# NCERT Solutions for Class 12 Physics Chapter 1 - Electric Charges and Fields

## NCERT Solutions for Class 12 Physics Chapter 1 Free PDF Download

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1. What is the force between two small charged spheres having charges of 2 × 10–7 C and 3 × 10–7 C placed 30 cm apart in air?

Sol. Let us consider two charges q1 and q2. We have q1 = 2 × 10–7C, q2 = 3 × 10–7C.

Distance between q1 and q2, r = 30 cm = 0.3 m. Using Coulomb’s law, the force between two charges is given by

$$\text{F}=\frac{1}{4\pi\epsilon}_{0}.\frac{q_{1}q_{2}}{r^{2}}$$

[The charges are placed in air, so w
e have neglected dielectric constant k, became in air
k = 1]

$$\text{Putting the values of}\space\frac{1}{4\pi\epsilon}_{0},q_1,q_2\space\text{and r, we get}\\\text{F}=\frac{9×10^{9}×2×10^{\normalsize-7}×3×10^{\normalsize-7}}{0.3×0.3}\\=\frac{9×2×3×10^{\normalsize-5}}{3×3×10^{\normalsize-2}}=6×10^{\normalsize-3}\text{N}$$

As the two charges q1 and q2 both are positive in nature. So, they repel each other.

2. The electrostatic force on a small sphere of charge 0.4 mC due to another small sphere of charge – 0.8 mC in air is 0.2N.

(a) What is the distance between the two spheres?

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

Sol. According to the question, q1 = 0.4 µC, q2 = –0.8 µC.

Let the distance between two charges be d. Force on charge q1(0.4 µC) due to another charge q2(–0.8 µC) is

F = 0.2N

(a) Using Coulomb’s law, the force between two charges is:

$$\text{F}=\frac{1}{4\pi\epsilon}_{0}.\frac{q_{1}q_{2}}{d}\\\text{0.2 = 9 ×10}^{9}×\frac{0.4×10^{\normalsize-6}×0.8×10^{\normalsize-6}}{d^{2}}$$

d2 = 16 × 9 × 10–4

d = 4 × 3 ×10–2

d = 12 × 10–2m

d = 12cm

Here, the charge q2 is negative in nature and q1 is positive in nature. So, the force between q1and q2 will be attractive in nature.

(b) The force of attraction on the second sphere due to the first sphere as the force between the two charges i.e. 0.2N. The electrostatic force between two charges is interactive force that means force on q1 due to q2 is same as force on q2 due to q1. Electrostatic force between two charges obeys the Newton’s third law.

3. Check that the ratio ke2/Gmemp is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?

$$\textbf{sol.}\space\text{In the ratio}\frac{Ke^{2}}{Gm_{e}m_{p}},k=\frac{1}{4\pi\epsilon_{0}}\text{(constant)}\\\text{where,\space G = gravitational constant}\\\text{m}_e \text{= mass of an electron}\\\text{m}_{p}=\text{mass of a proton}\\\text{By Coulomb’s law}\\\text{F}∝\frac{q_1q_2}{r^{2}}\\\Rarr\space k=\frac{\text{Fr}^{2}}{q_1q_2}\text{or k}=\frac{\text{Fr}^{2}}{q^{2}}\\\text{The dimensions of k}=\bigg(\frac{1}{4\pi\epsilon}_{0}\bigg)\\=\frac{[\text{MLT}^{\normalsize-2}][\text{L}^{2}]}{[\text{AT}][\text{AT}]}=[\text{ML}^{3}\text{T}^{\normalsize-4}\text{A}^{\normalsize-2}]$$

The dimensions of e = [AT]

$$\text{The dimensions of G =}\frac{[\text{MLT}^{\normalsize-2}][\text{L}^{2}]}{[\text{M}^{2}]}=[\text{M}^{\normalsize-1}\text{L}^{\normalsize3}\text{T}^{\normalsize-2}]\\\text{The dimensions of m}_{e}\space\text{or}\space\text{m}_{p}=[\text{M}]\\\text{The dimensions of}\space\frac{ke^{2}}{Gm_{e}m_{p}}\\=\frac{[\text{ML}^{3}\text{T}^{\normalsize-4}\text{A}^{\normalsize-2}][\text{A}^{2}\text{T}^{2}]}{[\text{M}^{\normalsize-1}\text{L}^{3}\text{T}^{\normalsize-2}][\text{M}^{2}]}=[\text{M}^{0}\text{L}^{0}\text{T}^{0}]\\\text{Thus, the given ratio is dimensionless.}\\\text{The value of}\space\frac{ke^{2}}{Gm_{e}m_{p}}\\=\frac{9 × 10^{9}×(1.6×10^{-19})^{2}}{6.67×10^{\normalsize-11}×9.1×10^{\normalsize-31}×1.67×10^{\normalsize-27}}$$

