Chapter 1 – Electric Charges And Fields Questions and Answers: NCERT Solutions for Class 12 Physics

1. What is the force between two small charged spheres of charges
2×10−7C2×10−7C and 3×10−7C3×10−7C placed 30cm30cm apart in air?

Ans: We are given the following information:
Repulsive force of magnitude,F=6×10−3NF=6×10−3N
Charge on the first sphere, q1=2×10−7Cq1=2×10−7C
Charge on the second sphere, q2=3×10−7Cq2=3×10−7C
Distance between the two spheres, r=30cm=0.3mr=30cm=0.3m
Electrostatic force between the two spheres is given by Coulomb’s law as,
F=14πε0q1q2r2F=14πε0q1q2r2
Where, ε0ε0is the permittivity of free space and,
14πε0=9×109Nm2C−214πε0=9×109Nm2C−2
Now on substituting the given values, Coulomb’s law becomes,
$\begin{align}
& F=\frac{9\times {{10}^{9}}\times 2\times {{10}^{-7}}\times 3\times {{10}^{-7}}}{{{\left( 0.3 \right)}^{2}}} \\
& \therefore F=6\times {{10}^{-3}}N \\
\end{align}$
Therefore, we found the electrostatic force between the given charged spheres to be F=6×10−3NF=6×10−3N. Since the charges are of the same nature, we could say that the force is repulsive.

2. The electrostatic force on a small sphere of charge 0.4μC0.4μC due to another small sphere of charge −0.8μC−0.8μC in air is 0.2N.
a) What is the distance between the two spheres?

Ans: Electrostatic force on the first sphere is given to be, F=0.2NF=0.2N
Charge of the first sphere is, q1=0.4μC=0.4×10−6Cq1=0.4μC=0.4×10−6C
Charge of the second sphere is, q2=−0.8μC=−0.8×10−6Cq2=−0.8μC=−0.8×10−6C
We have the electrostatic force given by Coulomb’s law as,
F=14πε0q1q2r2F=14πε0q1q2r2
⇒r=q1q24πε0F−−−−√⇒r=q1q24πε0F
Substituting the given values in the above equation, we get,
⇒r=0.4×10−6×8×10−6×9×1090.2−−−−−−−−−−−−−−−−√⇒r=0.4×10−6×8×10−6×9×1090.2
⇒r=144×10−4−−−−−−−−−√⇒r=144×10−4
∴r=0.12m∴r=0.12m
Therefore, we found the distance between charged spheres to be r=0.12mr=0.12m.

b) What is the force on the second sphere due to the first?

Ans: From Newton’s third law of motion, we know that every action has an equal and opposite reaction.
Thus, we could say that the given two spheres would attract each other with the same force.
So, the force on the second sphere due to the first sphere will be 0.2N0.2N.

3. Check whether the ratio ke2Gmempke2Gmempis dimensionless. Look up a table of physical constants and hence determine the value of the given ratio. What does the ratio signify?

Ans: We are given the ratio, ke2Gmempke2Gmemp.
Here, G is the gravitational constant which has its unit Nm2kg−2Nm2kg−2;
memeand mpmp are the masses of electron and proton in kgkg respectively;
ee is the electric charge in CC;
kk is a constant given by k=14πε0k=14πε0
In the expression for k, ε0ε0 is the permittivity of free space which has its unit Nm2C−2Nm2C−2.
Now, we could find the dimension of the given ratio by considering their units as follows:
ke2Gmemp=[Nm2C−2][C]2[Nm2kg−2][kg][kg]=M0L0T0ke2Gmemp=[Nm2C−2][C]2[Nm2kg−2][kg][kg]=M0L0T0
Clearly, it is understood that the given ratio is dimensionless.
Now, we know the values for the given physical quantities as,
e=1.6×10−19Ce=1.6×10−19C
G=6.67×10−11Nm2kg−2G=6.67×10−11Nm2kg−2
me=9.1×10−31kgme=9.1×10−31kg
mp=1.66×10−27kgmp=1.66×10−27kg
Substituting these values into the required ratio, we get,
ke2Gmemp=9×109×(1.6×10−19)26.67×10−11×9.1×10−3×1.67×10−22ke2Gmemp=9×109×(1.6×10−19)26.67×10−11×9.1×10−3×1.67×10−22
⇒ke2Gmemp≈2.3×1039⇒ke2Gmemp≈2.3×1039
We could infer that the given ratio is the ratio of electrical force to the gravitational force between a proton and an electron when the distance between them is kept constant.

4.
a) Explain the meaning of the statements ‘electric charge of a body is quantized’.

Ans: The given statement ‘Electric charge of a body is quantized’ means that only the integral number (1,2,3,…,n)(1,2,3,…,n) of electrons can be transferred from one body to another.
That is, charges cannot be transferred from one body to another in fraction.
b) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges?
Ans: On a macroscopic scale or large-scale, the number of charges is as large as the magnitude of an electric charge.
So, quantization is considered insignificant at a macroscopic scale for an electric charge and electric charges are considered continuous.

5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Ans: Rubbing two objects would produce charges that are equal in magnitude and opposite in nature on the two bodies.
This happens due to the reason that charges are created in pairs. This phenomenon is called charging by friction.
The net charge of the system however remains zero as the opposite charges equal in magnitude annihilate each other.
So, rubbing a glass rod with a silk cloth creates opposite charges of equal magnitude on both of them and this observation is found to be consistent with the law of conservation of charge.

6. Four point charges qA=2μCqA=2μC, qB=−5μCqB=−5μC, qC=2μCqC=2μCand qD=−5μCqD=−5μC are located at the corners of a square ABCD with side 10cm. What is the force on the 1μC1μC charge placed at the centre of this square?

