\overrightarrow { { E }_{ A } } \quad and\overrightarrow { { E }_{ B } }$ are respectively such that

0%
$\overrightarrow { { E }_{ A } } =\overrightarrow { { E }_{ B } }$
0%
$\overrightarrow { { E }_{ A } } =2\overrightarrow { { E }_{ B } }$
0%
$\overrightarrow { { E }_{ A } } =-2\overrightarrow { { E }_{ B } }$
0%
$\overrightarrow { { E }_{ A } } =\frac { 1 }{ 2 } \overrightarrow { { E }_{ B } }$

Explanation

It will be cleared Heat electric field dual to an electric dipole at axial point is

$2Kp/{ r }^{ 3 }$ while at a point on perpendicularity bisector is

$Kp/{ r }^{ 3 }$ . So, if electric field at point

$A$ is

$E$ then electric field at point

$B$ will be

$E/2$ .

Now,

${ E }_{ A }=-2{ E }_{ B }$

Figure shows two equipotential surface

$V_1$ and

$V_2$ . The component of electric field in

$x$ and

$y$ direction will be

0%
$E _ { x }i$ and $E _ { x } j$
0%
$-E _ { x }i$ and $E _ { x } j$
0%
$E _ { x }i$ and $-E _ { x } j$
0%
$-E _ { x }i$ and $-E _ { x } j$

Explanation

$\begin{array}{l} \overrightarrow { E } =-{ E_{ x } }\widehat { i } +{ R_{ y } }\widehat { j } \\ \, \because Both\, \, Ex\, \, \& \, \, \, Ey\, \, are\, \, equal \\ \therefore \overrightarrow { E } =-{ E_{ x } }\widehat { i } +{ R_{ y } }\widehat { j } \\ \therefore \, \, Option\, \, \left( B \right) \, \, is\, \, correct. \end{array}$

An electric field

$\xrightarrow [ E ]{ } =\left( 25\hat { i } +30\hat { j } \right) NC^{-1}$ exists in a region of space. If the potential at the origin is taken to be zero then potential at X = 2m,y = 2 m is

0%
- 110 V
0%
- 140 V
0%
- 120 V
0%
- 130 V

Explanation

A electric field

$\underset { E }{ \longrightarrow } \left( 25\hat { i } +30\hat { j } \right) N/C$

$X=2m$

$y=2m$

$E=25\hat { i } +30\hat { j }$

$\dfrac { dV }{ dx } =-E\Rightarrow \Delta V=-\int _{ 1 }^{ 2 }{ \overline { E } .d\overline { v } }$

$d\overline { v } =dx\hat { i } +dy\hat { j }$

${ V }_{ 2 }-{ V }_{ 1 }=-\int _{ 1 }^{ 2 }{ \left( 2\delta \hat { i } +30\hat { j } \right) } \left( dx\hat { i } +dy\hat { j } \right)$

$=-\int _{ 1 }^{ 2 }{ \left( -25dx+30dy \right) }$

${ V }_{ 2 }=0$ ,

$=-25{ \left[ x \right] }_{ 0 }^{ 2 }-30{ \left[ y \right] }_{ 0 }^{ 2 }$

${ V }_{ 2 }=\left( -25-30 \right) \times 2=-110V$

When a test charge is brought in from infinity along the perpendicular bisector of an electric dipole ,the work done is

0%
positive
0%
Zero
0%
Negative
0%
None of theses

Explanation

Test charge is brought in from infinity along the perpendicular bisector

Work done never

$(-Ve)$ in sign.

In this test

$+q$ and perpendicularity hence

Work done

$=+q\times E$

$=(+Ve)$

An electric charge produces an electric intensity of

$500 N/C$ at a point in air. If the air is replaced by a medium of dielectric constant

$2.5$ , then the intensity of the electric field due to the same charge at the same point will be:

0%
$100N/C$
0%
$150N/C$
0%
$200N/C$
0%
$300N/C$

Explanation

Electric intensity

$=500N/C$

$K=2.5$

Intensity of electric field due to the same charge at same point will be

$\phi =\dfrac { q }{ { \epsilon }_{ 0 } }$ (using Gauss law)

or,

$\phi =\dfrac { 500 }{ 2.5 } =\dfrac { 500 }{ 25 } \times 10=150N/C$

The magnetic induction field at the centre C of
the wire which is in the shape as shown in figure carrying current $'i'$ is

