 NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 1: Electric Charges and Fields (Part II)

NCERT Solutions for Class 12 Physics, Chapter 1, Electric Charges and Fields (Part 2) are available here. In this part, you will find solutions from question number 1.12 to 1.24 (of Class 12 Physics NCERT textbook). This part is a continuation of Part 1 where solutions from question number 1.1 to 1.11 are available. NCERT Solutions for Class 12 Physics, Chapter 1: Electric Charges and Fields are available here. This part is the continuation of Part 1. In Part 1, we have provided solutions from question number 1.1 to question number 1.11. Now, in Part 2, you will find solutions from question number 1.12 to question number 1.24. NCERT textbooks are prescribed in CBSE Syllabus 2017-18 and these NCERT Solutions are extremely important for Class 12 Physics board exam 2018.

NCERT Solutions for 12th Physics, Chapter 1: Electric Charges and Fields from question number 1.12 to question number 1.24 are given below:

Question 1.12: (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?

Solution1.12:

(a) The force is given by

|F| = (Q1 Q2)/(4 π εo r2 )

Substituting the respective values, we have

F = [(6.5x10‒7) × (6.5x10‒7) × (9x109)] / [50 x 10-2]2 = 0.015 N

(b) Now, when Q1 and Q2 will be doubled and r will be halved then new force will become,

|Fnew| = 16 F ⇒ |Fnew| = 16 × 0.015 N = 0.24 N.

NCERT Exemplar: CBSE Class 12 Physics – Chapter 1, Electric Charges and Fields

NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 1: Electric Charges and Fields (Part I)

Question 1.13: 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?

Solution 1.13:

When an uncharged sphere is brought in contact with sphere A, half of the charge will be shifted to third sphere. So the charge q1 on sphere A becomes 3.25x10‒7 C. Again on contact of sphere C with sphere B, redistribution of charge will happen and charge q2 on sphere B gets [{(6.5+3.25) × 10‒7}/2 =] 4.875x10‒7 C.

Then, by the same formula for force, we get

F = [(9x109).(q1.q2)] / r2 = 5.7x10‒3 N.

Question 1.14: Figure 1.33 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? Solution 1.14:

Charge 1 = negative

Charge 2 = negative

Charge 3 = positive.

Since mass of proton is more than mass of electron, charge 3 will have highest charge to mass ratio.

CBSE Class 12th Physics Notes: Electric Charges and Fields (Part - I)

CBSE Class 12th Physics Notes: Electric Charges and Fields (Part - II)

Question 1.15: Consider a uniform electric field E = 3 × 103 î 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 yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

Solution 1.15:

Given:

E = 3000 N/C

a = 10 cm

(a) Flux = E.ΔS = 3000 x 100 x 1/10000 = 30 Nm2/C

(b) Flux = E.ΔS.cos30o = 25.98 Nm2/C

Question 1.16: What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Solution 1.16:

Zero. All the flux entering will leave the cube.

Question 1.17: 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 Nm2/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?

Solution 1.17:

Given:

Outward flux= 8x103 Nm2/C

(a) We know, flux, = q/ϵo

So, net charge, q = (8 x 103) × (8.85 x 10‒12) = 7x10‒8 C

(b) No. The net charge is zero inside the box.

Question 1.18: A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. 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.) Solution 1.18:

Given in the question:

q = 10x10‒6 C

a = 10 cm

Considering the charge at the center of the imagined cube of dimension 10 cm, the flux through all the 6 faces of cube will be ϕ = q/ϵo. In case of square, there will be only one face, so, the flux through the square (or one face of cube is given by, ϕ’ = q/6ϵo.

= (10 × 10‒6) / [6 × (8.85x10‒12)]

= 1.88 × 105 Nm2/C.

Question 1.19: A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Solution 1.19:

Using the same formula used in previous question,

ϕ = q/ϵo = (2 × 10‒6) / (8.85 × 10‒12) = 2.2 × 105 Nm2/C.

Question 1.20: A point charge causes an electric flux of –1.0 × 103 Nm2/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?

Solution 1.20:

Given,

Flux = 103 Nm2/C

(a) Same flux will pass through the surface. It does not depend upon the dimension of Gaussian surface.

(b) From the formula, ϕ = q/ϵo q = ϕ × ϵo

Charge, q = (103) × (8.85 × 10-12) =  8.85 × 10‒9 C.

Question 1.21: 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 × 103 N/C and points radially inward, what is the net charge on the sphere?

Solution 1.21:

Given:

|E| = 1500 N/C

r = 20 cm

By the formula,

|E| = Q/(4πϵo r2 ) ⇒ Q = (|E|)× r2 × (4πϵo)

Q = [1500 × 0.22] / [9 × 109] = ‒ 6.67 × 10‒19 C.

Inward direction of electric field implies the concerned charge is negative.

Question 1.22: A uniformly charged conducting sphere of 2.4 m 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?

Solution 1.22:

Given in the question:

Surface charge density σ = 80 × 10‒6 C/m2

R = 1.2 m

(a) Using the formula

Q = σ × (4πR2)

Substituting the respective values, we have,

Q = 1.45 × 10‒3 C

(b) By using the formula,

ϕ = Q/ϵo, we have,

ϕ = [(1.45 × 10‒3)/(1.6 × 108)] NC‒1 m2.

Question 1.23: An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.

Solution 1.23:

Given,

E = 90000 N/C

r = 2 cm

Linear charge density, λ = E. (2 π ϵ r)

By putting values, we find,  λ = 10-7 C/m.

Question 1.24: 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 × 10–22 C/m2. 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?

Solution 1.24:

Given:

Surface charge density = 17 × 10‒22 C/m2

(a) Zero, as the field due to both plates will cancel out each other.

(b) Zero, with the above reason.

(c) E due to plate 1 (region between 2 plates)

|E1| = σ/2ϵo

= (17 × 10‒22)/[ 2 × 8.85 × 10‒12]

= 9.6 × 10‒11 N/C

Similarly, E2 = 9.6 × 10‒11 N/C

E1 will have outward direction towards plate 2, while E2 will have inward direction towards plate 2. So both the fields add up

Thus,

|E| = |E1| + |E2| = 2 |E1|.

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