Ans: Consider the square of side length 10cm10cm given below with four charges at its corners and let O be its centre.
From the figure we find the diagonals to be,
AC=BD=102–√cmAC=BD=102cm
⇒AO=OC=DO=OB=52–√cm⇒AO=OC=DO=OB=52cm
Now the repulsive force at O due to charge at A,
FAO=kqAqOOA2=k(+2μC)(1μC)(52√)2FAO=kqAqOOA2=k(+2μC)(1μC)(52)2…………………………………………… (1)
And the repulsive force at O due to charge at D,
FDO=kqDqOOD2=k(+2μC)(1μC)(52√)2FDO=kqDqOOD2=k(+2μC)(1μC)(52)2………………………………………….. (2)
And the attractive force at O due to charge at B,
FBO=kqBqOOB2=k(−5μC)(1μC)(52√)2FBO=kqBqOOB2=k(−5μC)(1μC)(52)2……………………………………………. (3)
And the attractive force at O due to charge at C,
FCO=kqCqOOC2=k(−5μC)(1μC)(52√)2FCO=kqCqOOC2=k(−5μC)(1μC)(52)2……………………………………………… (4)
We find that (1) and (2) are of same magnitude but they act in the opposite direction and hence they cancel out each other.
Similarly, (3) and (4) are of the same magnitude but in the opposite direction and hence they cancel out each other too.
Hence, the net force on charge at centre O is found to be zero.

7.
a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

Ans: An electrostatic field line is a continuous curve as the charge experiences a continuous force on being placed in an electric field.
As the charge doesn’t jump from one point to the other, field lines will not have sudden breaks.

b) Explain why two field lines never cross each other at any point?

Ans: If two field lines are seen to cross each other at a point, it would imply that the electric field intensity has two different directions at that point, as two different tangents (representing the direction of electric field intensity at that point) can be drawn at the point of intersection.
This is however impossible and thus, two field lines never cross each other.

8. Two point charges qA=3μCqA=3μCand qB=−3μCqB=−3μCare located 20cm apart in vacuum.
a) What is the electric field at the midpoint O of the line AB joining the two charges?

Ans: The situation could be represented in the following figure. Let O be the midpoint of line AB.
(Image Will Be Updated Soon)
We are given:
AB=20cmAB=20cm
AO=OB=10cmAO=OB=10cm
Take E to be the electric field at point O, then,
The electric field at point O due to charge +3μC+3μCwould be,
E1=3×10−64πε0(AO)2=3×10−64πε0(10×10−2)2NC−1E1=3×10−64πε0(AO)2=3×10−64πε0(10×10−2)2NC−1along OB
The electric field at point O due to charge −3μC−3μCwould be,
E2=∣∣3×10−64πε0(OB)2∣∣=3×10−64πε0(10×10−2)2NC−1E2=|3×10−64πε0(OB)2|=3×10−64πε0(10×10−2)2NC−1along OB
The net electric field,
⇒E=E1+E2⇒E=E1+E2
⇒E=2×9×109×3×10−6(10×10−2)2⇒E=2×9×109×3×10−6(10×10−2)2
⇒E=5.4×106NC−1⇒E=5.4×106NC−1
Therefore, the electric field at mid-point O is E=5.4×106NC−1E=5.4×106NC−1 along OB.

b) If a negative test charge of magnitude 1.5×10−19C1.5×10−19C is placed at this point, what is the force experienced by the test charge?

Ans: We have a test charge of magnitude 1.5×10−9C1.5×10−9C placed at mid-point O and we found the electric field at this point to be E=5.4×106NC−1E=5.4×106NC−1.
So, the force experienced by the test charge would be F,
⇒F=qE⇒F=qE
⇒F=1.5×10−9×5.4×106⇒F=1.5×10−9×5.4×106
⇒F=8.1×10−3N⇒F=8.1×10−3N
This force will be directed along OA since like charges repel and unlike charges attract.

9. A system has two charges qA=2.5×10−7CqA=2.5×10−7Cand qB=−2.5×10−7CqB=−2.5×10−7Clocated at points A:(0,0,−15cm)A:(0,0,−15cm) and B:(0,0,+15cm)B:(0,0,+15cm) respectively. What are the total charge and electric dipole moment of the system?

Ans: The figure given below represents the system mentioned in the question:
The charge at point A, qA=2.5×10−7CqA=2.5×10−7C
The charge at point B, qB=−2.5×10−7CqB=−2.5×10−7C
Then, the net charge would be, q=qA+qB=2.5×10−7C−2.5×10−7C=0q=qA+qB=2.5×10−7C−2.5×10−7C=0
The distance between two charges at A and B would be,
d=15+15=30cmd=15+15=30cm
d=0.3md=0.3m
The electric dipole moment of the system could be given by,
P=qA×d=qB×dP=qA×d=qB×d
⇒P=2.5×10−7×0.3⇒P=2.5×10−7×0.3
∴P=7.5×10−8Cm∴P=7.5×10−8Cm along the
+z+z
axis.
Therefore, the electric dipole moment of the system is found to be 7.5×10−8Cm7.5×10−8Cm and it is directed along the positive
zz
-axis.

10. An electric dipole with dipole moment 4×10−9Cm4×10−9Cm is aligned at 30∘30∘ with direction of a uniform electric field of magnitude 5×104NC−15×104NC−1. Calculate the magnitude of the torque acting on the dipole.

