Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

A charge is distributed along an infinite curved line in space with linear charge distribution λ. What will be the amount of force on a point charge q kept at a certain distance from the line?(a) q∫\(\frac {\lambda}{r^2}\hat{r}\)dl(b) q∫\(\frac {\lambda}{r^2}\hat{r}\)dr(c) q∫\(\frac {\lambda}{r^3}\hat{r}\)dl(d) q∫\(\frac {\lambda}{r^2}\)I have been asked this question in an internship interview.Query is from Continuous Charge Distribution topic in section Charges and Fields of Physics – Class 12

Answer»

Correct CHOICE is (a) q∫\(\frac {\lambda}{r^2}\hat{r}\)dl

The BEST EXPLANATION: Let the point charge is situated at a distance r from a small part dl on the LINE. The charge stored in stat small part=λ.dl. The force due to that small part will be DIRECTED towards the unit vector \(\hat{r}\). Therefore, force on that charge due to the entire linear charge distribution=q∫\(\frac {\lambda}{r^2}\hat{r}\)dl.

2.

What is the dimension of the dipole moment?(a) [L T I](b) [L T I^2](c) [M L T I](d) [L T I^-1]This question was posed to me in an online interview.The doubt is from Electric Dipole topic in section Charges and Fields of Physics – Class 12

Answer»

Right option is (a) [L T I]

For explanation I would SAY: The dipole moment is DEFINED as the product of a charge and distance. The dimension of charge (current*time) is [I T] and the dimension of distance is [L]. Therefore the dimension of dipole moment is [L T I]. Its unit in the CGS and the SI system are esu*cm and C*m RESPECTIVELY.

3.

Which among the following molecule is not a dipole?(a) NH3(b) H2O(c) HCl(d) CH4The question was posed to me during a job interview.This key question is from Electric Dipole in chapter Charges and Fields of Physics – Class 12

Answer»

Right answer is (d) CH4

For explanation I would say: Though water and ammonia are covalent molecules, they have a net dipole moment due to their DISTORTED STRUCTURE than the ideal one. Hydrochloric acid has a linear molecular structure but its net dipole moment is towards the HYDROGEN ATOM. But methane has a symmetrical tetrahedral structure and its net dipole moment is zero hence it doesn’t behave as a permanent dipole. Methane SOLVENTS are therefore non-polar solvents.

4.

The direction of electric field created by a negative charge is ___________(a) Directed outwards(b) Directed towards the charge(c) Maybe outwards or towards the charge(d) Circular in shapeI have been asked this question in exam.This question is from Electric Field in section Charges and Fields of Physics – Class 12

Answer»

Correct choice is (b) DIRECTED towards the CHARGE

Easy EXPLANATION: If a unit positive charge is KEPT NEAR a negative charge, the unit positive charge will be attracted towards the negative charge. That means the electric field is towards the negative charge. But in case of positive charge, the field is directed away from the charge.

5.

An electric field E is present perpendicular to a square plate of side length L. What is the flux passing through the plate?(a) LE(b) L^2E(c) \(\frac {L^2}{E}\)(d) ZeroThe question was posed to me by my college professor while I was bunking the class.The query is from Gauss’s Law in section Charges and Fields of Physics – Class 12

Answer» RIGHT option is (b) L^2E

Best explanation: We KNOW that electric flux is equal to the product of the electric field and the surface AREA of a surface. In this CASE, the plate has surface area L*L = L^2 and hence the net flux passing through the surface is = L^2E.
6.

Let B be the midpoint of AC. Two point charges Q are placed at A and C. What should be the value of charge placed at B so that the system remains at equilibrium?(a) –Q/2(b) –Q/4(c) +Q/2(d) +Q/4I have been asked this question in my homework.My question comes from Coulomb’s Law in section Charges and Fields of Physics – Class 12

Answer» RIGHT option is (b) –Q/4

For EXPLANATION I would say: Whatever the value of q be, it will undergo equal force on both sides, hence q will be steady.

Now, for stability of Q, NET force on Q must be ZERO.

\(\FRAC {1}{4\pi\epsilon_o} \, \frac {Q^2}{d^2} + \frac {1}{4\pi\epsilon_o} \, \frac {Q.q}{(\frac {d}{2})^2}\) = 0

⇒ \(\frac {Q^2}{d^2} = \frac {-Q q}{\frac {d^2}{4}}\)

⇒ Q = -49 ⇒ q = –\(\frac {Q}{4}\)
7.