0%
$\cfrac {\mu_o i}{4 \pi r}(1 + \pi)$
0%
$\cfrac {\mu_o i}{2 \pi r}(1 + \pi)$
0%
$\cfrac {\mu_o i}{\pi r}(1 + \pi)$
0%
$\cfrac {\mu_o i}{r}(1 + \pi)$

Explanation

$\left| \overline { B } \right| =\dfrac { { \mu }_{ 0 }I }{ 2\pi r } \left( 1+\pi \right)$

${ \mu }_{ 0 }$

$I=2A$

$\left| \overline { B } \right| =\dfrac { { \mu }_{ 0 } }{ 2\pi .3 } =\dfrac { { \mu }_{ 0 }i }{ r } \left( 1+\pi \right)$

1244639_1312780_ans_30d25d6956c8407aa00403df45658672.jpg

The force of attraction between two charges

$8\mu C$ and

$0.2 \,N$ . Find the distance of separation.

0%
$412$ meter
0%
$1.2$ meter
0%
$120$ meter
0%
$0.12$ meter

Explanation

$F = \dfrac{K \,q_1q_2}{r^2}$

$r^2 = \dfrac{9 \times 10^9 \times 8 \times 10^{-6} \times 4 \times 10^{-6}}{r^2}$

$r^2 = \dfrac{9 \times 8 \times 4}{2} \times 10^{-3}$

$r^2 = \dfrac{9 \times 8 \times 4}{2} \times 10^{-2}$

$r = 12 \times 10^{-1}$ meter

$= 1.2 \,m$

Determine the electric dipole moment of the system of three charges,placed on the vertices of an equilateral triangle ,as shown in the figure:

0%
$(ql)\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}$
0%
$\sqrt{3}ql\dfrac{\hat{j}-\hat{i}}{\sqrt{2}}$
0%
$-\sqrt{3}ql\hat{j}$
0%
$2ql\hat{j}$

Explanation

$\textbf{Step 1:Breaking the Charge -2q}$

Make the dipoles out of the given system of charges by breaking up charge

$-2q$ as

$-q$ and

$-q$ as shown in Figure

$\textbf{Step 2: Dipole moments of the dipole formed}$

We know that the direction of a dipole is from

$-q$ to

$q$ .

Now we have two dipole vectors

$\vec {P}_{1}$ and

$\vec {P}_{2}$ as shown in Figure.

Magnitude of dipoles:

$P_{1} = ql$ ,

$P_{2} = ql$

$\therefore P_1=P_2=P=ql$

$\textbf {Step 3: Resultant dipole moment }$

As sine components of

$P_1$ and

$P_2$ would cancel each other.

Net dipole moment

$=P\cos 30^o(-\hat{j})+ P \cos 30^o(-\hat{j})$

$=-2ql\cos 30^o \hat{j}= -\sqrt{3}ql \hat{j}$

Hence, Option (C) is correct.

2110657_1333326_ans_1712de9ee2024ff4bb720717426736c3.png

Which potential graph describes the electric field shown ?

The electric flux over the surface of a sphere if it is charged with

$10 \mu C$ is

0%
$36\pi \times { 10 }^{ 4 }{ \quad Nm }^{ 2 }/{ C }$
0%
$36\pi \times { 10 }^{ -4 }{ \quad Nm }^{ 2 }/{ C }$
0%
$36\pi \times { 10 }^{ 6 }{ \quad Nm }^{ 2 }/{ C }$
0%
$36\pi \times { 10 }^{ -6 }{ \quad Nm }^{ 2 }/{ C }$

Explanation

The electric flux over the surface of a sphere

$Q=10\mu C$

Now,

$\phi =\dfrac { 10\times { 10 }^{ -6 } }{ 8.854\times { 10 }^{ -12 } }$

$=\dfrac { { 10 }^{ 1-6+12 } }{ 8.854 } =36\pi \times { 10 }^{ 4 }N{ m }^{ 2 }/C$

A charge is placed at the edge of a cube of each side

$'a'.$ Calculate the electric flux through each face of the cube

$.$

0%
$\frac{Q}{{4{e_0}}}$
0%
$\frac{Q}{{24{e_0}}}$
0%
$\frac{Q}{{2{e_0}}}$
0%
$\frac{Q}{{15{e_0}}}$

Explanation

Charge placed at edge will be shared by

$4$ cubes

$.$

$,$ if we arrange

$4$ cubes charge will be at centre

hence uniform closing surface so according to gauss law total flux

$=\dfrac{Q}{{{e_0}}}$

$,$ flux through each cube

$\dfrac{Q}{{4{e_0}}}$ flux through each face

$= \dfrac{Q}{{24{e_0}}}$

Hence,

option

$(B)$ is correct answer.