= 2.29 × 1039

This means the electrostatic force between an electron and a proton is 2.29 × 1039 times the gravitational force between an electron and a proton.

4. (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.

(b) Why can one ignore quantization of electric charge when dealing with macroscopic, i.e., large scale charges?

Sol. (a) If the charge on a body can occur in some particular values only then the electric charge of a body is quantized Charge on any body is the integral multiple of electronic charge because the charge of an electron is the elementary charge in nature. The charge on any body can be expressed by the formula

q = ± ne

where, n = number of electrons transferred and e = charge on one electron.

The cause of quantization is that only integral number of electrons can be transferred from one body to other.

(b) We can ignore the quantization of charge when dealing with macroscopic charges because the charge on one electron is very small as compared to the large scale change i.e., 1.6 × 1019 C.

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.

Sol. According to the law of conservation of charge. “charge can neither be created nor be destroyed but can be transferred from one body to another body”.

Before rubbing the two bodies they both are neutral that means net charge of the system is zero. When the glass rod is rubbed with a silk cloth, some electrons from glass rod are transferred to silk cloth hence glass rod attains positive charge due to loss of electrons and silk cloth attains some negative charge due to gain of electrons.

But the total charge of the system remain zero, i.e., the charge before rubbing is same as the charge after rubbing. This is consistent with the law of conservation of charge.

6. Four point charges qA = 2µC, qB = –5µC, qC = 2µC, and qD = – 5µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1µC placed at the centre of the square?

Sol. The charge placed at the centre O is 1µC.

AB = BC = CD = DA = 10 cm

$$\text{AC}=\text{BD}=10\sqrt{2}\space\text{cm}\\\text{AO = BO = CO = DO}=\frac{10\sqrt{2}}{2}\\=5\sqrt{2}\space\text{cm}\\\text{Force on q due to q}_{\text{A}}\\\text{F}_{A}=\frac{1}{4\pi\epsilon}_{0}.\frac{\text{qq}_{A}}{(\text{OA})^{2}}\\=\frac{9×10^{9}×1×10^{\normalsize-6}×2×10^{\normalsize-6}}{(5\sqrt{2}×10^{\normalsize-2})^{2}}\\=\frac{9×2×10^{\normalsize-3}}{25×2×10^{\normalsize-4}}=\frac{90}{25}=\frac{18}{5}$$

= 3.6N(towards O to C)

Force on q due to qC

$$\text{F}_{c}=\frac{1}{4\pi\epsilon_{0}}.\frac{\text{qq}_{c}}{(\text{OC})^{2}}\\=\frac{9×10^{9}×1×10^{\normalsize-6}×2×10^{\normalsize-6}}{(5\sqrt{2}×10^{\normalsize-2})^{2}}\\=\frac{9×2×10^{\normalsize-3}}{25×2×10^{\normalsize-4}}=\frac{90}{25}=\frac{18}{5}$$

= 3.6N(towards O to A)

Force FA and FC are of same magnitude and opposite in direction. So, the resultant force of FA and FC is zero.