Ans: We are given the following:
Electric dipole moment, p→=4×10−9Cmp→=4×10−9Cm
Angle made by p→p→ with uniform electric field, θ=30∘θ=30∘
Electric field, E→=5×104NC−1E→=5×104NC−1
Torque acting on the dipole is given by
τ=pEsinθτ=pEsin⁡θ
Substituting the given values we get,
⇒τ=4×10−9×5×104×sin30∘⇒τ=4×10−9×5×104×sin⁡30∘
⇒τ=20×10−5×12⇒τ=20×10−5×12
∴τ=10−4Nm∴τ=10−4Nm
Thus, the magnitude of the torque acting on the dipole is found to be 10−4Nm10−4Nm.

11. A polythene piece rubbed with wool is found to have a negative charge of 3×10−7C3×10−7C
a) Estimate the number of electrons transferred (from which to which?)

Ans: When polythene is rubbed against wool, a certain number of electrons get transferred from wool to polythene.
As a result of which wool becomes positively charged on losing electrons and polythene becomes negatively charged on gaining them.
We are given:
Charge on the polythene piece, q=−3×10−7Cq=−3×10−7C
Charge of an electron, e=−1.6×10−19Ce=−1.6×10−19C
Let n be the number of electrons transferred from wool to polythene, then, from the property of quantization we have,
q=neq=ne
⇒n=qe⇒n=qe
Now, on substituting the given values, we get,
⇒n=−3×10−7−1.6×10−19⇒n=−3×10−7−1.6×10−19
∴n=1.87×1012∴n=1.87×1012
Therefore, the number of electrons transferred from wool to polythene would be1.87×10121.87×1012.

b) Is there a transfer of mass from wool to polythene?

Ans: Yes, during the transfer of electrons from wool to polythene, along with charge, mass is transferred too.
Let mm be the mass being transferred in the given case and meme be the mass of the electron, then,
m=me×nm=me×n
⇒m=9.1×10−31×1.85×1012⇒m=9.1×10−31×1.85×1012
∴m=1.706×10−18kg∴m=1.706×10−18kg
Thus, we found that a negligible amount of mass does get transferred from wool to polythene.

12.
a) Two insulated charged copper spheres AA and BB have their centres separated by a distance of 50cm50cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−7C6.5×10−7C? The radii of A and B are negligible compared to the distance of separation.

Ans: We are given:
Charges on spheres AA and BB are equal,
qA=qB=6.5×10−7CqA=qB=6.5×10−7C
Distance between the centres of the spheres is given as,
r=50cm=0.5mr=50cm=0.5m
It is known that the force of repulsion between the two spheres would be given by Coulomb’s law as,
F=qAqB4πε0r2F=qAqB4πε0r2
Where, εoεo is the permittivity of the free space
Substituting the known values into the above expression, we get,
F=9×109×(6.5×10−7)2(0.5)2=1.52×10−2NF=9×109×(6.5×10−7)2(0.5)2=1.52×10−2N
Thus, the mutual force of electrostatic repulsion between the two spheres is found to beF=1.52×10−2NF=1.52×10−2N.

b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Ans: It is told that the charges on both the spheres are doubled and the distance between the centres of the spheres is halved. That is,
qA′=qB′=2×6.5×10−7=13×10−7CqA′=qB′=2×6.5×10−7=13×10−7C
r′=12(0.5)=0.25mr′=12(0.5)=0.25m
Now, we could substitute these values in Coulomb’s law to get,
F′=qA′qB′4πε0r′2F′=qA′qB′4πε0r′2
⇒F=9×109×(13×10−7)2(0.25)2⇒F=9×109×(13×10−7)2(0.25)2
⇒F=0.243N⇒F=0.243N
The new mutual force of electrostatic repulsion between the two spheres is found to be 0.243N0.243N.

13. Suppose the spheres AA and BB in question 1212 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 AA and BB?

Ans: We are given the following:
Distance between the spheres AA and BB is r=0.5mr=0.5m
The charge on each sphere initially is found to be qA=qB=6.5×10−7CqA=qB=6.5×10−7C
Now, when an uncharged sphere CC is made to touch the sphere AA, a certain amount of charge from AA will get transferred to the sphere CC, making both AA and CC to have equal charges in them. So,
qA′=qC=12(6.5×10−7)=3.25×10−7CqA′=qC=12(6.5×10−7)=3.25×10−7C
Now, when the sphere CC is made to touch the sphere BB, there is a similar transfer of charge making both CC and BB to have equal charges in them. So,
qC′=qB′=3.25×10−7+6.5×10−72=4.875×10−7CqC′=qB′=3.25×10−7+6.5×10−72=4.875×10−7C
Thus, the new force of repulsion between the spheres AA and BB would now become,
F′=qA′qB′4πε0r2F′=qA′qB′4πε0r2
⇒F′=9×109×3.25×10−7×4.875×10−7(0.5)2⇒F′=9×109×3.25×10−7×4.875×10−7(0.5)2
⇒F′=5.703×10−3N⇒F′=5.703×10−3N

14. Figure below shows tracks taken by three charged particles in a uniform electrostatic field. Give the signs of the three charges and also mention which particle has the highest charge to mass ratio?

Ans: From the known properties of charges, we know that the unlike charges attract and like charges repel each other.
So, the particles 1 and 2 that move towards the positively charged plate while repelling away from the negatively charged plate would be negatively charged and the particle 3 that moves towards the negatively charged plate while repelling away from the positively charged plate would be positively charged.
Now, we know that the charge to mass ratio (which is generally known as emf) is directly proportional to the displacement or the amount of deflection for a given velocity.
Since the deflection of particle 3 is found to be maximum among the three, it would have the highest charge to mass ratio.

