Case Study MCQ’s chapter 1 class 12 physics

 Electric charges and fields 

Case Study type Questions of chapter 1 physics 

Case Study MCQ’s class 12 physics chapter 1 Electric charges and fields

Term -1 CBSE Boards contains McQs based on case study. I have bought some examples so that you can get good practice of case based mcqs for  your term 1 board exams. This blog contains Mcqs from class 12 physics chapter 1 Electric charges and fields… 

*Answers and Explanation At Last 

1) Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of charges positive and negative charges. Also, like charges repel each other whereas unlike charges attract each other.

1) Charge on a body which carries 200 excess electrons is:

A) -3.2 x 10^-18 C 

B) 3.2 x 10^18 C 

C) -3.2 x 10^-17 C 

D) 3.2  x 10^-17 C

2) Charge on a body which carries 10 excess electrons is : 

A) - 1.6 x 10^ -18 C 

B) 1.6 x 10^-18 C 

C) 2.6 x 10^-18 C

D) 1.6 x 10^-21 C

3) Charge is -

A) Transferable

 

        B) associated with mass

   C) conserved

        D) All of the above

4) A body is positively charged, it implies that:

 

A) there is only a positive charge in the body

        B) there is positive as well as negative charge in the

body but the positive charge is more than negative

charge

        C) there is equally positive and negative charge in the

body but the positive charge lies in the outer regions

 

      D) the negative charge is displaced from its position

 

5) On rubbing, when one body gets positively

charged and other negatively charged, the

electrons transferred from positively charged body

to negatively charged body are:

A) valence electrons only

B) electrons of inner shells

C) both valence electrons and electrons of the inner

shell.

D) None of the above

 

2) Electric field strength is proportional to the density of lines

of force i.e., electric field strength at a point is proportional to

the number of lines of force cutting a unit area element

placed normal to the field at that point. As illustrated in given

figure, the electric field at P is stronger than at  Q :

 1) Electric lines of force about a positive point

charge are:

A) Radially inwards

B) Circular clockwise

C) Radially outwards

D) Parallel straight lines

 

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

 

A) They always start from positive charge and terminate

on negative charges.

B) They are always perpendicular to the surface of a

charged conductor.

C) They always form closed loops.

D) They are parallel and equally spaced in a region of

uniform electric field.

 

3) Electric field lines are curved:

 

A) in the field of a single positive or negative

charge

B) in the field of two equal and opposite

charges.

C) in the field of two like charges.

D) Both B and C

 

4) Which one of the following patterns of

electric line of force is not possible in field due

to stationary charges?

 

5) The figure below shows the electric field

lines due to two positive charges. The

magnitudes EA, EBand EC of the electric fields

at point A, B and C respectively are related as

 

(a) EA>EB>EC

(b) EB>EA>EC

(c) EA=EB>EC

(d) EA>EB=EC

 

3) Smallest charge that can exist in nature is the

charge of an electron. During friction it is only the

transfer of electron which makes the body charged.

Hence net charge on any body is an integral multiple

of charge of an electron (1.6 x 10^-19 C) i.e., q=±ne

where r= 1, 2, 3, 4 ….

Hence no body can have a charge represented as 1.8e, 2.7e, 2e/

5, etc.

Recently, it has been discovered that elementary particles such

as protons or neutrons are elemental units called quarks.

 

1) Which of the following properties is not satisfied

by an electric charge?

 

A) Total charge conservation.

B) Quantization of charge

C) Two types of charge

D) Circular line of force

 

2) Which one of the following charges is

possible?

A) 5.8 x 10^-18 C

B) 3.2 x 10^-18 C

C) 4.5 x 10^-19 C

D) 8.6 x 10^-19 C

 

3) If a charge on a body is 1 nC, then how many

electrons are present on the body?

A) 6.25 x 10^27

B) 1.6 x 10^19

C) 6.25 x 10^28

D) 6.25 x 10^9

 

4) If a body gives out 10^9 electrons every second,

how much time is required to get a total charge of

1C from it?