Electric field due to a uniformly charged hollow sphere at a distance of r (where r is greater than the radius of the sphere) is __________(a) Proportional to r(b) Inversely proportional to r(c) Proportional to r^2(d) Inversely proportional to r^2The question was asked in an interview for job.My question is based upon Electric Field in division Charges and Fields of Physics – Class 12

Answer»

Correct option is (d) INVERSELY proportional to r^2

Easy explanation: If the total charge of the SPHERE is q then the electric field at a distance of r is equal to \(\FRAC {q}{4\pi\varepsilon_o r^2}\).Therefore the electric field is proportional is \(\frac {1}{r^2}\) (if r > radius of the sphere). But if r < radius of the sphere the electric field will be zero i.e. electric field inside a hollow sphere is always zero.

8.

What is the dimension of volume charge density?(a) [MLAT^-2](b) [M^0 L^-3 A T](c) [M L^-3 A T](d) [M L^-2 A T]I had been asked this question in an online quiz.This is a very interesting question from Basic Properties of Electric Charge in section Charges and Fields of Physics – Class 12

Answer»

Correct OPTION is (B) [M^0 L^-3 A T]

The best I can EXPLAIN: Volume charge density = \(\frac {charge}{volume}=\frac {CURRENT*time}{volume}=\frac {A*T}{L^3}\)=[M^0 A T L^-3]. But in case of surface charge density, the dimension will be [M^0 L^-2 A T] because surface charge density means an electric charge in a unit area of the surface.

9.

An electric field \(3\hat{j} + 4\hat{k}\) N/C is present at a place. What will be the electric flux passing through a square plate if the plate is placed in the y-z plane?(a) 7 Nm^2C^-1(b) 25 Nm^2C^-1(c) Zero(d) 17 Nm^2C^-1I have been asked this question by my college director while I was bunking the class.This intriguing question comes from Gauss’s Law topic in chapter Charges and Fields of Physics – Class 12

Answer»

The correct answer is (c) Zero

The EXPLANATION: The electric field is present in the y-z plane. So, if we place a square surface in the y-z plane, the area vector of the plane will be PERPENDICULAR to the electric field and hence cos⁡θ becomes 0 as θ is 90 degrees. So, E*surface area*cosθ becomes zero i.e. no flux through the surface.

10.

Net inward and outward flux of a closed surface are φ1 and φ2. What is the charge enclosed inside the surface?(a) Zero(b) -(∅1-∅2)εo(c) (∅1-∅2)εo(d) \(\frac {(\emptyset_1 + \emptyset_2)}{\varepsilon_o}\)This question was addressed to me during an internship interview.This is a very interesting question from Gauss’s Law in division Charges and Fields of Physics – Class 12

Answer»

Correct option is (b) -(∅1-∅2)εo

The best explanation: The net outward flux from the surface is (∅2-∅1) = -(∅1-∅2). We KNOW, according to GAUSS’s Law, net outward flux coming out of a CLOSED surface is \(\frac {1}{\varepsilon_o}\) times the CHARGE stored in it. Therefore charge stored in the surface = εo*flux coming out of the surface = (∅2-∅1)εo.

11.

What is the dimension of electric flux?(a) [M L^3 T^-3 I^-1](b) [M L^2 T^-3 I^-1](c) [M L^3 T^-3 I^1](d) [M L^3 T^3 I^-1]This question was posed to me by my college director while I was bunking the class.The doubt is from Electric Flux topic in division Charges and Fields of Physics – Class 12

Answer» RIGHT option is (a) [M L^3 T^-3 I^-1]

The best I can EXPLAIN: Electric flux=electric FIELD intensity* area. The dimension of field intensity is [M L T^-3 I^-1] and the dimension of the area is [L^2]. Therefore, the dimension of flux = [M L^3 T^-3 I^-1]. This can ALSO be justified that flux=potential*length. By PUTTING the dimensions of potential and length, we can get the same result.
12.

A ball of 80mg mass and a 2*10^-8 charge is hung with a thread in a uniform horizontal electric field of 2*10^4V/m. What is the angle made by the thread with vertical?(a) 27 degree(b) 30 degree(c) 45 degree(d) 0 degreeThis question was posed to me in semester exam.This intriguing question originated from Electric Field topic in section Charges and Fields of Physics – Class 12

Answer» CORRECT answer is (a) 27 degree

To elaborate: ACCORDING to the diagram, Tsin⁡θ = Eq; Tcos⁡θ = mg. Where T is the TENSION in the thread, m is the mass of the ball, E is the electric field and q is the charge of the ball. Dividing these two, we get tanθ = \(\frac {Eq}{mg}\). Now substituting the value of E=2*10^4V/m, m=80mg, q=2*10^-8C, g=9.8m/s^2, we get tanθ = \(\frac {25}{49}\). Therefore θ=27 degree.
13.