15. Consider a uniform electric field E=3×103i^N/CE=3×103i^N/C.
a) Find the flux of this field through a square of side 10cm10cmwhose plane is parallel to the y-z plane.

Ans: We are given:
Electric field intensity, E→=3×103i^N/CE→=3×103i^N/C
Magnitude of electric field intensity, ∣∣∣E→∣∣∣=3×103N/C|E→|=3×103N/C
Side of the square, a=10cm=0.1ma=10cm=0.1m
Area of the square, A=a2=0.01m2A=a2=0.01m2
Since the plane of the square is parallel to the y-z plane, the normal to its plane would be directed in the x direction. So, angle between normal to the plane and the electric field would be, θ=0∘θ=0∘
We know that the flux through a surface is given by the relation,
ϕ=|E||A|cosθϕ=|E||A|cos⁡θ
Substituting the given values, we get,
⇒ϕ=3×103×0.01×cos0∘⇒ϕ=3×103×0.01×cos⁡0∘
∴ϕ=30Nm2/C∴ϕ=30Nm2/C
Thus, we found the net flux through the given surface to be ϕ=30Nm2/Cϕ=30Nm2/C.

b) What would be the flux through the same square if the normal to its plane makes 60∘60∘ angle with the x-axis?

Ans: When the plane makes an angle of 60∘60∘ with the x-axis, the flux through the given surface would be,
ϕ=|E||A|cosθϕ=|E||A|cos⁡θ
⇒ϕ=3×103×0.01×cos60∘⇒ϕ=3×103×0.01×cos⁡60∘
⇒ϕ=30×12⇒ϕ=30×12
⇒ϕ=15Nm2/C⇒ϕ=15Nm2/C
So, we found the flux in this case to be, ϕ=15Nm2/Cϕ=15Nm2/C.

16. What is the net flux of the uniform electric field of exercise 1.151.15 through a cube of side 20cm20cm oriented so that its faces are parallel to the coordinate planes?

Ans: We are given that all the faces of the cube are parallel to the coordinate planes.
Clearly, the number of field lines entering the cube is equal to the number of field lines entering out of the cube. As a result, the net flux through the cube would be zero.

17. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0×103Nm2/C8.0×103Nm2/C.
a) What is the net charge inside the box?

Ans: We are given that:
Net outward flux through surface of the box,
ϕ=8.0×103Nm2/Cϕ=8.0×103Nm2/C
For a body containing of net charge qq, flux could be given by,
ϕ=qε0ϕ=qε0
Where, ε0=8.854×10−12N−1C2m−2=ε0=8.854×10−12N−1C2m−2= Permittivity of free space
Therefore, the charge qq is given by
q=ϕε0q=ϕε0
⇒q=8.854×10−12×8.0×103⇒q=8.854×10−12×8.0×103
⇒q=7.08×10−8⇒q=7.08×10−8
⇒q=0.07μC⇒q=0.07μC
Therefore, the net charge inside the box is found to be 0.07μC0.07μC.

b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

Ans: No, the net flux entering out through a body depends on the net charge contained within the body according to Gauss’s law.
So, if the net flux is given to be zero, then it can be inferred that the net charge inside the body is zero.
However, the net charge of the body being zero only implies that the body has equal amount of positive and negative charges and thus, we cannot conclude that there were no charges inside the box.

18. A point charge +10μC+10μC is a distance 5cm5cm directly above the centre of a square of side 10cm10cm, as shown in Figure below. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10cm10cm)

Ans: Consider the square as one face of a cube of edge length 10cm10cm with a charge qq at its centre, according to Gauss’s theorem for a cube, total electric flux is through all its six faces.
ϕtotal=qε0ϕtotal=qε0
The electric flux through one face of the cube could be now given by,
ϕ=ϕtotal6ϕ=ϕtotal6
.
ϕ=16qε0ϕ=16qε0
ε0=8.854×10−12N−1C2m−2=ε0=8.854×10−12N−1C2m−2= Permittivity of free space
The net charge enclosed would be, q=10μC=10×10−6Cq=10μC=10×10−6C
Substituting the values given in the question, we get,
ϕ=16×10×10−68.854×10−12ϕ=16×10×10−68.854×10−12
∴ϕ=1.88×105Nm2C−1∴ϕ=1.88×105Nm2C−1
Therefore, electric flux through the square is found to be 1.88×105Nm2C−11.88×105Nm2C−1.

19. A point charge of 2.0μC2.0μC is kept at the centre of a cubic Gaussian surface of edge length 9cm9cm. What is the net electric flux through this surface?

Ans: Let us consider one of the faces of the cubical Gaussian surface considered (square).
Since a cube has six such square faces in total, we could say that the flux through one surface would be one-sixth the total flux through the gaussian surface considered.
The net flux through the cubical Gaussian surface by Gauss’s law could be given by,
ϕtotal=qε0ϕtotal=qε0
So, the electric flux through one face of the cube would be,
ϕ=ϕtotal6ϕ=ϕtotal6
⇒ϕ=16qε0⇒ϕ=16qε0……………………………….. (1)
But we have,
ε0=8.854×10−12N−1C2m−2=ε0=8.854×10−12N−1C2m−2= Permittivity of free space
Charge enclosed, q=10μC=10×10−6Cq=10μC=10×10−6C
Substituting the given values in (1) we get,
ϕ=16×10×10−68.854×10−12ϕ=16×10×10−68.854×10−12
⇒ϕ=1.88×105Nm2C−1⇒ϕ=1.88×105Nm2C−1
Therefore, electric flux through the square surface is 1.88×105Nm2C−11.88×105Nm2C−1.