Force of attraction between two point charges

$Q$ and

$-Q$ separated by

$d$ meter is

$F_e$ . When these charges are placed two identical sphere of radius

$R=0.3\ d$ whose centries are

$d$ meter apart the force of attraction between them is

0%
Greater than $F_{e}$
0%
Equal to $F_{e}$
0%
Less than $F_{e}$
0%
None of these

Explanation

Force of attraction between two point charges

$Q$ and

$-Q$ separated by

$d$ .

Force

${ F }_{ e }$

radius

$=R=3d$

That is also equal to

${ F }_{ e }$ because the distance is same hence force is also same.

A system of charge as shown in figure. The magnitude net electric dipole moment of the system of charges will be

0%
$2 ql$
0%
$4 ql$
0%
$3 ql$
0%
zero

An electric dipole is situated in an electric field of uniform intensity E whose dipole moment is p and moment of inertia is I. if the dipole is displaced then the angular frequency of its oscillation is

0%
$( \dfrac {pE}{I} )^{1/2}$
0%
$( \dfrac {pE}{I} )^{3/2}$
0%
$( \dfrac {I}{pE} )^{1/2}$
0%
$( \dfrac {p}{IE} )^{1/2}$

Explanation

$\textbf{Step 1: Draw figure}$

$\textbf{[Refer Fig.]}$

$\textbf{Step 2: Torque}$

When we rotate dipole by small angle

$d\theta$ then expression torque on dipole is given by

$\vec{\tau}=\vec{P}\times \vec{E}$

$=-PE\sin d\theta$ [Restoring torque so negative sign]

$\Rightarrow\ \ \tau = -PE \ d\theta$ [

$\sin d \theta\simeq d\theta$ if

$d\theta$ is very small]

$\Rightarrow$

$I\alpha =-PE\ d\theta$ [Where

$I$ is moment of Inertia and

$\alpha$ is the angular acceleration]

$\Rightarrow$

$\alpha=\dfrac{-PE}{I}d\theta$

$....(1)$

$\textbf{Step 3: Angular frequency}$

For Angular Oscillations (angular SHM), we know that:

$\alpha= -\omega^{2} d\theta$ (Where

$\omega$ is the angular frequency of SHM)

$....(2)$

Comparing equations

$(1)$ and equation

$(2)$

$\Rightarrow$

${\omega}^{2}=\dfrac{PE}{I}$

$\therefore$

$\omega= \sqrt{\dfrac{PE}{I}}$

Hence, Option (A) correct

2114230_1371085_ans_990ff5fed6454adabddb8ae5d7c12df9.png

An electric dipole of moment P is placed along the lines of force of electric field E and is in equilibrium. the work done in deflecting it through an angle of

$180^0$ is equal to-

0%
$PE$
0%
$+2PE$
0%
$-2PE$
0%
$ZERO$

Explanation

Moment

$=P$

electric field

$=F$

work done

$=pE\cos\theta$

This is the working formula.

now,

$\theta ={ 180 }^{ 0 }$

$W=pE\cos{ 180 }^{ 0 }$

$=pE{ cos0 }^{ 0 }$

$=pE$

Which of the following is not a unit of charge ?

0%
coulomb
0%
ampere-second
0%
microcoulomb
0%
ampere/second

Explanation

(i) The charge unit coulomb

(ii)

$q=it=$ ampere

$\times$ second

(iii) micro coulomb (because

$s$ prefix only is added)

(iv) ampere/second

$\rightarrow$ it is not unit of charge.