Force on q due to qB

$$\text{F}_{B}=\frac{1}{4\pi\epsilon_{0}}.\frac{\text{qq}_{B}}{(\text{OB})^{2}}\\=\frac{9×10^{9}×10^{\normalsize-6}×2×10^{\normalsize-6}}{(5\sqrt{2}×10^{-\normalsize2})^{2}}\\=3.6\text{N}\\\text{(Towards O to B)}\\\text{Force on q due to q}_\text{D}\\\text{F}_{D}=\frac{1}{4\pi\epsilon_{0}}.\frac{qq_{D}}{(\text{OD})^{2}}\\=\frac{9×10^{9}×1×10^{\normalsize-6}×2×10^{\normalsize-6}}{(5\sqrt{2}×10^{\normalsize-2})^{2}}$$

= 3.6N

(Towards O to B)

Force on q due to qD

$$\text{F}_{D}=\frac{1}{4\pi\epsilon_{0}}.\frac{qq_{D}}{(\text{OD})^{2}}\\=\frac{9×10^{9}×1×10^{\normalsize-6}×2×10^{\normalsize-6}}{(5\sqrt{2}×10^{\normalsize-2})^{2}}$$

3.6N (towards O to D)

Here, the forces FB and FD are of same magnitude and opposite in direction. So, the resultant force of FD and FB is zero.

Thus, the net resultant force on 1µC is zero at all the forces balances each other.

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

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

Sol. (a) An electrostatic field line represents the actual path covered by a unit positive charge in an electric field. If the line have sudden breaks then the unit positive test charge jumps from one place to another which is not possible. So, the field line cannot have any sudden breaks.

(b) If two field lines intersect each other, then we can draw two tangents at the point of intersection which indicates that there are two directions of electric field at a particular point, which is not possible at the same instant. Thus, two electric field lines never cross each other at any point.

8. Two point charges qA = 3µC and qB = –3µC are located 20 cm apart in vacuum.

(a) What is the electric field at the mid-point O of the line AB joining the two charges?

(b) If a negative test charge of magnitude 1.5 × 10–9C is placed at this point, what is the force experienced by the test charge.

Sol. (a) AB = 20 cm

AO = OB = 10 cm = 0.1m

qA = 3µC = 3 × 10–6 C

qB = –3µC = –3 × 10–6C

Electric field due to qA at O

$$\text{E}_{\text{A}}=\frac{1}{4\pi\epsilon_{0}}.\frac{q_A}{(\text{AO})^{2}}\\=\frac{9×10^{9}×3×10^{\normalsize-6}}{(0.1)^{2}}\\=\frac{27×10^{3}}{0.1×0.1}$$

= 2.7×106 N/C

Here, EA and EB both are in the same direction. So, the resultant electric field at O.

E = EA + EB = 2.7 × 106 + 2.7 × 106 = 5.4 × 106 N/C

The direction of E (resultant electric field) will be from O to B or towards B.

(b) Let us consider, (a) the charge q is placed at the mid-point O. As shown in figure

q = – 1.5 × 10–9C

The net electric field at point O

$$\text{E}=\frac{\text{F}}{\text{q}}\\\therefore\space\text{F = qE,}$$

F = – 1.5 × 10–9 × 5.4 × 106

= – 8.1 ×10–3N

The direction of force is opposite to the direction of field i.e., O to A.

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

Sol. Two point charges qA and qB located at points A (0, 0, – 15 cm) and B (0, 0, 15 cm) respectively.
The distance between A and B is the length of dipole.

AB = (2l) = length of the dipole

= 30 cm = 30 × 10–2 m.

The total charge q on the dipole will be zero

q = qA + qB = 2.5 × 10–7 –2.5 × 10–7C

= 0

The electric dipole moment

p = qA × 2l = 2.5 × 10–7 × 30 × 10–2

p = 7.5 × 10–8 C–m

The direction of the dipole moment is always from negative charge to positive charge i.e. along B to A.

10. An electric dipole with dipole moment 4 × 10–9 C–m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 N/C. Calculate the magnitude of the torque acting on the dipole.

Sol. Given, dipole moment p = 4 × 10–9 C–m. Electric field E = 5 ×104 N/C.

Torque applied on a dipole

τ = p × E = pE sin θ

or

τ = 4 × 10–9 × 5 × 104 sin 30°

$$=\frac{20×10^{\normalsize-5}}{2}=10^{\normalsize-4}\space\text{N-m}$$

The direction of torque is perpendicular to both electric field and dipole moment.

11. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10–7C.

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

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

Sol. (a) The number of electrons transferred

$$n=\frac{\text{Total charge(q)}}{\text{Charge of electron(e)}}\\\text{n}=\frac{-3×10^{\normalsize-7}}{-1.6×10^{\normalsize-19}}=1.875×10^{12}$$

Thus, the number of electrons transferred is 1.875 × 1012. Electrons will be transferred from wool to polythene because polythene attains the negative charge i.e., it gains the electrons.

