Two thin infinite parallel sheets have uniform surface densities of charge $$+\sigma$$ and $$-\sigma .$$ The value of $$E$$ in the space outside the sheets is:

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

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

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

MCQ Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 1

Five ball numbered

$1$ to

$5$ are suspended using separate threads. Pairs

$(1,2)$ ,

$(2,4)$ and

$(4,1)$ show electrostatic attraction while pairs

$(2,3)$ ,

$(4,5)$ show repulsion. Therefore, ball

$1$ must be

0%
neutral
0%
metallic
0%
positively charged
0%
negatively charged

The magnitude of charges of electric dipole is

$3.2 \times {10^{ - 5}}$ and distance between them is

$2.4A.$ If it is placed in an electric field

$4 \times {10^5}\,V/m$ then its diple moment is (in C-m)

0%
$9.6 \times {10^{ - 5}}$
0%
$12.8 \times {10^{ - 14}}$
0%
$7.68 \times {10^{ - 15}}$
0%
$30.72 \times {10^{ - 24}}$

The electric field in a region is given by

$\vec { E } =a\hat { i } +b\hat { j }$ . Here a and b are constants. Find the net flux passing through a square area of side

${ L }_{ 0 }$ parallel to y-z plane

0%
$a{ L }_{ 0 }^{ 3 }$
0%
$2a{ L }_{ 0 }^{ 2 }$
0%
$a{ L }_{ 0 }^{ 2 }$
0%
$\left( a+b \right) { L }_{ 0 }^{ 2 }$

The nature of charge on the object can be detected by:

0%
Microscope
0%
Electroscope
0%
Electrometer
0%
Telescope

A cone of base radius

$R$ and height

$h$ is located in a uniform electric field

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

0%
$4 E h R$
0%
$\displaystyle \dfrac{1}{2}$ $E h R$
0%
$2 E h R$
0%
$E h R$

Explanation

We know that,

$\displaystyle \phi = E.A Cos \theta$
Flux crossing through the area of cone parallel to the electric field which is nothing but a right angle triangle having base

$2R$ and height

$h$
therefore,

$\displaystyle \phi= {EhR}$

The electric field in a region of space is given by

$\vec E=E_0\hat i+2 E_o\hat j$ where

$E_o=100 N/C$ . The flux of this field through a circular surface of radius

$0.02$ m parallel to the Y-Z plane is nearly :

0%
$0.125 Nm^2/C$
0%
$0.02 Nm^2/C$
0%
$0.005 Nm^2/C$
0%
$3.14 Nm^2/C$

Explanation

Answer is A.
We know that,

$\phi= E.A Cos\theta$ .
Since surface area is parallel to Y-Z plane therefore, y component of electric field will not contribute to flux through the circular surface.
So,

$\phi= \pi r^{2} E_o = 3.14\times 0.0004\times 100= 0.125 Nm^{2}/C$

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about

$150 N/C$ , directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be : [Given

$\epsilon_0=8.85\times 10^{-12} C^2/N-m^2, R_E=6.4\times 10^6m]$

0%
$+660 kC$
0%
$-660 kC$
0%
$-680 kC$
0%
$+680 kC$

Explanation

From Gauss' Law,
the electric field entering the Earth's atmosphere is E. The flux of this field is,

$\int E.ds$ =

$E \times 4 \pi {R}^{2}$
This should be equal to the charge enclosed inside the Earth divided by

$\epsilon$
Hence,

$E \times 4 \pi {R}^{2}$ =

$\dfrac{q}{\epsilon}$
Hence, putting back the value of

$3$ knowns, the value of

$q = -680 \times {10}^{3} C$

......... Charge is produced by friction

0%
Stationary
0%
No
0%
Attractive
0%
All

A few electric field lines for a system of two charges

$Q_{1}$ and

$Q_{2}$ fixed at two different points on the

$x$ -axis are shown in the figure. These lines suggest that

0%
$|Q_{1}|>|Q_{2}|$
0%
$|Q_{1}|<|Q_{2}|$
0%
at a finite distance to the left of $Q_{1},$ the electric field is zero
0%
at a finite distance to the right of $Q_{2},$ the electric field is zero

Explanation

Number of Electric field lines emerging from

$Q_1$ are more than the number of lines terminating at

$Q_2$ .
So ,

$Q_1$ is a positive charge of higher magnitude and

$Q_2$ is a negative charge of smaller magnitude .
Hence , the point at which the electric field is zero is on the right of the negative charge

$Q_2$

Consider an electric field

$\overline{E}=E_{0}\hat{x}$ , where

$E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is:

0%
$2E_{0}a^{2}$
0%
$\sqrt{2}E_{0}a^{2}$
0%
$E_{0}a^{2}$
0%
$\displaystyle \dfrac{E_{0}a^{2}}{\sqrt{2}}$

Explanation

$Flux\ through\ area = Electric\ field \times Area\ projected$

$\phi =EA cos45^o$

$= E_0(\sqrt{2}a^2) \times \dfrac{1}{\sqrt2}$

$= E_0a^2$

A point charge

$+Q$ is placed just outside an imaginary hemispherical surface of radius

$R$ as shown in the figure. Which of the following statements is/are correct?