The field potential inside a charged ball of radius R and centre at o depends only on distance from its center as

$V(r) = \alpha r^2 + \beta$ when

$\alpha , \beta$ are +ve constant. now choose correct options

0%
electric field inside the ball $E_r = -2 \alpha r$
0%
electric flux passing through an imaginary sphere of radius r centre at O will be $- 2 \pi \alpha r^4$
0%
volume charge density p(r) inside ball is $-6 \varepsilon_0 \alpha$
0%
electric energy of charged ball will be $\frac {48}{5} \pi \varepsilon_0 \alpha^2 R^5$

Explanation

Radius

$=R$

$V\left( r \right) ={ ar }^{ 2 }+\beta$ when,

$\alpha$ and

$\beta$ are

$(+Ve)$ constant electric field inside the ball

${ F }_{ r }=-2ar$

because,

$-\dfrac { d{ v }_{ r } }{ dr } =+2\alpha r={ E }_{ r }$

or,

${ E }_{ r }=-2\alpha r$ (derivative with respect to

$r$ )

A problem of practical interest is to make a beam of electrons turn at

$90^{\circ}$ corner. This can be done with the parallel plates shown in figure. An electron with kinetic energy

$8.0\times 10^{-17} J$ enters through a small hole in the bottom plate. The strength of electric field that is needed if the electron is to emerge from an exit hole

$1.0\ cm$ away from the entrance hole, traveling at right angles to its original direction is _______

$\times 10^{5} N/C$ .

0%
$3$
0%
$1$
0%
$6$
0%
$2.5$

A soap bubble is given a negative charge, then its radius

0%
Decreasing
0%
Increasing
0%
Remains unchanged
0%
Nothing can be predicated as information is insufficient

For the following electric field lines electric field is

0%
Non uniform
0%
uniform
0%
Both uniform & non uniform
0%
can't be said

Explanation

$\because$ the direction of

$\vec E$ is changing continuously

so, its non uniform.

Hence,

Option

$A$ is correct answer.

An electric field exists in the space between two charged metal plates.
which graph shows the variation of electric field strenth E with distance d from X along the line XY ?

Explanation

Strength

$=E$

distance

$=d$

$X$ along the line

$XY$

$E$

$\alpha$

$\dfrac { 1 }{ { d }^{ 2 } }$ in this context, the graph is the exact like this.

1244262_1404294_ans_072cbe381b944e378e77df60765c79fd.jpg

(i) Obtain the expression for the torque

$\overrightarrow r$ experienced by an electric dipole of dipole moment

$\overrightarrow p$ in a uniform electric field,

$\overrightarrow E$ (ii) what will happen if the field were not uniform? state the condition under which the dipole is in

0%
stable equilibrium
0%
unstable equilibrium
0%
electric dipole
0%
magnetic dipole

In the given figure, electric lines of force diagram is shown. Then:

0%
${E}_{A}<{E}_{B}$
0%
${E}_{A}>{E}_{B}>{E}_{C}$
0%
${E}_{C}>{E}_{B}$
0%
${E}_{A}={E}_{B}={E}_{C}$

Explanation

As field lines are denser at point B in comparison of A, therefore, the Electric field at point A is less than that at point B. Whereas at point C the field density is again reduced. So, the electric field at point C will also be less than the field at point B.

So option

$A$ is correct.

The flux through the hemispherical surface in the figure shown below is

0%
$\pi R^2E$
0%
$2\pi R^2E$
0%
$\pi RE$
0%
Zero

Explanation

$\textbf{Hint}$ : Flux can be calculated by the product of electrostatic field intensity and the area.

$\textbf{Step 1}$ :

Here there is no charge inside the hemisphere. We know Flux

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

Since

$q=0$ , flux

$=0$

Flux through the circular part+flux through the hemispherical part

$=0$

Flux through the hemispherical part

$=-$ Flux through the circular part

Flux through the hemispherical part is

$\phi=\bar{E}.d\bar{s}$

$\phi=|-E(Area \quad of \quad circular \quad region)|$

$\phi={E. \pi r^2}$

Option A is correct

There are three concentric thin sphere of radius a, b, c ( a > b > c ) . the total surface charges densities on their surfaces are

$\sigma , - \sigma , \ sigma$ respectively. the magnitude of electric field at r ( distance from centre ) such that a > r > b is :

0%
0
0%
$\dfrac { \sigma }{ \varepsilon_0 r^2 } (b^2-c^2)$
0%
$\dfrac { \sigma }{ \varepsilon_0 r^2 } (a^2+b^2)$
0%
none of these

Explanation

Sphere of radius

$a,b,c$

$a>b>c$ densities of charges surface

$=\sigma ,-\sigma$ sigma respectively.