A) 190.19 years

B) 150.12 years

C) 198.19 years

D) 188.21 years

 

5) A polythene piece rubbed with wool is

found to have a negative charge of 3.2 x

10^-7C . Calculate the number of electrons

transferred.

A) 2 x 10^12

B) 3 x 10^12

C) 2 x 10^14

D) 5 x 10^14

 

4) When electric dipole is placed in uniform electric field, its

two charges experience equal and opposite forces, which

cancel each other and hence net force on electric dipole in

uniform electric field is zero. However these forces are not

collinear, so they give rise to some torque on the dipole. Since

net force on electric dipole in uniform electric field is zero, so

no work is done in moving the electric dipole in uniform electric

field. However some work is done in rotating the dipole against

the torque acting on it.

 

1) The dipole moment of a dipole in a uniform

external field Ē is p. Then the torque τ acting on

the dipole is:

A) τ=p x E

B) τ = P. Ē

C) τ = 2(p + Ē)

D) τ = (P + E)

 

2) An electric dipole consists of two opposite

charges, each of magnitude 1.0 μC separated by a

distance of 2.0 cm. The dipole is placed in an

external field of 10^5N/C -1. The maximum torque on

the dipole is:-

A) 0.2 x 10^-3 Nm

B) 1 x 10^-3 Nm

C) 2 x 10^-3 Nm

D) 4 x 10^-3 Nm

 

3) Torque on a dipole in uniform electric field is

minimum when θ is equal to :-

A) 0°

B) 90°

C) 180°

D) Both A and C

 

4) When an electric dipole is held at an angle in

a uniform electric field, the net force F and

torque τ on the dipole are:-

A) F= 0, τ = 0

B) F≠0, τ≠0

C) F=0, τ ≠ 0

D) F≠0, τ=0

 

5) An electric dipole of moment p is placed in an electric

field of intensity E. The dipole acquires a position such that

the axis of the dipole makes an angle with the direction of

the field. Assuming that potential energy of the dipole to

be zero when 0 = 90°, the torque and the potential energy

of the dipole will respectively be:-

A) pEsinθ, -pEcosθ

B) pEsinθ, -2pEcosθ

C) pEsinθ, 2pEcosθ

D) pEcosθ, – pEsinθ

 

5) Net electric flux through a cube is the sum of fluxes

through its six faces. Consider a cube as shown in figure,

having sides of L= 10.0cm. The electric field is uniform, has a

magnitude E = 4.00 x 10^3 N/C and is parallel to the xyplane

at an angle of 37° measured from the x-axis towards

the +y- axis .

 

1) Electric flux through surface S6 is

A) -24 N m²/C

B) 24 N m²/C

C ) 32 N m²/C

D) -32 N m²/C

 

2) Electric flux passing through surface S1 is :

A) -24 N m²/C

B) 24 N m²/C

C) 32 N m²/C

D) -32 N m²/C

 

3) The surfaces that have zero flux are :

A) S1 and S3

B) S5 and S6

C) S2 and S4

D) S1 and S2

 

4) The total net electric flux through all faces of

the cube is :

A) 8 N m²/C

B) -8 N m²/C

C) 24 N m²/C

D) zero

 

5) The dimensional formula of surface integral

E.ds is :

A) [ ML² T-² A^-1]

B) [ ML^3 T^-3]

C) [M^-1L^3 T^-3A]

D) [ M L^3 T^-3A^-1]

 

6) Figure (a) shows an uncharged metallic sphere on an insulating

metal stand. Bring a negatively charged rod close to the metallic

sphere, as shown in Fig. (b). As the rod is brought close to the sphere,

the free electrons in the sphere move away due to repulsion and start

piling up at the farther end. The near end becomes positively charged

due to deficit of electrons. This process of charge distribution stops

when the net force on the free electrons inside the metal is zero.