(b) If the electrons are transferred from wool to polythene, the mass is also transferred.

Mass transferred from wool to polythene

= Number of electrons × Mass of one electron

= 1.875 × 1012 × 9.1 × 10–31 = 1.8 × 10–18 kg

Thus, 1.8 × 10–18 kg mass is transferred from wool to polythene.

12. (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10–7 C ? 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?

Sol. (a)

According to the question,

qA = 6.5 × 10–7C

qB = 6.5 × 10–7C

r = distance between A and B = 50 cm = 50 × 10–2 m

From the Coulomb’s law, the force between the two spheres,

$$\text{F}=\frac{1}{4\pi\epsilon_{0}}.\frac{q_{A}q_{B}}{r^{2}}\\=\frac{9×10^{9}×6.5×10^{\normalsize-7}×6.5×10^{\normalsize-7}}{(50×10^{2})^{3}}\\=\frac{9×6.5×6.5×10^{\normalsize-5}}{50×50×10^{\normalsize-4}}=1.521×10^{\normalsize-2}\text{N}$$

Thus, the force between A and B is 1.521 × 10–2 N.

(b) If the charge is doubled

qA' = 2qA and qB' = 2qB

$$\text{and distance between them is halved i.e., r' =}\frac{r}{2}\\\text{Now, the force between the two spheres will become}\\\text{F'}=\frac{1}{4\pi\epsilon_{0}}.\frac{q'_{A}q'_{B}}{r'}\\=\frac{1}{4\pi\epsilon_{0}}.\frac{(2q_{A})(2q_{B})}{\bigg(\frac{r}{2}\bigg)^{2}}\\=\frac{1}{4\pi\epsilon_{0}}.\frac{4q_{A}q_{B}}{\frac{r^{2}}{4}}=16\frac{1}{4\pi\epsilon_{0}}.\frac{q_{A}q_{B}}{r^{2}}$$

16F = 16 × 1.521 × 10–2 = 0.24N

13. Suppose the spheres A and B in Q.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? Sol. When the sphere C comes in contact with A, the charges will be divided equally on both spheres. Now, charge on A is $$q'\text{A}=\frac{q_{A} + q_{C}}{2}\\q'_{A}=\frac{6.5×10^{\normalsize-7} + 0}{2}=3.25×10^{7}\text{C}$$ And the charge on C will also be q'C = 3.25 ×10–7C Now, the sphere C comes in contact with B.

$$\text{Now, charge on B is q'B}=\frac{q_{B} + q'_{c}}{2}\\=\frac{6.5×10^{\normalsize-7} + 3.25 × 10^{\normalsize-7}}{2}\\q'_{B}=4.875×10^{\normalsize-7}\text{C}$$

Final charge on C is q’C= 4.875 × 10–7C

Final charge on A is q'A = 3.15 × 10–7C

The charge on B is q'B = 4.875 × 10–7C

From the Coulomb’s law, the force between two spheres is

$$\text{F}=\frac{1}{4\pi\epsilon_{0}}.\frac{q'_{A}.q'_{B}}{r^{2}}\\=\frac{9×10^{9}×3.25×10^{\normalsize-7}×4.875×10^{\normalsize-7}}{(50×10^{\normalsize-2})^{2}}\\=\frac{9×3.25×4.875×10^{\normalsize-5}}{50×50×10^{\normalsize-4}}=5.7×10^{\normalsize-3}\text{N}$$

14. The given figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Sol. We know that a positively charged particle is attracted towards the negatively charged plate and vice-versa.

Here, particle 1 and particle 2 are attracted towards positive plate. Thus, particle 1 and particle 2 are negatively charged. Particle 3 is attracted towards negatively charged plate so it has positive charge.

As the deflection in the path of a charged particle is directly proportional to the charge/mass ratio.

$$y∝\frac{q}{m}\\i.e$$

Here, the deflection in a particle 3 is maximum, therefore, the charge to mass ratio of particle 3 is maximum.

15. Consider a uniform electric field E = 3 × 103 N/C.

(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the Y-Z plane?

(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the X-axis?

Sol.