0%
The electric flux passing through the curved surface of the hemisphere is $-\dfrac{Q}{2\in _0} \left(1-\dfrac{1}{\sqrt{2}}\right)$
0%
The component of the electric field normal to the flat surface is constant over the surface
0%
Total flux through the curved and the flat surface is $\dfrac{Q}{\in_0}$
0%
The circumference of the flat surface is an equipotential

Explanation

$\phi = \dfrac{q}{2 \epsilon_0}(1-cos \phi)$

$Here, \phi=\dfrac{\pi}{4}$

So,

$\phi=\dfrac{q}{2\epsilon_0}(1-\dfrac{\pi}{4})$

$\phi=\dfrac{-q}{2\epsilon_0}(1-\dfrac{1}{\sqrt{2}})$

Also the circumference of the flat surface is an equipotential.

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 correct option(s):

0%
The net electric flux crossing the plane x = +a/2 is equal to the net electric flux crossing the plane x = -a/2
0%
The net electric crossing the plane y = +a/2 is more than the net electric flux crossing the plane y = -a/2
0%
The net electric flux crossing the entire region is $\frac { q }{ { \varepsilon }_{ o } }$
0%
The net electric flux crossing the piane z = +a/ 2 is equal to the riet electric flux crossing the piane x = +a/2

Explanation

$\text{Net flux through cube}$ =

$\displaystyle \frac{-q + 3q -q}{\epsilon_{o}} = \frac{q}{\epsilon_{o}}$

$\text{Due to the same area flux passing through}$

$\displaystyle x= \frac{-a}{2}, y= \frac{a}{2}$

$and$

$\displaystyle z=\frac{a}{2}$

$\text{is same}$

In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is however not constant, but increases uniformly along the positive z-direction at the rate of

$10^{5}N\ C^{-1}m^{-1}$ . The force experienced by the system having a total dipole moment equal to

$10^{-7}C\ m$ in the negative

$z-direction$ is

0%
$-10^{-2}N$
0%
$10^{-2}N$
0%
$10^{-4}N$
0%
$-10^{-4}N$

Explanation

Consider an electric dipole with

$-q$ charge at

$A$ and

$+q$ charge at

$B$ , placed along z-axis, such that its dipole moment is in negative

$z$ direction.
i.e.,

$P_{z} = 10^{-7}C\ m$
The electric field is along positive direction of z-axis, such that

$\dfrac {dE}{dz} = 10^{5} N\ C^{-1} m^{-1}$
From,

$F = qdE$

$= (q\times dz) \times \dfrac {dE}{dz} = p \dfrac {dE}{dz}$
Force experienced by the system in the negative z-direction,

$F = p\times \left (-\dfrac {dE}{dz}\right ) = 10^{-7} \times (-10^{5}) = -10^{-2}N$ .

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
0%
Assertion is correct but Reason is incorrect.
0%
Both Assertion and Reason are incorrect.

Explanation

To apply gauss law, the there must exist a symmetric distribution of charges, such that the gaussian surface could have a single value of electric field for the given surface in the equation

$\dfrac{q}{\epsilon_0}=\int \vec{E}.d\vec{A}$

Electric dipoles have no surface that have a symmetric distribution of electric field around, thus Gauss' Law cannot be used for the task.

Electric field outside a long wire carrying charge q is proportional to :

0%
$\dfrac { 1 }{ r }$
0%
$\dfrac { 1 }{ { r }^{ 2 } }$
0%
$\dfrac { 1 }{ { r }^{ \frac { 1 }{ 3/5 } } }$
0%
$\dfrac { 1 }{ { r }^{ \frac { 1 }{ 3/2 } } }$

Explanation

Using Gauss Law across the cylindrical Gaussian surface,

$\displaystyle\int E.dS = \dfrac{q_{enc}}{\epsilon_0}$

$\therefore E\times 2\pi r \times l = \dfrac{\lambda\times l}{\epsilon_0}$

$\therefore E = \dfrac{\lambda}{2\pi r\epsilon_0}$

Here

$\lambda$ is the linear charge density on the wire.

