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NCERT Exemplar Class 12 Physics solutions chapter 1 have been prepared by experts to help students understand the basic terms and different elements associated with the practical uses of electricity. NCERT Exemplar Class 12 Physics chapter 1 solutions revolve around field, force and potential surfacing out of charges that are not moving or changing with time i.e. static charges. The overall chapter deals with static charges in different situations which helps to understand their fields, forces, positions and potentials with the help of different examples given.
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For the purpose of learning and further use, students can avail NCERT Exemplar Class 12 Physics solutions chapter 1 PDF download, that have been prepared by experts after thorough study of the concepts.
Also, read - NCERT Class 11 Physics Syllabus
Question:1
In Fig.1.1, two positive charges q2 and q3 fixed along the y axis, exert a net electric force in the + x direction on a charge q1 fixed along the x axis. If a positive charge Q is added at (x, 0), the force on q1
A. shall increase along the positive x-axis.
B. shall decrease along the positive x-axis.
C. shall point along the negative x-axis.
D. shall increase but the direction changes because of the intersection of Q with q2 and q3.
Answer:
The answer is the option (a)Question:2
A point positive charge is brought near an isolated conducting sphere (Fig. 1.2). The electric field is best given by
A. Fig (i)
B. Fig (ii)
C. Fig (iii)
D. Fig (iv)
Answer:
The answer is the option (a)Question:3
The electric flux through the surface
A. in fig. 1.3 (iv) is the largest
B. in fig. 1.3 (iii) the least
C. in fig 1.3 (ii) is same as (iii) but samaller han (iv)
D. is same for all figure
Answer:
Answer: The answer is the option (d)Question:4
Five charges q1, q2, q3, q4, and q5 are fixed at their positions, as shown in Fig. 1.4. S is a Gaussian surface. Gauss's law is given by
Which of the following statements is correct?
A. E on the LHS of the above equation will have a contribution from q1, q5 and q3 while q on the RHS will have a contribution from q2 and q4 only.
B. E on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q2 and q4 only.
C. E on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q1, q3 and q5 only.
D. Both E on the LHS and q on the RHS will have contributions from q2 and q4 only..
Which of the following statements is correct?
Answer:
The answer is the option (b)Question:5
Figure 1.5 shows electric field lines in which an electric dipole p is placed as shown. Which of the following statements is correct ?
A. The dipole will not experience any force.
B.The dipole will experience a force towards right.
C.The dipole will experience a force towards left.
D.The dipole will experience a force upwards.
Answer:
The answer is the option (c)Question:6
A point charge +q, is placed at a distance d from an isolated conducting plane. The field at a point P on the other side of the plane is
A. directed perpendicular to the plane and away from the plane.
B. directed perpendicular to the plane but towards the plane.
C. directed radially away from the point charge.
D. directed radially towards the point charge.
Answer:
The answer is the option (a)Question:7
A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed.
A. perpendicular to the diameter
B. parallel to diameter
C. at an angle tilted towards the diameter
D. at an angle tilted away towards the diameter
Answer:
Explanation: In a uniformly positively charged hemisphere, the electric field should be perpendicular on a point away from the center towards the diameter.NCERT Exemplar Class 12 Physics Solutions Chapter 1 MCQII
Question:8
If over a surface, then
A. The electric field inside the surface and on it is zero.
B. The electric field inside the surface is necessarily uniform.
C. The number of flux lines entering the surface must be equal to the number of flux lines leaving it
D. All charge must necessarily be outside the surface.
Answer:
The correct answers are the options (c,d)Question:9
The Electric field at a point is
A. always continuous.
B. continuous if there is no charge at that point.
C. discontinuous only if there is negative charge at that point.
D. discontinuous if there is a charge at that point.
Answer:
The correct answers are the options (b,d)Question:6
If there were only one type of charge in the universe, then
A. on any surface
B. if the charge is outside the surface.
C. could not be defined.
D. if charge of magnitude q were inside the surface.
Answer:
The correct answers are the options (b,d)Question:11
Consider a region inside which there are various types of charges, but the total charge is zero. At points outside the region
A. The electric field is necessarily zero.
B. The electric field is due to the dipole moment of the charge distribution only.
C. The dominant electric field is , for large r, where r is the distance from an origin in this region.
D. The work done to move a charged particle along a closed path, away from the region, will be zero.
Answer:
The correct answers are the options (c,d)Question:13
A positive charge Q is uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring (Fig. 1.7). Then
A. If q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre.
B. If q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring.
C. If q < 0, it will perform SHM for small displacement along the axis.
D. q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0.
Answer:
Answer: The correct answers are the options (a, b, d)Question:12
Refer to the arrangement of charges in Fig. 1.6 and a Gaussian surface of radius R with Q at the centre. Then
A. total flux through the surface of the sphere is
B. field on the surface of the sphere is .
C. flux through the surface of sphere due to 5Q is zero.
D. field on the surface of sphere due to –2Q is same everywhere.
Answer:
The correct answers are the options (a,c)Question:14
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
Answer:
According to Gauss' law
where q= enclosed charge. So, +q and -q will cancel each other in a dipole and the sum of the charges will be zero.
Question:15
Answer:
Q is kept at the center of the spherical cavity. –Q charge is induced at the inner surface and +Q on the outer surface.Question:18
Sketch the electric field lines for a uniformly charged hollow cylinder shown in Fig 1.8.