Two point charges of the same polarities are hung with the help of two threads and kept close. The angle between the threads will be _________ if the system is taken to space.(a) 180 degree(b) 90 degree(c) 45 degree(d) 60 degreeThis question was posed to me during a job interview.My question is from Electric Field topic in section Charges and Fields of Physics – Class 12

Answer»

Correct ANSWER is (a) 180 degree

To explain: There is gravitational field on earth, so if we hang the two same charges there will be an interaction of vertical gravitational field and horizontal electric field. The SYSTEM will achieve equilibrium by creating a certain angle between the THREADS and hence the vertical and horizontal components of forces will balance. But in space, there is no gravity. So the charges will be at 180-degree separation.

14.

According to Coulomb’s Law force between two charges is k*q1*\(\frac {q_2}{r^2}\). What will happen if one of the charges is zero?(a) No attraction or repulsion(b) Always attraction(c) Always repulsion(d) Cannot predict the nature of the forceI got this question in an interview for job.I want to ask this question from Coulomb’s Law topic in division Charges and Fields of Physics – Class 12

Answer» RIGHT answer is (b) Always ATTRACTION

The best I can explain: Though it seems from the formula that the force must be zero, an OPPOSITE CHARGE is induced on the uncharged body and then the TWO opposite charges will attract each other. This also proves that induction occurs before attraction.
15.

Which among the following cannot be the charge of a charged body?(a) 4.8*10^-14 Coulomb(b) 6.4*10^-15 Coulomb(c) 5*10^-14 Coulomb(d) 3.2*10^-10 CoulombI got this question during a job interview.Asked question is from Electric Charges in division Charges and Fields of Physics – Class 12

Answer»

Correct answer is (C) 5*10^-14 Coulomb

For explanation I would say: Charge of a BODY must be an integral multiple of charge of an ELECTRON.5*10^-14 is not an integral multiple of (1.602*10^-19 Coulomb) which is equal to the charge of an electron. But we can SEE that 4.8, 6.4, 3.2 are integral multiples of 1.6.

16.

Electric flux coming out of a single Na^+ ion is _________ Nm^2C^-1(a) 1.8*10^-8(b) 1.8*10^-10(c) 5.4*10^-8(d) 3.6*10^-8I had been asked this question in final exam.Asked question is from Application of Gauss’s Law in chapter Charges and Fields of Physics – Class 12

Answer»

Right ANSWER is (a) 1.8*10^-8

For explanation I would say: A single Na^+ ION MEANS a single sodium atom without an electron. The charge of the ion is the same as the charge of an electron but positive i.e. +1.602*10^-19C. Therefore, applying Gauss’s LAW, electric flux coming out of the ion = \(\frac {charge \, of \, the \, ion}{\varepsilon}\). Substituting the VALUES, we get flux = 1.8*10^-8 Nm^2C^-1.

17.

The term ‘volume charge density’ is valid for a solid conducting sphere.(a) True(b) FalseThe question was posed to me in examination.I want to ask this question from Continuous Charge Distribution topic in chapter Charges and Fields of Physics – Class 12

Answer»

The correct answer is (b) False

Easiest explanation: We know that in the case of any conducting MATERIAL, the charge is always distributed on its OUTER surface instead of distribution THROUGHOUT the material. Therefore, the word volume charge density doesn’t MAKE any sense in the case of the conducting sphere. Instead, we can use the term surface charge density.

18.

1uC and -1uC are placed at a distance of 5 cm forming a dipole. What is the amount of torque required to place the dipole perpendicularly to an electric field of 3*10^5 N/C?(a) 5*10^-3 N. m(b) 15*10^-3 N. m(c) 1*10^-3 N. m(d) 10*10^-3 N. mThis question was addressed to me during an online exam.My question is based upon Dipole in a Uniform External Field topic in portion Charges and Fields of Physics – Class 12

Answer»

The CORRECT choice is (b) 15*10^-3 N. m

Easy explanation: Dipole moment of the dipole = 1*10^-6*5*10^-2 C. m=5*10^-8 C. m. Given the electric field-intensity is E=3*10^5 N/C. Therefore required TORQUE will be P*E*sinθ where θ is the angle between the electric field and the dipole moment = 90 degrees. Therefore torque required in this case will be 5 *10^-8*3*10^5*sin 90 N. m=15*10^-3 N. m.

19.

