CBSE Class 12 Physics Chapter-1 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 1 – Electric Charges and Fields 5 Mark Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-1 Important Questions
CBSE Class 12 Physics Important Questions Chapter 1 – Electric Charges and Fields
5 Mark Questions
1. (1) The electric field 
due to a point change at any point near to it is defined as:
where q is the test charge and 
is the force acting on it. What is the significance of lim 
in this expression?
(2) Two charges each 
but opposite in sign forms a system. These charges are located at points A (0,0, -10) cm and B (0,0, +10) cm respectively. What is the total charge and electric dipole moment of the system?
Ans. (1) The Significance of writing lim means the test charge should be vanishingly small so that it should not disturb the presence of source charge.
(2) 
(i) Total charge of the system
= 
= zero. 
(ii) 
Direction of Dipole moment -Along negative z-axis.
2. (a) Sketch electric lines of force due to (i) isolated positive change (ie q>0) and (ii) isolated negative change (ie q<0)
(b) Two point changes q and –q are placed at a distance 2a apart. Calculate the electric field at a point P situated at a distance r along the perpendicular bisector of the line joining the charges. What is the field when r >> a?
Ans.
(b)
can be calculated by using a parallelogram law of vector addition. 
3. (a) What is an equi-potential surface? Show that the electric field is always directed perpendicular to an equi-potential surface.
(b) Derive an expression for the potential at a point along the axial line of a short electric dipole?
Ans. (a) The surface which has same potential through out is called an equipotential surface.
Since work done is moving a test charge along an equipotential surface is always zero.
Or
(b) Consider an electric dipole of dipole length 2a and point P on the axial line such that OP= r
where O is the center of the dipole.
Electric Potential at point P due to the dipole
For a short electric dipole (a) can be neglected
4. Check that the ratio 
is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Ans. The given ratio is
.
Where,
G = Gravitational constant
Its unit is 
= Masses of electron and proton.
Their unit is kg.
e = Electric charge.
Its unit is C.
= Permittivity of free space
Its unit is 
Hence, the given ratio is dimensionless.
G = 
Hence, the numerical value of the given ratio is
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.
5. Four point charges 
= 
C, 
= –
C, 
= C, and 
= –C are located at the corners of a square ABCD of side 10 cms. What is the force on a charge of 1 
C placed at the centre of the square?
Ans. The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
Where,
(Sides) 
(Diagonals) 
s
A charge of amount 
C is plasced at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 C charge at centre O is zero.
6. Two point charges 
C and 
are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 
is placed at this point, what is the force experienced by the test charge?
Ans. (a) The situation is represented in the given figure. O is the mid-point of line AB.
Distance between the two charges, 
Net electric field at point 
Electric field at point O caused by +
charge,
= 
along OB
Where,
= Permittivity of free space
Magnitude of electric field at point O caused by – charge,
= 
= 
along OB
[since the values 
of are same, the value is multiplied with2]
Therefore, the electric field at mid-point O is 
(b) A test charge of amount 
is placed at mid-point O.
Force experienced by the test charge = F
The force is directed along line 
. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 
along OA.
7. A system has two charges 
C and 
C located at points A: 
and 
respectively. What are the total charge and electric dipole moment of the system?
Ans. Both the charges can be located in a coordinate frame of reference as shown in the given figure.
At A, amount of charge, 
At B, amount of charge, 
Total charge of the system,
Distance between two charges at points A and B,
Electric dipole moment of the system is given by,
along positive z-axis
Therefore, the electric dipole moment of the system is 
m along positive z-axis.
8. (a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is
s? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Ans. (a) Charge on sphere 
Charge on sphere 
Distance between the spheres, 
Force of repulsion between the two spheres,
Where,
Free space permittivity
=
∴
Therefore, the force between the two spheres is 
(b) After doubling the charge, charge on sphere Charge on sphere 
The distance between the spheres is halved.
∴
Force of repulsion between the two spheres,
s
Therefore, the force between the two spheres is 
9. Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Ans. Distance between the spheres, 
Initially, the charge on each sphere, 
When sphere A is touched with an uncharged sphere C, amount of charge from A will transfer to sphere
C. Hence, charge on each of the spheres, A and C, 
.
When sphere C with charge is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given as,
Each sphere will share each half. Hence, charge on each of the spheres, C and B, is
.
Force of repulsion between sphere A having charge and sphere B having charge
=
=
Therefore, the force of attraction between the two spheres is
.
10. 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 and of magnitude 
What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?
Ans. The situation is represented in the following figure.
A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.
Charge density of plate A, 
Charge density of plate B, 
In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the respective plates.
Electric field E in region II is given by the relation,
Where,
=Permittivity of free space =
∴
Therefore, electric field between the plates is .
11. An oil drop of 12 excess electrons is held stationary under a constant electric field of 
in Millikan’s oil drop experiment. The density of the oil is 1.26 g
. Estimate the radius of the drop. 
