Cbse Ncert Solutions of Class 12 Physics: Electric Charges and Fields (Q 17-24)

 


Chapter 1, Electric Charges and Fields

NCERT Solutions of 12th CBSE Physics Textbook Exercises


Question 1.17: Careful measurement of the electric field at the surface of the black box indicates that the net outward flux through the surface of the box is 8.0 x 103 Nm2/C
(i) What is the net charge inside the box?
(ii) 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?
Answer:
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(ii) We cannot conclude that the net electric charge inside the box is zero if the outward flux through the surface of black box is zero because there might be equal amounts of positive and negative charges cancelling each other and thus making the resultant charge equal to zero. Thus, we can only conclude that the net charge inside the box is zero.
Question 1.18: A point charge + 10µC is 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.)
Answer: The given square ABCD can be imagined as one of the side faces of a cube of side 0.10 m. The given charge can be imagined to be at the centre of this cube at a distance of 5 cm.
Here, q = + 10µC = 10-5 C.
Then, according to the Gauss’s Theorem, the total electric flux through all the 6 faces of the cube is given by -
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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?
Answer: Here, charge at the centre of the Gaussian surface,
q = 2µC = 2 x 10-6 C.
ɛo = 8.854 x 10-12 C2 N-1 m-2
Ф = electric flux through it = ?
According to Gauss’s Theorem, the electric flux through the six faces of the cubes i.e., through the Gaussian surface is given by -
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Question 1.20: A point charge causes an electric flux of -1.0 x 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 ?
Answer: Here, Ф = electric flux through the spherical Gaussian surface = -1.0 x 103 Nm2 C-1.
r = radius of Gaussian spherical surfaces = 10 cm.
Let, q = charge enclosed at its centre.
(a) According to Gauss’s law, the electric flux through a Gaussian surface depends upon the charge enclosed inside the surface and not upon its size. Thus, the electric flux will remain unchanged i.e., -1.0 x 103 Nm2 C-1 through the spherical Gaussian surface of double radius i.e. of 20 cm as it also encloses the same amount of charge.
(b) q = point charge = ?, ɛo= 8.854 x 10-12 Nm-2C2.





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 x 103 N/C and points radially inward, what is the net charge on the sphere ?
Answer: Here, R = radius of the conducting sphere = 0.10 m
r = distance of the point from centre of sphere = 20 cm = 0.2 m
Clearly r > R
E = electric field at a point at a distance of 20 cm from the sphere = 1.5 x 103 NC-1 acting inward.
q = net charge on the sphere = ?
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?
Answer: Here, σ = surface charge density = 80.0 µC/m2= 80 x 10-6 C/m2
    R = radius of the charged sphere = 2.4/2 = 1.2 m.
(a)  q = charge on the sphere = ?
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Question 1.23: An infinite line charge produces a field of 9 x 104 N/C at a distances of 2 cm. Calculate the linear charge density.
Answer: Here, E = electric field produced by infinite line charge = 9 x 104 NC-1.
r = distance of the point where E is produce = 2 cm = 0.02 m.
λ = linear charge density = ?
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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 x 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?
Answer:
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The arrangement of the plates is shown in the figure where, σ = surface density of plates = 17.0 x 10-22 Cm-2, ɛo = 8.854 x 10-12 C2 N-1m-2.
(a) E in the outer region of the first plate = E to the left of the first plate = ? The region I is to the left of the first plate thus, electric field E1 in this region due to the two plates is given by the following equation:
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(b) E in the outer region of second plate = E to the right of second plate, i.e., in region III =?. The E for the region III is given by -
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(c) E between the plates = E in II region = EII =? EII is given by -
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 Class XII CBSE Physics - Chapter 1, Electric Charges and Fields - NCERT Solutions


# CBSE Physics Guide - Electric Charges and Fields - Study materials, Additional Objective Questions, Short Questions & Long Questions, Numerical Problems with Solutions important for CBSE 12th Board Exam



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