If an electric dipole is placed in a uniform electric field ______ will act on the dipole.(a) A force but no torque(b) Both force and torque(c) Torque but no force(d) No torque or forceI have been asked this question in an online quiz.My question is from Dipole in a Uniform External Field topic in chapter Charges and Fields of Physics – Class 12

Answer»

Correct option is (c) Torque but no force

To elaborate: Dipole is the COMBINATION of two EQUAL but opposite charges, kept at a certain distance. If it is placed in a uniform electric field, both the charges will suffer the same but opposite forces on them. As a result, the net force on the dipole BECOMES zero, but due to equal and opposite forces ACTING on two DIFFERENT points, there is a net torque acting on the dipole.

20.

It is observed that N number of field lines come out perpendicularly from the unit surface area of the sphere. What is the amount of charge inside the sphere? ∈ is the dielectric constant of the medium.(a) -4πR^2 N∈(b) 4πR^2 N∈(c) \(\frac {4\pi R^2}{N\epsilon}\)(d) –\(\frac {4\pi R^2N}{\epsilon}\)I had been asked this question by my college professor while I was bunking the class.My question comes from Electric Field Lines in portion Charges and Fields of Physics – Class 12

Answer»

The correct ANSWER is (b) 4πR^2 N∈

Easy explanation: Field lines come out from the surface. That means the charge must be POSITIVE. According to Maxwell’s principle, field lines COMING out from a charge q is equal to \(\frac {q}{\EPSILON}\) where ∈ is the relative permittivity. The total number of field lines coming out of the SPHERE is 4πR^2N. This must be equal to \(\frac {q}{\epsilon}\). Therefore q=\(\frac {4\pi R^2}{N\epsilon}\).

21.

Which one of the following is similar between electrostatic force and gravitational force?(a) Force can be attractive or repulsive(b) The force depends on the medium between the bodies(c) Both the forces are strong forces(d) Force is inversely proportional to the distance between the bodiesThis question was posed to me in an interview for internship.This intriguing question originated from Coulomb’s Law in division Charges and Fields of Physics – Class 12

Answer»

The correct option is (d) Force is inversely proportional to the distance between the bodies

Best EXPLANATION: Gravitational force cannot be REPULSIVE and it is a very WEAK force. Gravitational force does not depend on the medium. But both the FORCES are inversely proportional to the distance between them.

22.

1 Coulomb = ________ Electro Static Unit.(a) 3*10^9(b) 3*10^8(c) 1.602*10^-19(d) 2*10^9The question was asked during an interview.The question is from Electric Charges in portion Charges and Fields of Physics – Class 12

Answer» CORRECT OPTION is (a) 3*10^9

The explanation is: 1 Coulomb means3*10^9 Electro Static Unit of CHARGE. Coulomb and ESU are two different units of charge and their conversion FORMULA is necessary. Coulomb is the unit of charge in the SI system and esu is the unit of charge in the CGS system.
23.

Flux coming out from a balloon of radius 10 cm is 1.0*10^3 N.m^2.C-1^-1. If the radius of the balloon is doubled, the flux coming out from the balloon will be _______(a) 0.5 times(b) 2 times(c) Same(d) 4 timesThe question was posed to me during an internship interview.This interesting question is from Electric Flux in division Charges and Fields of Physics – Class 12

Answer»

Right OPTION is (c) Same

For explanation I would say: If the radius of the BALLOON is 2 times the initial value, the surface area of the balloon will be 4 times the initial value because of surface area=4π(radius)^2. But in this case, flux COMING out from the balloon is dependent only upon the charge inside the balloon, not on the area of the surface. As charge enclosed in the balloon is the same in both the cases, the flux will also be the same.

24.

If the force acting on a point charge kept on the axis of an electric dipole is F, what will be the amount of force if the distance of the point charge is doubled from the dipole?(a) 2F(b) \(\frac {F}{8}\)(c) \(\frac {F}{2}\)(d) \(\frac {F}{4}\)I had been asked this question in quiz.My question is from Electric Dipole in section Charges and Fields of Physics – Class 12

Answer» RIGHT option is (b) \(\frac {F}{8}\)

Easiest explanation: If the distance of a point CHARGE, kept on the axis of a dipole, is sufficiently large than the SEPARATION between the charges of a dipole, then it can be shown that electric field E at a distance r from the dipole varies as E α \(\frac {1}{r^3}\). Now if the distance becomes 2R, electric field intensity will decrease by a factor of 2^3=8 times. Therefore the force F will now become \(\frac {F}{8}\).
25.