Connect the sphere to the ground by a conducting wire. The electrons

will flow to the ground while the positive charges at the near end will

remain held there due to the attractive force of the negative charges

on the rod, as shown in Fig. (c). Disconnect the sphere from the

ground. The positive charge continues to be held at the near end.

Remove the electrified rod. The positive charge will spread uniformly

over the sphere as shown in Fig. (e) . In this experiment, the metal

sphere gets charged by the process of induction and rod does not

loose any of its charge.

 

1) What do you call the process of charging a

conductor by bringing it near another charged

object?

A) Induction

B) Polarisation

C) Neutralization

D) Conduction

 

2) The negatively charged balloon is brought near the two

cans. What happens?

 

A) the negative charges in can B move towards the balloon

B) the negative charges in can A move away from the

balloon

C) the positive charges in can B move towards the balloon

D) the positive charges in can A move away from the

balloon

 

3) Transferring a charge without touching is-

A) Conduction

B) Induction

C) Grounding

D) Friction

 

4) Due to electrostatic induction in aluminium

rod due to charged plastic rod, the total charge

on the aluminium rod is

A) Zero

B) Positive

C) Negative

D) None of these

 

5) If we bring charged plastic rod near neutral

aluminium rod, then rods will -

A) Repel each other

B) Attract each other

C) Remain their position

D) Exchange charges

 

7) When a glass rod is rubbed with silk, the rod acquires one kind of

charge and the silk acquires second of the charge. This is true for

any pair of objects that are rubbed to be electrified. Now if the

electrified glass rod is brought in contact with silk, with which it was

rubbed, they no longer attract each other. They also do not attract or

repel other light objects as they did on being electrified. Thus, The

charges acquired after rubbing are lost when the charged bodies

are brought in contact. What can you conclude from these

observations? It just tells us that unlike charges acquired by the

objects neutralise or nullify each others effect. Therefore the charges

were named as positive and negative by the

American scientist Benjamin Franklin. We know that when we add a

positive number to a negative number of the same magnitude, the

sum is zero. This might have been the philosophy in naming the

charges as positive and negative. By convention, the charge on a glass

rod or cats fur is Called positive and that on plastic rod or silk is termed

negative. If an object possesses and electric charge, it is said to be

electrified or charged. When it has no charge it is said to be neutral.

 

1) When you charge a balloon by rubbing it on

your hair this is an example of what method of

charging?

A) Friction

B) Conduction

C) Grounding

D) Induction

 

2) Neutral atoms contain equal numbers of

positive_ and negative_

A) Electrons and protons

B) Protons and electrons

C) Neutrons and electrons

D) Protons and neutrons

 

3)Which particle in an atom can you physically

manipulate?

A) Protons

B) Electrons

C) Neutrons 

D) you can't manipulate any particle in an atom

 

4) If a negatively charged rod touches a

conductor, the conductor will be charged by

what method?

A) Friction

B) Conduction

C) Induction

D) Convection

 

5) A negatively charged rod is touched to the

top of an electroscope, which one is correct in

the given figure :-

 

8) Gauss’s law and coulombs law, although expressed in different

forms, are equivalent ways of describing the relation between

charge and electric field in static conditions. Gauss’s law is ε0 Φ =

qencl. , when qencl. is the net charge inside an imaginary closed

surface called Gaussian surface. ∮E.dA gives the electric flux

through the Gaussian surface. The two equations hold only when

the net charge is in vacuum or air.

 

1) If there is only one type of charge in the

universe, then -

A) ∮ E.dA not equal to 0

B) ∮E.dA could not be defined

C) ∮E.dA = infinity if charge is inside

D) ∮E.dA = 0 if charge is outside, ∮ E.dA = q/

ε0 if charge is inside

 

2) What is the nature of Gaussian surface

involved in Gauss law of electrostatic?

