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

Class 12 Physics NCERT book solutions for Chapter 1 - Electric Charges And Fields Questions and Answers.

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Class 12 Physics NCERT book solutions for Chapter 1 - Electric Charges And Fields Questions and Answers.

2×10−7C2×10−7C and 3×10−7C3×10−7C placed 30cm30cm apart in air?

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.

a) What is the distance between the two spheres?

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.

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.

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.

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

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.

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.

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.

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

As the charge doesn’t jump from one point to the other, field lines will not have sudden breaks.

This is however impossible and thus, two field lines never cross each other.

a) What is the electric field at the midpoint O of the line AB joining the two charges?

(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.

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.

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.

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=pEsinθ

Substituting the given values we get,

⇒τ=4×10−9×5×104×sin30∘⇒τ=4×10−9×5×104×sin30∘

⇒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.

a) Estimate the number of electrons transferred (from which to which?)

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.

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.

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.

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.

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.

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

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.

a) Find the flux of this field through a square of side 10cm10cmwhose plane is parallel to the y-z plane.

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|cosθ

Substituting the given values, we get,

⇒ϕ=3×103×0.01×cos0∘⇒ϕ=3×103×0.01×cos0∘

∴ϕ=30Nm2/C∴ϕ=30Nm2/C

Thus, we found the net flux through the given surface to be ϕ=30Nm2/Cϕ=30Nm2/C.

ϕ=|E||A|cosθϕ=|E||A|cosθ

⇒ϕ=3×103×0.01×cos60∘⇒ϕ=3×103×0.01×cos60∘

⇒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.

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.

a) What is the net charge inside the box?

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.

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.

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.

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.

a) If the radius of the Gaussian surface were doubled, how much flux could pass through the surface?

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.

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.

(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.

a) Find the charge on the sphere.

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.

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.

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.

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.

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.

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=PEsinθ

τ=PEsin180∘=0τ=PEsin180∘=0

Therefore, it can be concluded that the torque experienced by the system is zero.

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)

Now, let

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.

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.

Such a closed metallic body provides hindrance to electrostatic fields and thus can be used as a shield.

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.

(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, dEcosθdEcosθ is the perpendicular component and dEsinθdEsinθ is the parallel component.

When the whole wire is considered, the component dEsinθdEsinθ gets cancelled and only the perpendicular component dEcosθdEcosθ 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=λdxcosθ4πε0(l2+x2) ….(1)

Now, in ΔAZOΔAZO, we have

tanθ=xltanθ=xl

x=ltanθ……(2)x=ltanθ……(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)cosθ

⇒dE1=λcosθdθ4πε0l…..(5)⇒dE1=λcosθdθ4πε0l…..(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=∫−π2π2λ4πε0lcosθdθ

⇒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

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.

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.

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.

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.

Compare this motion with motion of a projectile in the gravitational field discussed in section 4.10 of class XI textbook of Physics.

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.

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.

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