Answer:
The side view and the top view of the electric field lines for a uniformly charged hollow cylinder are given below:Question:19
What will be the total flux through the faces of the cube (Fig. 1.9) with side of length a if a charge q is placed at
(a) A: a corner of the cube.
(b) B: mid-point of an edge of the cube.
(c) C: centre of a face of the cube.
(d) D: mid-point of B and C.
Answer:
a) The charge is equally distributed amongst the 8 cubes of a divided cube with sides 2a when the charge is placed at the center of the large cube. Therefore, the total flux through the faces of the cubeQuestion:21
Answer:
Question:22
Fig. 1.10 represents a crystal unit of cesium chloride, CsCl. The cesium atoms, represented by open circles are situated at the corners of a cube of side 0.40nm, whereas a Cl atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the Cl atom carries an excess electron.
(i) What is the net electric field on the Cl atom due to eight Cs atoms?
(ii) Suppose that the Cs atom at the corner A is missing. What is the net force now on the Cl atom due to seven remaining Cs atoms?
Answer:
a) As the chlorine atoms are distributed equally from all the eight corners of the cube from the centre, the net electric field at the centre is therefore zero.Question:24
Fig. 1.11 shows the electric field lines around three-point charges A, B and C.
(a) Which charges are positive?
(b) Which charge has the largest magnitude? Why?
(c) In which region or regions of the picture could the electric field be zero? Justify your answer.
(i) near A, (ii) near B, (iii) near C, (iv) nowhere.
Answer:
i) A and C are positive charges as the electric field lines are protruding out of these two charges.Question:25
Five charges, q each are placed at the corners of a regular pentagon of side 'a' (Fig. 1.12).
(a)
(i) What will be the electric field at O, the center of the pentagon?
(ii) What will be the electric field at O if the charge from one of the corners (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by –q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its corners?
Answer:
i) The net charge at the centre will be zero as all the charges are equidistant from the centre and they will cancel out each other.Question:26
In 1959, Lyttleton and Bondi suggested that the expansion of the universe could be explained if matter carried a net charge. Suppose that the universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = – (1 + y)e where e is the electronic charge.
(a) Find the critical value of y such that expansion may start.
(b) Show that the velocity of expansion is proportional to the distance from the centre.
Answer:
a) Let the radius of the Universe be R and assume that the hydrogen atoms are uniformly distributed. The expansion will only occur if the Coulomb repulsion is larger than the gravitational attraction at a distance RQuestion:31
Total charge –Q is uniformly spread along the length of a ring of radius R. A small test charge +q of mass m is kept at the centre of the ring and is given a gentle push along the axis of the ring.
(a) Show that the particle executes a simple harmonic oscillation.
(b) Obtain its time period.
Answer:
Let the charge q is displaced slightly by along the axis of ring. Let force on the charge q will be towards O. The motion of charge q will be simple harmonic, it the force on charge q must be proportion to z and is directed towards O.Question:30
Answer:
Let the charge q is displaced slightly by perpendicular to the line joining the two fixed charges. The net force on the charge q will be towards O.The motion of charge -q to be simple harmonic if the force on charge q must be proportional to its distance from the centre O and is directed towards O.Question:29
where the distance r is measured in , F in dynes and the charges in electrostatic units (es units), where 1es unit of charge
The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by . An approximate value of c then is
(i) Show that the coloumb law in cgs units yields
1 esu of charge = 1 (dyne)1/2 cm.
Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L.
(ii) Write 1 esu of charge where x is a dimensionless number. Show that this gives
Answer:
Question:28
Two fixed, identical conducting plates , each of surface area S are charged to –Q and q, respectively, where . A third identical plate (), free to move is located on the other side of the plate with charge q at a distance d (Fig 1.13). The third plate is released and collides with the plate β. Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst
(a) Find the electric field acting on the plate before the collision.
(b) Find the charges on β and after the collision.
(c) Find the velocity of the plate after the collision and at a distance d from the plate β.
Answer:
(a) Net electric field field at plate before collision is vector sum of electric field at plate due to plate .Question:27
Consider a sphere of radius R with charge density distributed as
a) Find the electric field at all points r.
(b) Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero? Assume that the introduction of the proton does not alter the negative charge distribution.
Answer:
a) The charge density distribution expression in the sphere tells that the electric field is radial.The NCERT Exemplar solutions for Class 12 Physics chapter 1 provided here are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE Main exams.
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It is important that students cover some key topics from this chapter for exams. Here are a few of them:
· NCERT Exemplar Class 12 Physics solutions chapter 1 enlightens about the different properties of electric charges, and how there are positive and negative charges in different things as well as the human body that attract and repel each other and how these charges can be transferred to different devices through different methods.
· The chapter also covers a variety of numericals that help to calculate the force between two point charges as well as forces between multiple charges with the application of Coulomb’s law.
· The presence of electric field around any charge and its electric flux is also explained in the chapter and its application with the help of Gauss’s law.
Also read NCERT Solution subject wise
Must read NCERT Notes subject wise
Also Check NCERT Books and NCERT Syllabus here:
We have solved all 31 questions from the main exercise that is mentioned in the NCERT book in NCERT exemplar Class 12 physics solutions chapter 1.
Yes, these solutions are highly helpful to clear both theoretical understandings of the chapter, and also get an idea of how to solve numerical questions in an exam.
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