The magnitude of electric field at

$=r$ (distance from centre)

$a>r>b$

Using Gauss law,

$E=\dfrac { \sigma }{ { \epsilon }_{ 0 }{ r }^{ 2 } } \left( { a }^{ 2 }+{ b }^{ 2 } \right)$

${ V }_{ A }={ V }_{ C }$

$\left( a-b+c \right) =\dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ c } +C$

$a-b=\dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ r }$

Now,

$E=\dfrac { \sigma }{ { \epsilon }_{ 0 }{ r }^{ 2 } } \left( { a }^{ 2 }+{ b }^{ 2 } \right)$

An infinite large sheet has charge density

$\sigma C/m^2$ . Find electric field at a distance d perpendicular to the sheet.

0%
$E=\dfrac{\sigma}{2\varepsilon_0}$
0%
$E=\dfrac{\sigma}{\varepsilon_0}$
0%
$E=\dfrac{2\sigma}{\varepsilon_0}$
0%
None of these

Explanation

$E=\dfrac{\sigma}{2\varepsilon_0}$ it does not depend on d.
1277397_1631051_ans_ba614877cc9b4e85a84a30aee20585e8.jpg

A charged oil drop is suspended in a uniform field of 3 x 10⁴ v/m so that it neither falls nor rises. The charge on the drop will (Take the mass of the charge = 9.9 x 10^-15 kg and g = 10 m/s² )

0%
$3.3\times 10^{-18} N/C$
0%
$2.25\times 10^{10} N/C$
0%
$5\times 10^{10} N/C$
0%
None

Explanation

$\begin{array}{l} At\, \, equilibrium,electric\, \, force\, \, of\, \, drop\, \, balances\, \, weight\, \, of\, drop. \\ qE=mg \\ q=\dfrac { { mg } }{ E } \\ Now, \\ q=\dfrac { { 9.9\times { { 10 }^{ -15 } } } }{ { 3\times { { 10 }^{ 4 } } } } \\ =3.3\times { 10^{ -18 } }C \end{array}$

If potential (in volts) in a region is expressed as V(x , y , z) = 6xy - y+ 2yz, the electric field (in N/C) at point (1 , 1 , 0) is

0%
$-(2\hat{i} + 3\hat{j} + \hat{k})$
0%
$-(6\hat{i} + 9\hat{j} + \hat{k})$
0%
$-(3\hat{i} + 5\hat{j} + 3\hat{k})$
0%
$-(6\hat{i} + 5\hat{j} + 2\hat{k})$

Explanation

$\begin{array}{l} As\, \, we\, \, learnt\, \, in \\ In\, \, space- \\ { E_{ x } }=\frac { { -dv } }{ { dx } } \\ { E_{ y } }=\frac { { -dv } }{ { dy } } \\ { E_{ z } }=\frac { { -dv } }{ { dz } } \\ Thus, \\ we\, \, get\, \, the\, \, value\, \, is-\left( { 6\widehat { i } +5\widehat { j } +2\widehat { k } } \right) \\ Option\, \, D\, \, is\, \, correct. \end{array}$

Two charges

$q$ and

$-q$ placed at

$(a, 0)$ and

$(-a, 0)$ on the x-axis. Another charge

$2q$ is taken from

$(0,0)$ to

$(0, a)$ , then the work done to do so

0%
$\cfrac{q}{2\pi\varepsilon_0 a}$
0%
$\cfrac{q}{4\pi\varepsilon_0 a}$
0%
Zero
0%
Infinite

Explanation

When

$2q$ is at

$(0,0)$

$V = 0$ because both

$-q$ and

$+q$ cancel each other.

Similarly when

$2q$ is at

$(a,0)$ both cancel each other.

$\begin{array}{l} { V_{ i } }=0 \\ { V_{ f } }=0 \\ \therefore \, No\, \, work\, \, is\, \, done. \end{array}$

Hence,

option

$(C)$ is correct answer.

1238651_1494019_ans_afae5b1a50394d0fb827bff8c997078b.png

Which of the following is false for electric lines of force ?