If a charge is placed outside a closed surface, flux due to that charge inside the surface will be ________(a) Positive(b) May be positive or negative, depending on the nature of the charge(c) Negative(d) ZeroThis question was addressed to me during a job interview.The query is from Electric Flux in chapter Charges and Fields of Physics – Class 12

Answer»

The correct OPTION is (d) Zero

Easy explanation: According to Gauss’s principle, if there is no charge bound INSIDE a surface, net electric FLUX coming out of the surface will always be 0. In this case, the charge is kept outside the surface, so it will GENERATE no field LINES and hence no flux inside the surface. The situation will be the same if an electric dipole is placed inside a surface as dipole has equal positive and negative charges; the total charge inside the surface becomes 0.

26.

An electric charge q is placed at one of the corner points of a cube of side a. What will be the net outwards flux coming out of the cube due to the point charge?(a) \(\frac {q}{\varepsilon_o}\)(b) \(\frac {q}{2\varepsilon_o}\)(c) \(\frac {q}{4\varepsilon_o}\)(d) \(\frac {q}{8\varepsilon_o}\)The question was asked by my school principal while I was bunking the class.This key question is from Gauss’s Law topic in portion Charges and Fields of Physics – Class 12

Answer»

The correct CHOICE is (d) \(\frac {q}{8\varepsilon_o}\)

To explain: We can assume 8 imaginary cubes, all of side a placed together, sharing a common CORNER point and forming a bigger cube of side 2A. The point charge is placed at the common point of the cubes. ACCORDING to Gauss’s Law, net electric flux coming out of the bigger cube=\(\frac {q}{\varepsilon_o}\). THEREFORE, flux coming out from each of the smaller cubes i.e. cubes of side a is \(\frac {q}{8\varepsilon_o}\).

27.

A non-conducting sphere has uniform charge density in it. The electric field at a point inside the sphere will be _______(a) Zero(b) Only due to the charge inside that point(c) Only due to the charge outside that point(d) Due to the entire charge of the sphereThis question was addressed to me in examination.My question is based upon Gauss’s Law in chapter Charges and Fields of Physics – Class 12

Answer»

Correct answer is (b) Only due to the charge INSIDE that point

The best explanation: In the case of a non-conducting sphere, an ELECTRIC charge is uniformly distributed THROUGHOUT the sphere. Inside the sphere, the electric field at a point will be only due to the electric charge inside that point. There will be no effect of the REMAINING charge outside that point. In the case of the conducting sphere, the electric field at every point inside the sphere is zero.

28.

How many electrons must be added to an isolated spherical conductor of radius 20cm to produce an electric field 1000N/C just outside the surface?(a) 2.77*10^20(b) 2.77*10^10(c) 1.77*10^10(d) 5.4*10^10I have been asked this question at a job interview.I'd like to ask this question from Continuous Charge Distribution topic in portion Charges and Fields of Physics – Class 12

Answer»

Correct choice is (b) 2.77*10^10

The explanation is: The electric field E = k \(\frac {Q}{R^2}\) where k=9*10^9 N m^2 C^-2 and Q is the charge of the sphere, r is the radius of the sphere. Therefore Q = \(\frac {E*r^2}{k}=\frac {1000*0.2^2}{9*10^9}\) = 4.44*10^-9. NUMBER of electrons REQUIRED for this amount of charge is = \(\frac {4.44*10^{-9}}{1.602*10^{-19}}\) = 2.77*10^10.

29.

If a non-polar substance is placed in an electric field, what will happen?(a) A net dipole moment will be observed(b) The substance will oscillate(c) The substance will orient itself perpendicular to the electric field(d) It will conduct electricityI got this question by my college professor while I was bunking the class.The query is from Dipole in a Uniform External Field in division Charges and Fields of Physics – Class 12

Answer»

Right CHOICE is (a) A net DIPOLE moment will be observed

The best I can explain: A non-polar substance consists of a HUGE number of dipoles in it, but they are oriented randomly and hence the net dipole moment of the substance becomes 0 and it ACTS as a non-polar substance. But if it is placed in an electric field, the dipoles present in it will ORIENT themselves in the direction of the field and hence a net dipole moment will be observed, known as induced dipole moment. No current flow or oscillation of the substance will be observed.

30.

If electric field intensity at a certain distance from a dipole on its axis is E1 and at the same distance but on the perpendicular bisector of the dipole is E2, then __________(a) E1=E2(b) E1=2 E2(c) E1=4 E2(d) E1=3 E2The question was posed to me in a national level competition.Enquiry is from Electric Dipole in chapter Charges and Fields of Physics – Class 12

Answer»

Correct CHOICE is (b) E1=2 E2

Explanation: If the dipole moment of a dipole is p, then electric field INTENSITY at a distance r from the dipole on its axis will be \(\frac {2p}{r^3}\). But electric field intensity at a distance r from the dipole on its perpendicular bisector is \(\frac {p}{r^3}\). Both the formulas are DERIVED by assuming r >> length of the dipole. THUS E1=\(\frac {2p}{r^3}\) and E2=\(\frac {p}{r^3}\) and hence E1=2 E2.