20. A point charge causes an electric flux of −1.0×103Nm2/C−1.0×103Nm2/C to pass through a spherical Gaussian surface of 10cm10cm radius centred on the charge.
a) If the radius of the Gaussian surface were doubled, how much flux could pass through the surface?

Ans: We are given:
Electric flux due to the given point charge, ϕ=−1.0×103Nm2/Cϕ=−1.0×103Nm2/C
Radius of the Gaussian surface enclosing the point charge,r=10.0cmr=10.0cm
Electric flux piercing out through a surface depends on the net charge enclosed by the surface according to Gauss’s law and is independent of the dimensions of the arbitrary surface assumed to enclose this charge.
Hence, if the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −103Nm2/C−103Nm2/C.

b) What is the magnitude of the point charge?

Ans: Electric flux could be given by the relation,
ϕtotal=qε0ϕtotal=qε0
Where,q=q= net charge enclosed by the spherical surface
ε0=8.854×10−12N−1C2m−2=ε0=8.854×10−12N−1C2m−2= Permittivity of free space
⇒q=ϕε0⇒q=ϕε0
Substituting the given values,
⇒q=−1.0×103×8.854×10−12=−8.854×10−9C⇒q=−1.0×103×8.854×10−12=−8.854×10−9C
⇒q=−8.854nC⇒q=−8.854nC
Thus, the value of the point charge is found to be −8.854nC−8.854nC.

21. A conducting sphere of radius 10cm10cm has an unknown charge. If the electric field at a point 20cm20cm from the centre of the sphere of magnitude 1.5×103N/C1.5×103N/C is directed radially inward, what is the net charge on the sphere?

Ans: We have the relation for electric field intensity EE at a distance
(d)(d)
from the centre of a sphere containing net charge qq is given by,
E=q4πε0d2E=q4πε0d2 ……………………………………………… (1)
Where,
Net charge, q=1.5×103N/Cq=1.5×103N/C
Distance from the centre, d=20cm=0.2md=20cm=0.2m
ε0=8.854×10−12N−1C2m−2=ε0=8.854×10−12N−1C2m−2= Permittivity of free space
14πε0=9×109Nm2C−214πε0=9×109Nm2C−2
From (1), the unknown charge would be,
q=E(4πε0)d2q=E(4πε0)d2
Substituting the given values we get,
⇒q=1.5×103×(0.2)29×109=6.67×10−9C⇒q=1.5×103×(0.2)29×109=6.67×10−9C
⇒q=6.67nC⇒q=6.67nC
Therefore, the net charge on the sphere is found to be6.67nC6.67nC.

22. A uniformly charged conducting sphere of 2.4m2.4m diameter has a surface charge density of 80.0μC/m280.0μC/m2.
a) Find the charge on the sphere.

Ans: Given that,
Diameter of the sphere, d=2.4md=2.4m.
Radius of the sphere, r=1.2mr=1.2m.
Surface charge density,
σ=80.0μC/m2=80×10−6C/m2σ=80.0μC/m2=80×10−6C/m2
Total charge on the surface of the sphere,
Q=Charge density ×Surface areaQ=Charge density × Surface area
⇒Q=σ×4πr2=80×10−6×4×3.14×(1.2)2⇒Q=σ×4πr2=80×10−6×4×3.14×(1.2)2
⇒Q=1.447×10−3C⇒Q=1.447×10−3C
Therefore, the charge on the sphere is found to be 1.447×10−3C1.447×10−3C.

b) What is the total electric flux leaving the surface of the sphere?

Ans: Total electric flux (ϕtotal)(ϕtotal) leaving out the surface containing net charge QQ is given by Gauss’s law as,
ϕtotal=Qε0ϕtotal=Qε0…………………………………………………. (1)
Where, permittivity of free space,
ε0=8.854×10−12N−1C2m−2ε0=8.854×10−12N−1C2m−2
We found the charge on the sphere to be,
Q=1.447×10−3CQ=1.447×10−3C
Substituting these in (1), we get,
ϕtotal=1.447×10−38.854×10−12ϕtotal=1.447×10−38.854×10−12
⇒ϕtotal=1.63×10−8NC−1m2⇒ϕtotal=1.63×10−8NC−1m2
Therefore, the total electric flux leaving the surface of the sphere is found to be 1.63×10−8NC−1m21.63×10−8NC−1m2.

23. An infinite line charge produces a field of magnitude 9×104N/C9×104N/C at a distance of 2cm2cm. Calculate the linear charge density.

Ans: Electric field produced by the given infinite line charge at a distance ddhaving linear charge densityλλ could be given by the relation,
E=λ2πε0dE=λ2πε0d
⇒λ=2πε0Ed⇒λ=2πε0Ed…………………………………….. (1)
We are given:
d=2cm=0.02md=2cm=0.02m
E=9×104N/CE=9×104N/C
Permittivity of free space,
ε0=8.854×10−12N−1C2m−2ε0=8.854×10−12N−1C2m−2
Substituting these values in (1) we get,
⇒λ=2π(8.854×10−12)(9×104)(0.02)⇒λ=2π(8.854×10−12)(9×104)(0.02)
⇒λ=10×10−8C/m⇒λ=10×10−8C/m
Therefore, we found the linear charge density to be 10×10−8C/m10×10−8C/m.

24. 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 17.0×10−22Cm−217.0×10−22Cm−2. What is EE in the outer region of the first plate? What is EE in the outer region of the second plate? What is E between the plates?