An electric line of force in the

$xy-plane$ is given by equation

$x^{2} + y^{2} = 1$ . A particle with unit positive charge, initially at rest at the point

$x = 1, y = 0$ in the

$xy - plane$ .

0%
Not move at all
0%
Will move along straight line
0%
Will move along the circular line of force
0%
Information is insufficient to draw any conclusion

Explanation

Charge will move along the circular line of force because

$x^{2} + y^{2} = 1$ is the equation of circle in

$xy - plane$ .

Complete the following statements with an appropriate word /term be filled in the blank space(s).

The force of attraction or repulsion between two charges is given by ......................

0%
Gauss's law
0%
Kepler's law
0%
Coulomb's law
0%
Gravitational law

If the flux of the electric field through a closed surface is zero:

0%
the electric field must be zero everywhere on the surface
0%
the electric field must not be zero everywhere on the surface
0%
the charge inside the surface must be zero
0%
the charge in the vicinity of the surface must be zero

Explanation

Gauss law:

$\phi=$

$\oint E\cdot da=\dfrac {q_{enclosed}}{\varepsilon_0}$
Given that the flux through the closed surface is zero.

$\therefore q_{enclosed}=0$

Electric field intensity at a point due to an infinite sheet of charge having surface charge density $\sigma$ is $E$ . If the sheet were conducting, electric intensity would be

0%
$\dfrac{E}{2}$
0%
$E$
0%
$2 E$
0%
$4 E$

Explanation

Normal sheet

Apply gauss theorem

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

$E=\dfrac {\sigma}{2\varepsilon_0}$

Conducting sheet

We know inside a conductor E is zero.

$\therefore E\cdot (A)=\dfrac {A\cdot \sigma}{\varepsilon_0}$

$E=\dfrac {\sigma}{\varepsilon_0}$

$\therefore E_{conducting}=2 E_{normal sheet}$

Number of electric lines of force emanating from $1C$ of positive charge in vacuum is :

0%
$8.85\times10 ^{-12}$
0%
$9\times10 ^{9}$
0%
$1/4\pi \times 9\times 10^{9}$
0%
$1.13\times10 ^{11}$

Explanation

$\phi$ from

$1C$ of charge is

$= \dfrac{Q_{enclosed}}{\epsilon_o}$

$= \dfrac{1}{8.8501\times 10^{-12}}$

$= 1.129\times 10^{11}$

Lightning is a ________ phenomenon

0%
Electromagnetic
0%
Current electricity
0%
Electrochemical
0%
Electrostatic

An infinite plane sheet of a metal is charged to charge density

$\sigma\ C/m^{2}$ in a medium of dielectric constant K. Intensity of electric field near the metallic surface will be :

0%
$E=\dfrac{\sigma }{\varepsilon _{_{0}}K}$
0%
$E=\dfrac{\sigma }{2\varepsilon _{_{0}}}$
0%
$E=\dfrac{\sigma }{2\varepsilon _{_{0}}K}$
0%
$E=\dfrac{K\sigma }{2\varepsilon _{_{0}}}$

Explanation

We know inside a conductor

$E=0$

$\therefore$ applying gauss law we get

$E\times A=\dfrac {\sigma \times A}{K{\varepsilon_0}}$

$E=\dfrac {\sigma}{K{\varepsilon_0}}$

When we remove polyester or woollen clothes in dark, we can see a spark and hear a crackling sound. Which of the following is responsible for it?

0%
Static electricity.
0%
Current electricity.
0%
Reflection of light.
0%
Refraction of light.

Two charged objects are brought close to each other. Choose the most appropriate statement from the following options:

0%
they may attract
0%
they may repel
0%
they may attract or repel depending on the type of charges they carry
0%
there will be no effect

The value of $E$ in the space outside the sheets is:

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

$Debye$ is the unit of:

0%
electric flux
0%
electric dipole moment
0%
electric potential
0%
electric field intensity

Explanation

The

$debye$ (

$D$ ) is the CGS unit of electric dipole moment (

$= V\times d$ )

$1\ debye=1 \times 10^{-18}\ statcoulomb.cm$

What is the nature of electric charge?

0%
only positive
0%
only negative
0%
positive and negative both
0%
neutral.

State whether the following statements are True or False.

If E=0, at all points of a closed surface, The electric flux through the surface is zero.

0%
True
0%
False

Explanation

True. If

$E=0$ , at all points of a closed surface that means there is no electric field present. Therefore, electric flux should also be zero.

True Or False?
1)A proton repels an electron.
2)An electron repels a proton.

0%
False, False
0%
True, True
0%
True, False
0%
False, True.