0%
They always start from positive charges and terminate on negative charges
0%
They are always perpendicular to the surface of a charged conductors
0%
They always form closed loops
0%
They are parallel and equally spaced in a region of uniform electric field

If the surface normal vector

$\hat{n}$ makes an angle

$\theta$ withthe electric field

$\vec{E}$ , then the electric flux through asurface of area

$d S$ is given by

0%
$\phi=E d S \sin \theta$
0%
$\phi=E d S \cos \theta$
0%
$\phi=d S \cos \theta$
0%
$\phi=E \cos \theta$

When

${ 10 }^{ 14 }$ electrons are removed from a neutral metal sphere, the charges on the sphere become

0%
$16\mu C$
0%
$-16\mu C$
0%
$32\mu C$
0%
$-32\mu C$

Three charges

$+ 2 q , - q$ and

$- q$ are placed at pooul

$A$

$( a , 0 ) , B ( a , a )$ and

$C( - a ,$ a) respectively in

$x ^ { 2 } - y$ plane. The magnitude of net electric dipole moment of the system will be

0%
$\sqrt { 11 } q a$
0%
$q$
0%
3 $q a$
0%
$\sqrt { 4 } q a$

Explanation

Three charges

$=+2q,-q$ and

$-q$ .

poul

$A(a,0)$

$B(a,a)$

$C(-a,a)$ respectively

${ x }^{ 2 }-y$

The magnitude of net electric dipole moment

$\sqrt { { 9a }^{ 2 }+{ 3a }^{ 2 } } =\sqrt { 11 } a$

Now, the magnitude of net electric dipole moment of the system

$=\sqrt { 4q } a$ .

What is the Coulomb's force between two $\alpha -$ particles separated by a distance of $3.2 \times {10^{ - 15}}{\text{m}}$

0%
$90 N$
0%
$45N$
0%
$60 N$
0%
$75 N$

in which case charges would be produce???

0%
A plastic ruler is rubbed vigorously with a woollen cloth.
0%
A glass rod rubbed with a piece of silk is brought near a negatively charged paper cylinder.
0%
A positively charged rod is moved slowly towards an uncharged paper cylinder, which is suspended by a string, until the rod touches the cylinder.
0%
all of these.

A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences :

0%
both a net force and torque
0%
only a net force
0%
neither a net force nor a torque
0%
only a torque

Explanation

Since dipole is placed in a uniform electric field, so it experiences zero external force.

Also, dipole is placed to parallel to electric field.

Angle between

$P$ and

$E$ is zero i.e.

$\theta = 0^o$

Torque

$\tau = PE\sin \theta = PE\times \sin 0^o = 0$

In Region of Electric field Given by

$\vec{E} = (Ax + B) \hat{i}$ . Where

$A = 20$ unit and

$B = 10$ unit. If Electric potential at

$x = 1\,m$ is

$V_1$ and at

$x = -5 \,m$ is

$V_2$ . Then

$V_1 - V_2$ is equal to

0%
$150 \,V$
0%
$170 \,V$
0%
$140 \,V$
0%
$180 \,V$

Explanation

$v_f - v_i = -\int E.dx$

$v_1 - v_2 = \displaystyle \overset{1}{\underset{-5}{\int}} (Ax + B)dx$

$= -\left(\dfrac{Ax^2}{2} + Bx\right)^1$

$= -\left[\left(\dfrac{A}{2} + B\right) - \left(\dfrac{25A}{2} - 5B\right)\right]$

$= - \left[\dfrac{20}{2} + 10 - 250 +50\right]$

$= [10 + 10 + 50 - 250]$

$= + 180$

Which of the following charges can from an electric dipole ?

0%
$+1\mu\, C, \, -1 m \, C$
0%
$+1\times 10^{-3}m\, C, \, -1 \mu \, C$
0%
$+5 \times 10^{-3}\, C, \, -2\times 10^{-3} \,C$
0%
$-1 C, \, -1 C$

Electric lines of forces.

0%
May form closed path
0%
Must form closed path
0%
May be discontinuous
0%
Both (a) and (c) are correct

The electric intensity at a point

$A$ at distance

$x$ from uniformly charged non-conducting plane is

$E$ . Then

0%
The electric charge per unit area on the plane is $2 \varepsilon E$
0%
Magnitude $E/2$ of field is due to the charges at points within distance $2x$ from the point $'A'$
0%
Magnitude $E$ of field is due to the charges at points within distance $2x$ from point $A$
0%
Both (a) and (c) are correct

Two positive charges and an negative charge, all of equal magnitude, are set at the corners of an equilateral triangle.
Which diagram best represents the electric field surrounding the charges?

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 10 - 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 10 - MCQExams.com

0:0:3

Practice Class 12 Medical Physics Quiz Questions and Answers

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