31.

Electric field inside a hollow conducting sphere ______(a) Increases with distance from the center of the sphere(b) Decreases with distance from the center of the sphere(c) Is zero(d) May increase or decrease with distance from the centerThis question was posed to me in a national level competition.My query is from Electric Field topic in chapter Charges and Fields of Physics – Class 12

Answer»

The correct choice is (C) Is zero

To ELABORATE: According to Gauss’s law, if there is no charge inside a closed surface, the field inside the closed surface will always be zero. We know the charge is distributed on the OUTER surface of a conducting hollow sphere because the charges want to maintain maximum distance AMONG them due to repulsion. So there is no charge inside the sphere and hence no ELECTRIC field.

32.

Which among the following is false?(a) Coulomb force is a central force(b) The force between two charges depend on the medium between them(c) Coulomb force is a weak force(d) The net force on a charge is the vector sum of the forces acting on it due to several other chargesThis question was addressed to me during an online exam.This key question is from Forces Between Multiple Charges in division Charges and Fields of Physics – Class 12

Answer»

Right choice is (c) Coulomb force is a weak force

The best EXPLANATION: Coulomb force is a strong force. It can be shown that the Coulomb force between TWO electrons is 10^43 TIMES than GRAVITATIONAL force. All the other options are basic properties of Coulomb’s Law.

33.

Coulomb’s Law is valid for ______(a) Only point charge(b) For both point charge and distributed charge(c) Only distributed charges(d) Neither distributed nor point chargeThis question was addressed to me in my homework.My question is taken from Coulomb’s Law in division Charges and Fields of Physics – Class 12

Answer» RIGHT option is (a) Only point charge

Best explanation: COULOMB’s Law explains the force between TWO point charges only. A distributed charge can be considered as the sum of INFINITE point charges and thus the force between two distributed charge systems can be explained.
34.

What is the C.G.S. unit of charge?(a) Stat Coulomb(b) Coulomb(c) EMU(d) PascalI got this question during an interview for a job.I would like to ask this question from Coulomb’s Law in chapter Charges and Fields of Physics – Class 12

Answer»

Correct choice is (a) Stat COULOMB

For explanation: The C.G.S. UNIT of charge is Stat Coulomb or ESU but in the SI SYSTEM the unit is Coulomb. EMU is ANOTHER unit of charge in the SI system. Pascal is the unit of pressure.

35.

An electron of mass m is kept in a vertical electric field of magnitude E. What must be the value of E so that the electron doesn’t fall due to gravity?(a) m*g*e(b) \(\frac {e}{(m*g)}\)(c) \(\frac {(m*g)}{e}\)(d) \(\frac {1}{(m*g*e)}\)This question was addressed to me in quiz.The above asked question is from Electric Field in chapter Charges and Fields of Physics – Class 12

Answer»

The CORRECT OPTION is (c) \(\frac {(m*g)}{e}\)

Easiest explanation: Gravitational force on the electron is m*g (weight of the electron). Electrical force on the body is e*E. If the electron doesn’t FALL then these two forces BALANCE each other, so m*g=E*e. Therefore E= \(\frac {(m*g)}{e}\).

36.

The amount of force exerted on a unit positive charge in an electric field is known as _____(a) Electric field intensity(b) Electric flux(c) Electric potential(d) Electric lines of forceI had been asked this question during an online interview.The above asked question is from Electric Field in chapter Charges and Fields of Physics – Class 12

Answer»

The correct choice is (a) Electric field intensity

For explanation: The zone near a CHARGE where its attraction or repulsion force works, is known as the electric field of that charge. THEORETICALLY, it is up to INFINITE but practically it has limitations. If a unit POSITIVE charge is kept in that field, it will undergo some force which is known as electric field intensity at that POINT.

37.

Find the electric field intensity at 10cm away from a point charge of 100 esu.(a) 1 dyne/esu(b) 10 dyne/esu(c) 100 dyne/esu(d) 9*10^9 dyne/esuThe question was asked in examination.This interesting question is from Electric Field in section Charges and Fields of Physics – Class 12

Answer»

The CORRECT option is (a) 1 dyne/esu

To elaborate: We know that electric field intensity at r distance from a point charge q is DEFINED as \(\frac {q}{r^2}\) in the CGS SYSTEM. Here q=100 esu, r=10cm. Substituting the value we get E=1dyne/esu. We must remember that factor \(\frac {1}{4\pi\varepsilon}\) is CONSIDERED only in the SI system and its value is 9*10^9, but in the CGS system, we simply take the value of this constant as unity.