(a) In Y-Z plane, the area vector (Perpendicular to the square) is along X-axis.

Area S = 10 × 10 = 100 cm2 = 10–2 m2

$$\text{Area vector S = 10}^{\normalsize-2}\hat{i}\text{m}^{2}\\\text{Electric flux,}\phi=\vec{\text{E}}.\vec{\text{S}}=\text{ES cos}\space\theta\\=\text{ES}\space[\because\space\text{cos}\theta=1]\\= 3 × 10^{3} × 10^{\normalsize–2} = 30 \text{N-m}^{2}/\text{C}$$

(b) Now, the area vector makes an angle of 60° with X-axis.

i.e., θ = 60°

$$\text{Using the formula of electric flux}\space\phi =\vec{\text{E}}.\vec{\text{S}}\\\phi=\text{ES cos}\space\theta =3×10^{3}×10^{\normalsize-2}\text{cos}60\degree=3×10×\frac{1}{2}$$

= 15 N-m2/C.

16. What is the net flux of the uniform electric field of Q. 15 through a cube of side 20 cm oriented so that its faces are parallel to coordinate planes?

Sol. By the definition, the number of lines entering in the cube is the same as that the number of lines leaving the cube. So, the net flux over the cube is 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 × 103 N-m2/C.

(a) What is the net charge inside the box?

(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?

Sol. (a) Using the concept of Gauss’s theorem,

$$\text{Net flux,}\phi=\frac{\text{Charge}}{\epsilon_{0}}=\frac{q}{\epsilon_{0}}\\\Rarr\space q=\epsilon_{0}\phi=8.854×10^{\normalsize-12}×10^{3}\\q=0.07×10^{\normalsize-6}\text{C}=0.07\mu\text{C}$$

Thus, flux is outward and the charge is positive in nature.

(b) If net outward flux = 0

Then, we can conclude that the net charge inside the box is zero, i.e., the box may have either zero charge or have equal amount of positive and negative charges. Thus, we cannot conclude that there is no charge inside the box.

18. A point charge +10µC is at a distance 5 cm directly above the centre of a square of side 10 cm, as shown in figure. What is the magnitude of the electric flux through the square?

(Hint : Think of the square as one face of a cube with edge 10 cm.)

Sol. Let us imagine an enclosed cubical surface and Let the charge q is placed at the centre of cube. The total flux enclosed through the cube is

$$\phi=\frac{q}{\epsilon_{0}}\space...(\text{i})$$

Here, q = 10µC

The flux enclosed by one face.

i.e. flux linked with each square

$$\phi=\frac{\phi}{6}=\frac{1}{6}.\frac{q}{\epsilon_{0}}\space[\text{From Eq. (i)}]\\\phi'=\frac{1}{6}×\frac{10×10^{\normalsize-6}}{8.854×10^{\normalsize-12}}$$

= 1.88 × 105 N-m2/C

Thus, the flux linked with the square is 1.88 × 105 N-m2/C.

19. A point of 2.0mC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Sol. Consider a charge q is placed at the centre of a cubic Gaussian surface.

According to Gauss’s theorem, the net electric flux (f) through the surface is

$$\phi=\frac{q}{\epsilon_{0}}\\=\frac{2×10^{\normalsize-6}}{8.854×10^{\normalsize-12}}\\=2.26 × 10^{5}\space\text{N-m}^{2}/\text{C}$$

Thus, the net electric flux through the surface is 2.26 × 105 N-m2/C.

20. A point charge causes an electric flux of –1.0 × 103 N-m2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.

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

(b) What is the value of the point charge?

Sol. (a) Electric flux passing through any surface is independent of the radius of Gaussian surface, so if the radius of Gaussian surface were doubled, the electric flux linked with the surface unaffected. It depends only on charge.

$$\text{i.e.}\space\phi∝\frac{q}{\epsilon_{0}}$$

(b) Φ = –1.0 × 103 N-m2/C

Using Gauss’s theorem,

$$\text{Thus flux linked} \phi =\frac{q}{\epsilon_{0}}$$

q = Φε0 = – 1.0 × 103 × 8.854 × 10–12 = – 8.85 × 10–9C

Thus, the value of point charge is – 8.85 × 10–9 C.

21. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103N/C and point radially inwards, what is the net charge on the sphere?