The ampere-sec stands for the unit of :

0%
Power
0%
Current
0%
Energy
0%
Charge

Explanation

Charge is given by the relation,

$q = I*t$
so its unit is ampere - sec

Electroscope is an instrument used to find the quantity of charge on different bodies.

0%
True
0%
False

Explanation

Electroscope is an instrument used to measure the amount of charge on a body.

Hence, given statement is

$True$ .

Debye is the unit of

0%
Electric flux
0%
Electric dipole moment
0%
Electric potential
0%
Electric field intensity

Explanation

The debye (symbol: D) is a CGS unit (a non-SI metric unit) of electric dipole moment named in honour of the physicist Peter J. W. Debye. It is defined as

$1×10^{−18}$ statcoulomb-centimetre.

Which of the following charge is acquired by silk when the silk is rubbed with glass?

0%
Positive
0%
Negative
0%
Nuetral
0%
Can not be determined

The laws of forces that govern the force between two electric charges were discovered by :

0%
Faraday
0%
Ampere
0%
Ohm
0%
Coulumb

Explanation

Coulomb stated that the force (F) of attraction or repulsion between two points charges

$q_1$ and

$q_2$ separated by a distance

$r$ is proportional to the product of the magnitude of the two charges, and inversely proportional to the square of the distance of separation between them. i.e.,

$F \propto \dfrac{q_1 q_2}{r^2}$

The solid angle subtended by the total surface area of a sphere at the centre is :

0%
$4\pi$
0%
$2\pi$
0%
$\pi$
0%
$3\pi$

Explanation

A steradian is the solid angle subtended at the center of a sphere of radius r by a section of its surface area of magnitude equal to

${ r }^{ 2 }$ . Since the surface area is

$4\pi { r }^{ 2 }$ , there are

$4\pi$ steradians surrounding a point in space.

An electric dipole is a pair of _______ and ______ charges.

0%
equal, same
0%
equal, opposite
0%
different, same
0%
different, opposite

What is the net charge on an neutral atom?

0%
positive
0%
negative
0%
zero
0%
none of these

Explanation

If an atom has an equal number of protons and electrons, its net charge is $0$ .
In an atom, each proton has a charge of

$+1$ , each electron a charge of

$-1$ , and each neutron has no charge.
If the number of protons and electrons are not same in an atom, they are called ions.

What do you understand by the static or frictional electricity ?

0%
Discarging of two bodies on rubbing them together is called static or frictional electricity.
0%
Charging of two bodies on rubbing them together is called static or frictional electricity.
0%
Electricity which is static is called static or frictional electricity.
0%
Electricity on high friction bodies is called static or frictional electricity.

What happens when some charge is placed on a soap bubble?

0%
Its radius decreases
0%
Its radius increases
0%
The bubble collapses
0%
None of these

State whether true or false:

Like charges attract each other.

0%
True
0%
False

Which are the kinds of charges?

0%
Positive
0%
Negative
0%
Both A & B
0%
None

When a silk handkerchief is rubbed with glass the charge acquired by silk is :

0%
positive
0%
negative
0%
neutral
0%
none of these

Electric field lines of force

0%
exist everywhere
0%
exist only in the immediate vicinity of electric charges
0%
exist only when both positive and negative charges are near one another
0%
are imaginary

In a region of space having a uniform electric field

$E$ , a hemispherical bowl of radius

$r$ is placed. The electric flux

$\phi$ through the bowl is

0%
$2\pi RE$
0%
$4\pi {R}^{2}E$
0%
$2\pi {R}^{2}E$
0%
$\pi {R}^{2}E$

Explanation

$\phi =E(ds)\quad \cos { \theta } =E(2\pi { r }^{ 2 })\cos { { 0 }^{ o } } =2\pi { r }^{ 2 }E\quad$

State whether true or false:

Unlike charges attract each other.

0%
True
0%
False

A negatively charged ebonite rod is brought near the glass rod which has been robbed on silk. State your observation between them

0%
Attraction
0%
Repulsion
0%
No change
0%
cant say

Two uncharged bodies on rubbing, get charged due to:

0%
conduction
0%
friction
0%
induction
0%
none of these

When we remove polyester or woollen cloth in dark, we can see sparkand hear a cracking sound. These are due to:

0%
static electricity
0%
current electricity
0%
reflection of light
0%
refraction of light

State the kind of charge on a positive ion :

0%
Positive
0%
Negative
0%
zero
0%
cant say

Complete the following sentences:

A glass rod is rubbed with silk. The charge on glass rod is ...... and the charge on silk is ......

0%
$positive, positive$
0%
$negative, negative$
0%
$positive, negative$
0%
$negative, positive$

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 1 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 1 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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