Ans. Excess electrons on an oil drop, n = 12
Electric field intensity, E =
Density of oil, 
Acceleration due to gravity, g 
Charge on an electron, e 
Radius of the oil drop = r
Force (F) due to electric field E is equal to the weight of the oil drop (W)
F = W
Eq = mg
Ene 
Where,
q = Net charge on the oil drop = ne
m = Mass of the oil drop
= Volume of the oil drop × Density of oil
Therefore, the radius of the oil drop is 
12. 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 
per metre. What are the force and torque experienced by a system having a total dipole moment equal to 
Cm in the negative z-direction?
Ans. Dipole moment of the system, p = q × dl = 
Rate of increase of electric field per unit length,
Force (F) experienced by the system is given by the relation,
F = qE
The force is 
in the negative z-direction i.e., opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is 180°.
Torque 
is given by the relation,
= pE sin180°
= 0
Therefore, the torque experienced by the system is zero.
13. (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Ans. (a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.
Let q is the charge inside the conductor and 
is the permittivity of free space.
According to Gauss’s law,
Flux, 
Here, E = 0
Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.
(b) The outer surface of conductor A has a charge of amount Q. Another conductor B having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount – q will be induced in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is Q + q.
(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.
14. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is 
, where is the unit vector in the outward normal direction, and 
is the surface charge density near the hole.
Ans. Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.Let E is the electric field just outside the conductor, q is the electric charge, is the charge density, and is the permittivity of free space.
Charge 
According to Gauss’s law,
Therefore, the electric field just outside the conductor is
. This field is a superposition of field due to the cavity 
and the field due to the rest of the charged conductor. These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.
Therefore, the field due to the rest of the conductor is
.
Hence, proved.
15. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density 
without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Ans. Take a long thin wire XY (as shown in the figure) of uniform linear charge density.
Consider a point A at a perpendicular distance l from the mid-point O of the wire, as shown in the following figure.
Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x
Let q be the charge on this piece.
Electric field due to the piece,
The electric field is resolved into two rectangular components. 
is the perpendicular component and 
is the parallel component.
When the whole wire is considered, the component is cancelled.
Only the perpendicular component affects point A.
Hence, effective electric field at point A due to the element dx is dE1.
On differentiating equation (2), we obtain
From equation (2),
Putting equations (3) and (4) in equation (1), we obtain
The wire is so long that tends from 
to 
.
By integrating equation (5), we obtain the value of field 
as,
Therefore, the electric field due to long wire is 
16. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge 
e, and the ‘down’ quark (denoted by d) of charge (-1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
Ans.A proton has three quarks. Let there be n up quarks in a proton, each having a charge of
.
Charge due to n up quarks
Number of down quarks in a proton = 3 – n
Each down quark has a charge of 
.
Charge due to (3 – n) down quarks 
Total charge on a proton = + e
Number of up quarks in a proton, n = 2
Number of down quarks in a proton = 3 – n = 3 – 2 = 1
Therefore, a proton can be represented as
A neutron also has three quarks. Let there be n up quarks in a neutron, each having a charge of 
.
Charge on a neutron due to n up quarks 
Number of down quarks is 3 – n, each having a charge of 
.
Charge on a neutron due to 
down quarks = 
Total charge on a neutron = 0
Number of up quarks in a neutron, n = 1
Number of down quarks in a neutron = 3 – n = 2
Therefore, a neutron can be represented as
.
17. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Ans. A proton has three quarks. Let there be n up quarks in a proton, each having a charge of
.
Charge due to n up quarks
Number of down quarks in a proton = 3 – n
Each down quark has a charge of .
Charge due to (3 – n) down quarks 
Total charge on a proton = + e
Number of up quarks in a proton, n = 2
Number of down quarks in a proton = 3 – n = 3 – 2 = 1
Therefore, a proton can be represented as
A neutron also has three quarks. Let there be n up quarks in a neutron, each having a charge of 
.
Charge on a neutron due to n up quarks 
Number of down quarks is 3 – n, each having a charge of 
.
Charge on a neutron due to down quarks = 
Total charge on a neutron = 0
Number of up quarks in a neutron, n = 1
Number of down quarks in a neutron = 3 – n = 2
Therefore, a neutron can be represented as.
18. A particle of mass m and charge ( – q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is 
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.
Ans. Charge on a particle of mass m = – q
Velocity of the particle = vx
Length of the plates =L
Magnitude of the uniform electric field between the plates = E
Mechanical force, F = Mass (m) × Acceleration (a)
Therefore, acceleration, 
Time taken by the particle to cross the field of length L is given by,
t
In the vertical direction, initial velocity, u = 0
According to the third equation of motion, vertical deflection s of the particle can be obtained as,
Hence, vertical deflection of the particle at the far edge of the plate is
. This is similar to the motion of horizontal projectiles under gravity.