Sol. Let the value of unknown charge be q.

Electric field at 20 cm away E = 1.5 × 103 N/C

$$\text{Electric charge E =}\frac{1}{4\pi\epsilon_{0}}.\frac{q}{r^{2}}(r=\text{distance})\\1.5×10^{3}=\frac{9×10^{9}×q}{(20×10^{\normalsize-2})^{2}}\\\text{q}=\frac{1.5×10^{3}×20×20×10^{\normalsize-4}}{9×10^{9}}\\=6.67×10^{\normalsize-9}\text{C}$$

As, the electric field is radially inwards that means nature of unknown charge q is negative.

22. A uniformly charged conducting sphere of 2.4m diameter has a surface charge density of 80.0 µC/m2.

(a) Find the charge on the sphere.

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

Sol. Given, diameter of sphere = 2.4 m

$$\text{Radius of sphere, r =}\frac{2.4}{2}=1.2\text{m}$$

Surface charge density σ = 80 µC/m2

= 80 × 10–8 C/m2

$$\text{(a) Surface charge density =}\frac{\text{Charge}}{\text{Surface area}}$$

$$\sigma=\frac{q}{4\pi r^{2}}\\\therefore\space q = σ × 4\pi r^{2}$$

= 80 × 10–6 × 4 × 3.14 × 1.2 × 1.2

q = 1.4 × 10–3C

(b) Using Gauss’s theorem, total flux leaving the surface

$$\phi=\frac{\text{Total charge}}{ε_{0}}\\\phi=\frac{q}{\epsilon_{0}}=\frac{1.45×10^{\normalsize-3}}{8.854×10^{\normalsize-12}}$$

Φ = 1.6 × 108 N-m2/C

Thus, the flux leaving the surface of sphere is 1.6 × 108 N-m2/C.

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

Sol. Electric field due to infinite line charge,

$$\text{E}=\frac{\lambda}{2\pi\epsilon_{0}r},\text{where} \lambda\space\text{be the linear charge density}$$

$$\text{Dividing and multiplying by 2 to get}\space\frac{1}{4\pi\epsilon_{0}}\\\text{because, we have the value of}\space\frac{1}{4\pi\epsilon_{0}.}\\\text{E}=\frac{2}{2}×\frac{\lambda}{2\pi\epsilon_{0}r}=\frac{2\lambda}{4\pi\epsilon_{0}r}\\\text{Putting the values, we get}\\\text{9 × 10}^{4} =\frac{2 × 9 ×10^{9}×\lambda}{2×10^{\normalsize-2}}\\\lambda=\frac{9×10^{4}×2×10^{\normalsize-2}}{2×9×10^{9}}=10^{\normalsize-7}\text{C/m}$$

Thus, the linear charge density is 10–7 C/m.

24. Two large, thin metal plates are parallel and closed to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10-22 C/m2. What is E.

(a) in the outer region of the first plate?

(b) in the outer region of the second plate?

(c) between the plates?

Sol. There are two plates A and B having surface charge densities σA = 1.70 × 10–22 C/m2 on A and σB = – 17.0 × 10–22 C/m2 on B respectively.

(a) According to Gauss’s theorem, if the plates have same surface charge density but having opposite signs, then the electric field outer of first plate is zero.

$$\text{E}_{1}=\text{E}_{A}-\text{E}_{B}=\frac{\sigma}{2\epsilon_{0}}+\bigg(-\frac{\sigma}{2\epsilon_{0}}\bigg)=0$$

(b) Similarly, the electric field in outer region of the second plate is also zero.

$$\text{E}_{\text{II}}=\text{E}_{\text{A}}-\text{E}_{\text{B}}=\frac{\sigma}{2\epsilon_{0}} + \bigg(-\frac{\sigma}{2\epsilon_{0}}\bigg)=0$$

(c) Electric field between the plate

$$\text{E}_{\text{II}}=\text{E}_{\text{A}}+\text{E}_{\text{B}}=\frac{\sigma}{2\epsilon_{0}} + \frac{\sigma}{2\epsilon_{0}}\\=\frac{\sigma}{\epsilon_{0}}=\frac{17.0×10^{\normalsize-22}}{8.85×10^{-12}}$$

E = 1.92 × 10–10 N/C