Ans: The given nature of metal plates is represented in the figure below:
Here, A and B are two parallel plates kept close to each other. The outer region of plate A is denoted as II, outer region of plate B is denoted as IIIIII, and the region between the plates, A and B, is denoted as IIII.
It is given that:
Charge density of plate A, σ=17.0×10−22C/m2σ=17.0×10−22C/m2
Charge density of plate B, σ=−17.0×10−22C/m2σ=−17.0×10−22C/m2
In the regions IIandIIIIII, electric field E is zero. This is because the charge is not enclosed within the respective plates.
Now, the electric field EE in the region IIII is given by
E=|σ|ε0E=|σ|ε0
Where,
Permittivity of free space ε0=8.854×10−12N−1C2m−2ε0=8.854×10−12N−1C2m−2
Clearly,
E=17.0×10−228.854×10−12E=17.0×10−228.854×10−12
⇒E=1.92×10−10N/C⇒E=1.92×10−10N/C
Thus, the electric field between the plates is 1.92×10−10N/C1.92×10−10N/C.

25. An oil drop of 1212 excess electrons is held stationary under a constant electric field of 2.55×104NC−12.55×104NC−1 in Millikan’s oil drop experiment. The density of the oil is 1.26gm/cm31.26gm/cm3. Estimate the radius of the drop. (g=9.81ms−2,e=1.60×10−19C)(g=9.81ms−2,e=1.60×10−19C).

Ans: It is given that:
The number of excess electrons on the oil drop,
n=12n=12
Electric field intensity, E=2.55×104NC−1E=2.55×104NC−1
The density of oil, ρ=1.26gm/cm3=1.26×103kg/m3ρ=1.26gm/cm3=1.26×103kg/m3
Acceleration due to gravity, g=9.81ms−2g=9.81ms−2
Charge on an electron e=1.60×10−19Ce=1.60×10−19C
Radius of the oil drop =r=r
Here, the force (F) due to electric field E is equal to the weight of the oil drop (W).
Clearly,
F=WF=W
⇒Eq=mg⇒Eq=mg
⇒Ene=43πr2ρ×g⇒Ene=43πr2ρ×g
Where,
qq is the net charge on the oil drop =ne=ne
mm is the mass of the oil drop =Volume of the oil drop×Density of oil=Volume of the oil drop×Density of oil
=43πr3×p=43πr3×p
Therefore, radius of the oil drop can be calculated as
r=3Ene4πρg−−−−−√r=3Ene4πρg
⇒r=3×2.55×104×12×1.6×10−194×3.14×1.26×103×9.81−−−−−−−−−−−−−−−−−−−−−−−−−−−√⇒r=3×2.55×104×12×1.6×10−194×3.14×1.26×103×9.81
⇒r=946.09×10−21−−−−−−−−−−−−√⇒r=946.09×10−21
⇒r=9.72×10−10m⇒r=9.72×10−10m
Therefore, the radius of the oil drop is 9.72×10−10m9.72×10−10m.

26. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

a)

Ans: The field lines shown in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor which is a characterizing property of electric field lines.

b)

Ans: The lines shown in (b) do not represent electrostatic field lines because field lines cannot emerge from a negative charge and cannot terminate at a positive charge since the direction of the electric field is from positive to negative charge.

c)

Ans: The field lines shown in (c) do represent electrostatic field lines as they are directed outwards from positive charge in accordance with the property of electric field.

d)

Ans: The field lines shown in (d) do not represent electrostatic field lines because electric field lines should not intersect each other.

e)

Ans: The field lines shown in (e) do not represent electrostatic field lines because electric field lines do not form closed loops

27. In a certain region of space, the electric field is along the z-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105NC−1105NC−1 per meter. What are the force and torque experienced by a system having a total dipole moment equal to 10−7Cm10−7Cm in the negative z-direction?

Ans: We know that the dipole moment of the system, P=q×dl=−10−7CmP=q×dl=−10−7Cm.
Also, the rate of increase of electric field per unit length is given as
dEdl=105NC−1dEdl=105NC−1
Now, the force (F) experienced by the system is given by F=qEF=qE
F=qdEdl×dlF=qdEdl×dl
⇒F=PdEdl⇒F=PdEdl
⇒F=−10−7×105⇒F=−10−7×105
⇒F=−10−2N⇒F=−10−2N
Clearly, the force is equal to −10−2N−10−2N in the negative z-direction i.e., it is opposite to the direction of the electric field.
Thus, the angle between the electric field and dipole moment is equal to
180∘180∘
Now, the torque is given by τ=PEsinθτ=PEsin⁡θ
τ=PEsin180∘=0τ=PEsin⁡180∘=0
Therefore, it can be concluded that the torque experienced by the system is zero.

28.
a) A conductor A with a cavity as shown in the Fig. 1.36(a) is given a charge QQ. Show that the entire charge must appear on the outer surface of the conductor.
(Image Will Be Updated Soon)

Ans: Firstly, let us consider a Gaussian surface that is lying within a conductor as a whole and enclosing the cavity. Clearly, the electric field intensity E inside the charged conductor is zero.
Now, let
qq
be the charge inside the conductor and ε0ε0, the permittivity of free space.
According to Gauss’s law,
Flux is given by
ϕ=E→.ds=qε0ϕ=E→.ds=qε0
Here, ϕ=0ϕ=0 as E=0E=0 inside the conductor
Clearly,
0=q8.854×10−120=q8.854×10−12
⇒q=0⇒q=0
Therefore, the charge inside the conductor is zero.
And hence, the entire charge QQ appears on the outer surface of the conductor.

b) Another conductor B with charge qq is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q+qQ+q Fig.1.36(b)Fig.1.36(b).