A) Magnitude

B) Scalar

C) Vector

D) Electrical

 

3) A charge 10 microC is placed at the centre of a

hemisphere of radius R = 10cm as shown. The

electric flux through the hemisphere (in MKS units)

is -

A) 20 x 10^5

B) 10 x 10^5

C) 6 x 10^5

D) 2 x 10^5

 

4) The electric flux through a closed surface

area S enclosing charge Q is φ. If the surface

area is doubled, then the flux is -

A) 2φ

B) φ/2

C) φ/4

D) φ

 

5) A Gaussian surface encloses a dipole. The

electric flux through this surface is

A) q/ ε0

B) 2q/ ε0

C) q/ 2 ε0

D) zero

 

9) In practice, we deal with charges much greater in magnitude

than the charge on an electron, so we can ignore the quantum

nature of charges and imagine that the charge is spread in a

region in a continuous manner. Such a charge distribution is

known as continuous charge distribution. There are three types

of continuous charge distribution: i) Line charge distribution ii)

Surface charge distribution iii) Volume charge distribution.

 

1) Statement 1: Gauss’s Law can’t be used to

calculate electric field near an electric dipole.

Statement 2: Electric dipole don’t have symmetrical

charge distribution.

A) statement 1 & 2 are true

B) statement 1 is false but statement 2 is true

C) statement 1 is true but statement 2 is false

D) Both statements are false.

 

2) An electric charge of 8.85 x 10^-13 C is

placed at the centre of a sphere of radius 1m.

The electric flux through the sphere is -

A) 0.2 Nm²/C

B) 0.1 Nm²/C

C) 0.3 Nm²/C

D) 0.01 Nm²/C

 

3) The electric field within the nucleus is generally

observed to be linearly dependent on r. So 

 A) a=0

B) R/2

C) a=R

D) a= 2R/3

E is proportional to r

4) What charge would be required to electrify a

sphere of radius 25cm so as to get a surface

charge density of 3/Π C/m²?

A) 0.75C

B) 7.5C

C) 75C

D) zero

 

5) The SI unit of linear charge density is-

A) Cm

B) C/m

C) C/m²

D) C/m^3

 

10) Surface charge density is defined as charge per unit

surface area of surface charge distribution, i.e, sigma= dq/dS

. Two large, thin metal plates are parallel and close to each

other. On their inner faces, the plates have surface charge

densities of opposite signs having magnitude of 17.0 x

10^-22C/ m² as shown. The intensity of electric field at a

point is E = sigma/2 ε0, where ε0 = permittivity of free space.

 

1) E in outer region of the first plate is (I) :

A) 17 x 10^-22 N/C

B) 1.5 x 10^-25 N/C

C) 1.9 x 10^-10 N/C

D) zero

 

2) E in outer region of the second plate (III) :

A) 17 x 10^-22 N/C

B) 1.5 x 10^-25 N/C

C) 1.9 x 10^-10 N/C

D) zero

 

3) E between the plates (II) :

A) 17 x 10^-22 N/C

B) 1.5 x 10^-25 N/C

C) 1.9 x 10^-10 N/C

D) zero

 

4) The ratio of E from right side of B at a distance

2cm and 4 cm, respectively is

A) 1:2

B) 2:1

C) 1:1

D) 1: root 2

 

5) In order to estimate the electric field due to a

thin infinite plane metal plate, the Gaussian

surface considered is

A) Spherical

B) Cylindrical

C) Straight line

D) None of these

For Answers & Explanation click below-

https://youtu.be/n9YSV8HzRi4

 Answers - 

I)1. C, 2. A, 3. D, 4. B, 5. A  II) 1. C, 2. C, 3. D,4. C, 5 A  III) 1.D, 2. B, 3. D, 4. C, 5. A,  IV) 1. A, 2. C, 3. D, 4. C, 5. A   V) 1. D, 2. A, 3. C, 4. D, 5. D VI) 1. A, 2. B, 3. B, 4. A, 5. B VII) 1. A, 2. B, 3. B, 4. B, 5. C   VIII) 1. D, 2. C, 3. C, 4. D, 5. D  IX) 1. A, 2. B, 3. C, 4. A, 5. B  X) 1. D, 2. D, 3. C 4. C, 5. B

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