Case Study Question for Class 12 Physics Chapter 1 Electric Charges and Fields
There is Case Study Questions in class 12 Physics in session 2020-21. The first two questions in the board exam question paper will be based on Case Study and Assertion & Reason. Case Study Questions will have 5 MCQs out of which students will have to attempt any 4 questions. Here are the questions based on case study.
Case Study Question 1:
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.
(i) Electric lines of force about a positive point charge are
(a) radially outwards
(b) circular clockwise
(c) radially inwards
(d) parallel straight lines
(ii) 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.
(iii) Which one of the following patterns of electric line of force is not possible in field due to stationary charges?
(iv) 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)
(v) The figure below shows the electric field lines due to two positive charges. The magnitudes EA, EB and 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
Answer
Answers: (i) a(ii) (c)
(iii) (c)
(iv)(d)
(v) (a)
Case Study Question 2:
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.
(i) 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.
(ii) 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
(iii) If a charge on a body is 1 nC, then how many electrons are present on the body?
(a) 6.25 x 1027
(b) 1.6 x 1019
(c) 6.25 X 1028
(d) 6.25 X 109
(iv) If a body gives out 109 electrons every second, how much time is required to get a total charge of 1 from it?
(a) 190.19 years
(b) 150.12 years
(c) 198.19 years
(d) 188.21 years
(v) 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 1012
(b) 3 x 1012
(c) 2 x 1014
(d) 3 x 1014
Answer
Answers: (i) (d)(ii) (b)
(iii) (d)
(iv)(c)
(v) (a)
Case Study Question 3:
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.
(i) The dipole moment of a dipole in a uniform external field Ē is B. 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)
(ii) 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 105 NC-1. The maximum torque on the dipole is
(a) 0.2 x 10-3 Nm
(b) 1x 10-3 Nm
(c) 2 x 10-3 Nm
(d) 4x 10-3 Nm
(iii) 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)
(iv) 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
(v) 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
Answer
Answers: (i) (a)(ii) (c)
(iii) (d)
(iv)(c)
(v) (a)
Case Study Question 4:
A charge is a property associated with the matter due to which it experiences and produces an electric and magnetic field. Charges are scalar in nature and they add up like real number. Also, the total charge of an isolated system is always conserved. When the objects rub against each other charges acquired by them must be equal and opposite.
(i) The cause of a charging is:
(a) the actual transfer of protons.
(b) the actual transfer of electrons.
(c) the actual transfer of neutrons.
(d) none the above
Answer
Answer: (d) the actual transfer of electrons.(ii) Pick the correct statement.
(a) The glass rod gives protons to silk when they are rubbed against each other.
(b) The glass rod gives electrons to silk when they are rubbed against each other.
(c) The glass rod gains protons from silk when they are rubbed against each other.
(d) The glass rod gains electrons when they are rubbed against each other.
Answer
Answer: (b) The glass rod gives electrons to silk when they are rubbed against each other.(iii) If two electrons are each 1.5 × 10–10 m from a proton, the magnitude of the net electric force they will exert on the proton is
(a) 1.97 × 10–8 N (b) 2.73 × 10–8 N
(c) 3.83 × 10–8 N (d) 4.63 × 10–8 N
Answer
Answer: (a) 1.97 × 10–8 N(iv) A charge is a property associated with the matter due to which it produces and experiences:
(a) electric effects only
(b) magnetic effects only
(c) both electric and magnetic effects
(d) none of these.
Answer
Answer: (c) both electric and magnetic effects(v) The cause of quantization of electric charges is:
(a) Transfer of an integral number of neutrons.
(b) Transfer of an integral number of protons.
(c) Transfer of an integral number of electrons.
(d) None of the above.
Answer
Answer: (c) Transfer of an integral number of electrons.Case Study Question 5:
Surface Charge Density. Surface charge density is defined as the charge per unit surface area the
surface (Arial) charge symmetric distribution and follow Gauss law of electro statics mathematical term of surface charge density σ=ΔQ/ΔS
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite sign (± s). Having magnitude 8.8 × 10–12 cm–2 as shown here. The intensity of electrified at a point is E =σ/ε0 and flux is Φ=E.ΔS, where ΔS = 1 m2 (unit arial plate)
(i) E in the outer region (I) of the first (A) plate is
(a) 1.7 × 10–22 N/C
(b) 1.1 × 10–12 V/m
(c) Zero
(d) Insufficient data
Answer
Ans. (c) Zero(ii) E in the outer region (III) of the second plate (B) is
(a) 1 N/C
(b) 0.1 V/m
(c) 0.5 N/C
(d) zero
Answer
Ans. (d) Zero(iii) E between (II) the plate is
(a) 1 N/C
(b) 0.1 V/m
(c) 0.5 N/C
(d) None of these
Answer
Ans. (d) None of these(iv) The ratio of E from left side of plate A at distance 1 cm and 2 m respectively is
(a) 1 : 2
(b) 10 : 2
(c) 1 : 1
(d) 20 : 1
Answer
Ans. (c) 1 : 1(v) In order to estimate the electric field due to a thin finite plane metal plate the Gaussian surface considered is
(a) Spherical
(b) Linear
(c) Cylindrical
(d) Cybic
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