38.

Current carrier in conductors is ____(a) Electron(b) Proton(c) Neutron(d) PositronThis question was addressed to me during an interview for a job.This intriguing question comes from Conductors and Insulators topic in chapter Charges and Fields of Physics – Class 12

Answer»

Correct option is (a) Electron

Easy EXPLANATION: Conductors have LOTS of free electrons that can CARRY electricity if the potential difference is applied across them. Protons are positively CHARGED PARTICLE and Neutron are electrically neutral. They don’t carry electricity in conductors.

39.

A lightning arrester must have the following property.(a) Discontinuity(b) Poor conductivity(c) Needle end(d) Low melting pointI had been asked this question at a job interview.I need to ask this question from Basic Properties of Electric Charge topic in section Charges and Fields of Physics – Class 12

Answer»

Correct choice is (c) NEEDLE END

Explanation: Needle end can arrest the LIGHTNING easily. Besides, the device must provide a CONTINUOUS path to electricity so that current passes to ground. It must have a high melting point else it will meltdown due to the HEAT generated during carrying lightning current.

40.

A circular annulus of inner radius r and outer radius R has a uniform charge density a. What will be the total charge on the annulus?(a) a*(R^2-r^2)(b) π*a*(R^2-r^2)(c) a*(R-r)(d) π*a* R^2This question was addressed to me in final exam.My question comes from Continuous Charge Distribution topic in chapter Charges and Fields of Physics – Class 12

Answer»

Right answer is (B) π*a*(R^2-r^2)

For EXPLANATION I would say: Total surface area of the ANNULUS = π*(R^2-r^2) because it has outer radius R and inner radius r. We know surface CHARGE density is the amount of charge stored on the unit surface area. In this case, surface charge density is a. THEREFORE total charge on the annulus = π*a*(R^2-r^2).

41.

The CO2 molecule is a dipole. The statement is ______(a) True(b) FalseThis question was posed to me in an internship interview.Question is from Electric Dipole in section Charges and Fields of Physics – Class 12

Answer»

The correct answer is (a) True

To elaborate: If we observe the structure of the CO2 molecule, we will find that the two C=O bonds are not in a STRAIGHT line DUE to the PRESENCE of lone pair of electron of C atom. Therefore the dipole moments of the two C=O bonds don’t cancel out each other and hence there is a net dipole moment present in case of CO2 molecule. It means the center of POSITIVE and negative charge does not coincide in the case of polar molecules.

42.

Three equal positive charges are kept at the corners of an equilateral triangle. What will be the vector sum of the forces acting on the particles?(a) Directed towards the center(b) Directed radially outside(c) Acts along one of the sides of the triangle(d) ZeroThe question was posed to me in unit test.Query is from Forces Between Multiple Charges topic in division Charges and Fields of Physics – Class 12

Answer»

Correct choice is (d) ZERO

Explanation: The magnitude of the force acting on each of the charges will be the same and they will differ from each other by 120 degrees. THEREFORE their resultant force will ALWAYS be zero.

43.

1 dyne/esu = _____ N/C(a) 3*10^4(b) 10^5(c) 3*10^5(d) 9*10^4This question was posed to me in an online interview.This key question is from Electric Field Lines in section Charges and Fields of Physics – Class 12

Answer» RIGHT OPTION is (a) 3*10^4

Best explanation: 1 N/C = \(\frac {1 N}{1 C} = \frac {10^5 dyne}{3*10^9 esu} = \frac {1}{3*10^4}\)dyne/esu. Therefore 1 dyne/esu=3*10^4 N/C. Both of them are the UNIT of electric field INTENSITY. But N/C is the unit of electric field intensity in the SI system while dyne/esu is the unit of field intensity in the CGS system.
44.

A 1C charge is placed at the origin. Other infinite numbers of unit charges are placed at √2, √4, √8, √16,….(up to infinite) distances from the origin in a straight line. What will be the total force acting on the 1^st charge?(a) 9*10^9 N(b) 18*10^9 N(c) Infinite(d) 1 NThe question was posed to me in unit test.The origin of the question is Forces Between Multiple Charges topic in chapter Charges and Fields of Physics – Class 12

Answer»

Right choice is (a) 9*10^9 N

To explain: The FORCE on 1^st charge will be = 9*10^-9*(\(\frac {1}{\sqrt{2}^2} + \frac {1}{\sqrt{4}^2} + \frac {1}{\sqrt{8}^2}\) +⋯) = 9*10^-9*\(\frac {\frac {1}{2}}{1-\frac {1}{2}}\) = 9*10^-9N [INFINITE GP formula].