Ans: The outer surface of conductor A has a charge of QQ.
It is given that another conductor B, having a charge
+q+q
is kept inside conductor A and is insulated from conductor A.
Clearly, a charge of
−q−q
will get induced in the inner surface of conductor A and a charge of
+q+q
will get induced on the outer surface of conductor A.
Therefore, the total charge on the outer surface of conductor A amounts to Q+qQ+q.

c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Ans: A sensitive instrument can be shielded from a strong electrostatic field in its environment by enclosing it fully inside a metallic envelope.
Such a closed metallic body provides hindrance to electrostatic fields and thus can be used as a shield.

29. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is [σ2ε0]n∧[σ2ε0]n∧, where n∧n∧ is the unit vector in the outward normal direction, and σσ is the surface charge density near the hole.

Ans: Firstly, let us consider a conductor with a cavity or a hole as shown in the diagram below. It is known that the electric field inside the cavity is zero.
Let us assume E to be the electric field just outside the conductor, qq be the electric charge, σσ be the charge density, and ε0ε0, the permittivity of free space.
We know that charge |q|=σ×d|q|=σ×d
Now, according to Gauss’s law,
ϕ=E.ds=|q|ε0ϕ=E.ds=|q|ε0
⇒E.ds=σ×dε0⇒E.ds=σ×dε0
⇒E=σε0n∧⇒E=σε0n∧
where n∧n∧ is the unit vector in the outward normal direction.
Thus, the electric field just outside the conductor is σε0n∧σε0n∧.
Now, this field is actually a superposition of the field due to the cavity E1E1 and the field due to the rest of the charged conductor E2E2.
These electric fields are equal and opposite inside the conductor whereas equal in magnitude as well as direction outside the conductor.
Clearly,
E1+E2=EE1+E2=E
⇒E1=E2=E2=σ2ε0n∧⇒E1=E2=E2=σ2ε0n∧
Therefore, the electric field in the hole is σ2ε0n∧σ2ε0n∧.
Hence, proved.

30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λλ without using Gauss’s law. Hint:UseCoulomb′slawdirectlyandevaluatethenecessaryintegralHint:UseCoulomb′slawdirectlyandevaluatethenecessaryintegral

Ans: Firstly, let us take a long thin wire XY as shown in the figure below. This wire is of uniform linear charge density λλ.
(Image Will Be Updated Soon)
Now, consider a point A at a perpendicular distance l from the mid-point O of the wire as shown in the figure below:
(Image Will Be Updated Soon)
Consider E to be the electric field at point A due to the wire.
Also consider a small length element dxdx on the wire section with OZ=xOZ=x as shown.
Let qq be the charge on this element.
Clearly, q=λdxq=λdx
Now, the electric field due to this small element can be given as
dE=14πε0λdx(AZ)2dE=14πε0λdx(AZ)2
However, AZ=12+x2−−−−−−√AZ=12+x2
⇒dE=λdx4πε0(12+x2)⇒dE=λdx4πε0(12+x2)
Now, let us resolve the electric field into two rectangular components. Doing so, dEcosθdEcos⁡θ is the perpendicular component and dEsinθdEsin⁡θ is the parallel component.
When the whole wire is considered, the component dEsinθdEsin⁡θ gets cancelled and only the perpendicular component dEcosθdEcos⁡θ affects the point A.
Thus, the effective electric field at point A due to the element dxdx can be written as
dE1=λdxcosθ4πε0(l2+x2)dE1=λdxcos⁡θ4πε0(l2+x2) ….(1)
Now, in ΔAZOΔAZO, we have
tanθ=xltan⁡θ=xl
x=ltanθ……(2)x=ltan⁡θ……(2)
On differentiating equation (2), we obtain
dx=lsec2dθ……(3)dx=lsec2dθ……(3)
From equation (2)
x2+l2=l2+l2tan2θx2+l2=l2+l2tan2θ
⇒l2(1+tan2θ)=l2sec2θ⇒l2(1+tan2θ)=l2sec2θ
⇒x2+l2=l2sec2θ…..(4)⇒x2+l2=l2sec2θ…..(4)
Putting equations (3) and (4) in equation (1), we obtain
dE1=λlsec2dθ4πε0(l2sec2θ)cosθdE1=λlsec2dθ4πε0(l2sec2θ)cos⁡θ
⇒dE1=λcosθdθ4πε0l…..(5)⇒dE1=λcos⁡θdθ4πε0l…..(5)
Now, the wire is taken so long that it ends from −π2−π2 to +π2+π2.
Therefore, by integrating equation (5), we obtain the value of field E1E1 as
∫−π2π2dE1=∫−π2π2λ4πε0lcosθdθ∫−π2π2dE1=∫−π2π2λ4πε0lcos⁡θdθ
⇒E1=λ4πε0l×2⇒E1=λ4πε0l×2
⇒E1=λ2πε0l⇒E1=λ2πε0l
Thus, the electric field due to the long wire is derived to be equal to
λ2πε0lλ2πε0l