45.

An electric dipole will be in stable equilibrium if the angle between the axis of the dipole and the electric field is ________(a) 0 degree(b) 180 degree(c) 90 degree(d) 45 degreeThis question was posed to me in my homework.Origin of the question is Dipole in a Uniform External Field topic in section Charges and Fields of Physics – Class 12

Answer»

The correct choice is (a) 0 degree

Easy explanation: Torque ACTING on a DIPOLE is p*E*sinθ where E is the electric field. Now θ is 0 degree, so sinθ becomes 0 and hence no torque acts on the dipole. So in this case, no force or torque acts on the dipole. Therefore it will be the CONDITION of stable equilibrium. In the case of θ=180 degree, sinθ is ALSO 0 but the condition is known as unstable equilibrium i.e. if we rotate the dipole a bit, it will not come back to its initial position.

46.

The band gap between the valence band and conduction band is the measure of ______(a) The conductivity of the material(b) The resistivity of the material(c) Charge density(d) Ease of ionizationI had been asked this question in my homework.My question is based upon Conductors and Insulators topic in section Charges and Fields of Physics – Class 12

Answer»

Right choice is (a) The conductivity of the material

Easiest explanation: The more the BAND gap between the VALENCE band and conduction band, the worse is the conductivity of the material. For conductors, there are overlapping bands. So, conductors can carry ELECTRICITY. But there is a HUGE energy gap in the CASE of insulators. So, they don’t carry electricity at all.

47.

If a 10C charge is placed in a medium having relative permittivity 5.4, what will be the amount of flux coming out of that charge?(a) 1*10^11 Nm^2C^-1(b) 2.1*10^11 Nm^2C^-1(c) 2.1*10^10 Nm^2C^-1(d) 7.1*10^11 Nm^2C^-1I had been asked this question during an interview for a job.I need to ask this question from Application of Gauss’s Law in section Charges and Fields of Physics – Class 12

Answer» CORRECT answer is (B) 2.1*10^11 Nm^2C^-1

The explanation: GAUSS’s Law states that electric flux coming out of a charge is=\(\frac {AMOUNT \, of \, charge}{permittivity \, of \, the \, medium}\). In this case, the charge is 1C and permittivity of the medium=5.4*8.85*10^-12. Substituting the values, we get the total flux coming out of the charge is 2.1*10^11 Nm^2C^-1.
48.

In the CGS system, \(\frac {1}{\varepsilon_o}\) is replaced by _______ if k is the dielectric constant of the medium.(a) \(\frac {4\pi}{k}\)(b) \(\frac {2\pi}{k}\)(c) \(\frac {1}{k}\)(d) 1This question was posed to me in my homework.This question is from Gauss’s Law topic in division Charges and Fields of Physics – Class 12

Answer»

The correct choice is (a) \(\frac {4\pi}{k}\)

Best explanation: In the SI system, Gauss LAW states that flux= (\(\frac {1}{\varepsilon_o}\))*electric CHARGE. But is the CGS system, the law states that, flux= \((\frac {4\pi}{k})\)*electric charge [where k is the DIELECTRIC constant of the medium]. Therefore we have to replace (\(\frac {1}{\varepsilon_o}\)) with (\(\frac {4\pi}{k}\)).

49.

A dipole is placed in an electric field such that the axis of the dipole is parallel to the electric field. The torque acting on the dipole will be the maximum in this case. The statement is _____(a) True(b) FalseI got this question in an online quiz.I want to ask this question from Dipole in a Uniform External Field topic in division Charges and Fields of Physics – Class 12

Answer»

Right answer is (b) False

To elaborate: We know, the torque acting on a dipole of moment p and placed in a field E is equal to \(\VEC{p} \times \vec{E}\) which can also be written as p*E*sinθ following the CROSS product RULE. In this CASE, the field is parallel to the axis of the dipole. Therefore θ=0 and sinθ is also equal to 0. Hence, the torque acting on the dipole will be the minimum i.e. 0. The torque will be maximum if θ=90 DEGREE and so sinθ becomes 1 in that case.

50.

Flux linked to a surface is said to be positive if the flux lines are coming out of the surface. The statement is ______(a) True(b) FalseThe question was asked in unit test.Asked question is from Electric Flux in section Charges and Fields of Physics – Class 12

Answer» CORRECT answer is (a) True

The best I can explain: Depending on the DIRECTION of ELECTRIC flux lines i.e. electric field intensity, we can differentiate between POSITIVE and negative flux. If the flux lines are going inside a surface, the flux is SAID to be negative. But if the flux lines are coming out of the surface, the flux is said to be positive.