31. 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 uu)  of charge (+12)e(+12)e and the ‘down’ quark (denoted by dd) of charge −(13)e−(13)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: It is known that a proton has three quarks. Let us consider
nn
up quarks in a proton, each having a charge of +(23e)+(23e).
Now, the charge due to
nn
up quarks =(23e)n=(23e)n
The number of down quarks in a proton =3−n=3−n
Also, each down quark has a charge of −13e−13e
Therefore, the charge due to (3−n)(3−n) down quarks =(−13e)(3−n)=(−13e)(3−n)
We know that the total charge on a proton =+e=+e
Therefore,
e=(23e)n+(−13e)(3−n)e=(23e)n+(−13e)(3−n)
⇒e=(2ne3)−e+ne3⇒e=(2ne3)−e+ne3
⇒2e=ne⇒2e=ne
⇒n=2⇒n=2
Clearly, the number of up quarks in a proton, n=2n=2
Thus, the number of down quarks in a proton =3−n=3−2=1=3−n=3−2=1
Therefore, a proton can be represented as uuduud.
A neutron is also said to have three quarks. Let us consider
nn
up quarks in a neutron, each having a charge of +(23e)+(23e) .
It is given that the charge on a neutron due to
nn
up quarks =(+32e)n=(+32e)n
Also, the number of down quarks is (3−n)(3−n), each having a charge of =(−32)e=(−32)e
Thus, the charge on a neutron due to (3−n)(3−n) down quarks =(−13e)(3−n)=(−13e)(3−n)
Now, we know that the total charge on a neutron =0=0
Thus,
0=(23e)n+(−13e)(3−n)0=(23e)n+(−13e)(3−n)
⇒0=(2ne3)−e+ne3⇒0=(2ne3)−e+ne3
⇒e=ne⇒e=ne
⇒n=1⇒n=1
Clearly, the number of up quarks in a neutron, n=1n=1
Thus, the number of down quarks in a neutron =3−n=2=3−n=2
Therefore, a neutron can be represented as uddudd.

32.
a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where
E=0E=0
) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

Ans: Firstly, let us assume that the small test charge placed at the null point of the given setup is in stable equilibrium.
By stable equilibrium, it means that even a slight displacement of the test charge in any direction will cause the charge to return to the null point as there will be strong restoring forces acting around it.
This further suggests that all the electric lines of force around the null point act inwards and towards the given null point.
But by Gauss law, we know that the net electric flux through a chargeless enclosing surface is equal to zero. This truth contradicts the assumption which we had started with. Therefore, it can be concluded 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 at a certain distance apart.

Ans: When we consider this configuration setup with two charges of the same magnitude and sign placed at a certain distance apart, the null point happens to be at the midpoint of the line joining these two charges.
As per the previous assumption, the test charge, when placed at this mid-point will experience strong restoring forces when it tries to displace itself.
But when the test charge tries to displace in a direction normal to the line joining the two charges, the test charge gets pulled off as there is no restoring force along the normal to the line considered.
Since stable equilibrium prioritizes restoring force in all directions, the assumption in this case also gets contradicted.

33. A particle of mass mm and charge (−q)(−q) enters the region between the two charged plates initially moving along x- axis with speed vxvx (like particle 1 in Fig 1.33). The length of plate is LL and a uniform electric field EE is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL22mvx2qEL22mvx2.
Compare this motion with motion of a projectile in the gravitational field discussed in section 4.10 of class XI textbook of Physics.

Ans: It is given that:
The charge on a particle of mass m=−qm=−q
Velocity of the particle =vx=vx
Length of the plates =L=L
Magnitude of the uniform electric field between the plates =E=E
Mechanical force, F= Mass (m)×Acceleration (a)F= Mass (m)×Acceleration (a)
Thus, acceleration, a=Fma=Fm
However, electric force, F=qEF=qE
Therefore, acceleration, =qEm=qEm………(1)
Here, the time taken by the particle to cross the field of length LL is given by,
t=Length of the plateVelocity of the plate=Lvxt=Length of the plateVelocity of the plate=Lvx ……(2)
In the vertical direction, we know that the initial velocity, u=0u=0
Now, according to the third equation of motion, vertical deflection ss of the particle can be derived as
s=ut+12at2s=ut+12at2
⇒s=0+12(qEm)(Lvx)2⇒s=0+12(qEm)(Lvx)2
⇒s=qEL22mvx2⇒s=qEL22mvx2 …..(3)
Thus, the vertical deflection of the particle at the far edge of the plate is qEL22mvx2qEL22mvx2.
In comparison, we can see that this is similar to the motion of horizontal projectiles under gravity.

34. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx=2.0×106ms−1vx=2.0×106ms−1. If EEbetween the plates separated by 0.5cm0.5cm is 9.1×102N/C9.1×102N/C, where will the electron strike the upper plate? (|e|=1.6×10−19C,me=9.1×10−31kg|e|=1.6×10−19C,me=9.1×10−31kg )

Ans: We are given the velocity of the particle, vx=2.0×106ms−1vx=2.0×106ms−1.
Separation between the two plates, d=0.5cm=0.005md=0.5cm=0.005m
Electric field between the two plates, E=9.1×102N/CE=9.1×102N/C
Charge on an electron, e=1.6×10−19Ce=1.6×10−19C
mass of an electron, me=9.1×10−31kgme=9.1×10−31kg
Letssbe the deflection when the electron strikes the upper plate at the end of the plate LL, then, we have the deflection given by,
s=qEL22mvxs=qEL22mvx
⇒L=2dmvxqE−−−−−√⇒L=2dmvxqE
Substituting the given values,
⇒L=2×0.005×9.1×10−31×(2.0×106)21.6×10−19×9.1×102−−−−−−−−−−−−−−−−−−−√=0.025×10−2−−−−−−−−−−√=2.5×10−4−−−−−−−−√⇒L=2×0.005×9.1×10−31×(2.0×106)21.6×10−19×9.1×102=0.025×10−2=2.5×10−4
⇒L=1.6×10−2=1.6cm⇒L=1.6×10−2=1.6cm
Therefore, we found that the electron will strike the upper plate after travelling a distance of 1.6cm1.6cm.