Param Himalaya : NCERT Solutions For Class 12 Physics Chapter 1 Electric Charges And Fields is prepared and uploaded for reference by academic team of expert members of Param Himalaya.
Question 1.1 : What is the force between two small charged spheres having charges of 2 × 10-7C and 3 × 10-7C placed 30 cm apart in air?
Solution :
\[F = 9×10^{9} \frac{q_{1}q_{2}}{r^{2}}\]
\[F = \frac{9×10^{9}×2×10^{-7}×(3×10^{-7)}}{0.30×0.30}\]
\[F = 6×10^{-3} N\]
Question 1.2 : The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
Solution :
(a) Force on charge 1 due to charge 2 is given by the relation
\[F_{12} = 9×10^{9}\frac{q_{1}q_{2}}{r^{2}}\]
\[r^{2} = \frac{(9×10^{9})q_{1}q_{2}}{F_{12}}\]
\[r^{2} =\frac{(9×10^{9})(.8×10^{-6})(.4×10^{-6})}{.2}\]
\[r^{2} =\frac{2.88×10^{-3}}{2×10^{-1}}\]
\[r^{2}= 1.44 × 10^{-2}\]
\[r = \sqrt{1.44 × 10^{-2}}\]
\[r = .12m = 12 cm\]
(b) \[F_{21} = F_{12} = .2N Attractive\]
Question 1.3 :Check that the ratio \[ke^{2}/Gm_{e}m_{p}\] is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Solution 1.3 :
\[\frac{ke^{2}}{Gm_{e}m_{p}} = \frac{(9×10^{9})(1.6×10^{-19})^{2}(Nm^{2}C^{-2})(C^{2})}{(6.67×10^{-11})(9.1×10^{-31})(1.66×10^{-27})(Nm^{2}kg^{-2})(kg^{-2})}\]
\[\frac{ke^{2}}{Gm_{e}m_{p}} = 2.29×10^{39}\]
Thus , the given ratio is a number and is dimensionless. This ratio signifies that electrostatic force between electron and proton is very-very large as compared to the gravitational force between them.
Question 1.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantized’. (b) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges?
Solution :
a) Total charge on a body is equal to integral multiple of charge on an electron (e).
i.e, \[q =\pm ne\]
where n = 0,1,2,3.....
b) At macroscopic level , the quantisation of charges has no practical importance because the charge at macroscopic level is very large as compared to elementary charge e i.e 1.6+10-19 C.
for example, a small charge of 1μC has about 1013electronic charges. in such cases the charge may be treated as continuous and not quantised
Question 1.5 : 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.
Solution :
Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilates each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.
Question 1.6 : Four point charges qA = 2 μC, qB = –5 μC, qC = 2 μC, and qD = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Solution:
The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
Where, (Sides) AB = BC = CD = AD = 10 cm (Diagonals) AC = BD = 10√2 cm AO = OC = DO = OB = 5√2 cm
A charge of amount 1μC is placed at point O. Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1μC charge at centre O is zero.
Question 1.7 (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?
Solution :
(a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
(b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.
Question 1.8 Two point charges qA = 3 μC and qB = –3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge?
Solution :
(a) Electric field at the mid point of the separation between two equal and opposite charges is given by
\[\overrightarrow{E} = \overrightarrow{E_{1}} +\overrightarrow{E_{2}}\] \[i.e. \overrightarrow{E} = \overrightarrow{E_{1}} +\overrightarrow{E_{1}} =2\overrightarrow{E_{1}}\]
Question 1.9 A system has two charges qA = 2.5 × 10–7 C and qB = –2.5 × 10–7 C located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Solution :
Clearly , the given points are lying on z-axis.
Distance between charges , 2l
2l = 15+15 = 30 cm = .3 m
Total charge = 2.5 × 10-7 C – 2.5 × 10-7 C = 0
Dipole moment = q×2l = 2.5 × 10–7 × .3 = 7.5 × 10-8 C m along negative z-axis.
Question 1.10 : 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 NC-1 . Calculate the magnitude of the torque acting on the dipole.
Solution:
Using τ = pESinθ, we get
= (4 × 10-9 )(5×104)Sin30°
= 2 × 10-4× 1/2 = 10-4Nm
Question 1.11 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?
Solution :
(a) using q = ne , we get
\[n = \frac{q}{e} = \frac{-3.2×10^{-7}}{-1.6×10^{-19}}\]
\[n = 2×10^{12}\]
(b) Yes, but of negligible amount because mass of an electron is very-very small mass transferred,
\[= m_{e} × n = (9.1 × 10^{-31} ) ×(2×10^{12})\]
\[= 1.82 × 10^{-18} kg\]
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?
Solution :
(a) using,
\[F = \frac{9 ×10^{9} ×q ×q}{r^{2}} , we get\]
\[F = \frac{(9×10^{9})(6.5×10^{-7})(6.5×10^{-7})}{(50×10^{-2})^{2}}\]
\[=1.5×10^{-2}N\]
(b) \[F^{'} = \frac{(9×10^{9})(2q)(2q)}{(\frac{r}{2})^{2}}\]
\[= \frac{9+10^{9}+q+q+2+2}{(\frac{r^{2}}{4})} = 16 F\]
\[= 16 ×1.5 ×10^{-2} = 0.24 N\]
Questions 1.13. Figure 1.30 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 :
Unlike charges attract each other, therefore particle 1 and 2 negatively charged whereas particle 3 has positive charge . Particle 3 gets maximum deflection so it has highest charge (e) to mass ratio because deflection, y ∝ e/m
Question 1.14 : 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 :
(a) Electric flux through the square :
\[\phi = \vec{E}.\vec{dS}\]
\[= (3+10^{3}) \hat{i}(\frac{10}{100}+\frac{10}{100}\hat{i})\]
\[ = 30 Nm^{2}C^{-1}\]
(b) Again
\[\phi = \vec{E}.\vec{dS} = EdScos \Theta\]
\[= (3+10^{3})(\frac{10}{100}+\frac{10}{100}) cos60\]
\[= (3+10^{3})(10^{-2})(\frac{1}{2})\]
\[= 15 N m^{2}C^{-1}\]
Question 1.15 What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Solution :
Zero , because number of field lines entering the cube is equal to the number of field lines coming out of the cube.
Question 1.16 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 :
(a) Using , \[\phi = \frac{q}{\varepsilon_{0}}\]
\[we\;get \;q = \phi.\varepsilon_{0}\]
\[= (8×10^{3})(8.854×10^{-12})\]
\[= 70.8 × 10^{-9}C\]
\[= 0.07 μC\]
(b) No , it cannot be said so because there may be equal number of positive and negative elementary charges inside the box. it can only be said that net charge inside the box is zero.
Question 1.17 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. 1.31. 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 :
The charge can be assumed to be placed as shown in the fiqure.
\[\phi = \frac{q}{\varepsilon_{0}}\\ the\;flux\;through\; one \;square \;surface \\ =\frac{1}{6}\frac{q}{\varepsilon_{0}}\\=\frac{1}{6}×\frac{10^{-5}}{8.854×10^{-12}}\\=1.88×10^{5} Nm^{2}C^{-1}\]
Question 1.18 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 :
\[Using\;\phi = \frac{q}{\varepsilon_{0}}\]
\[=\frac{2×10^{-6}}{8.854×10^{-12}}\]
\[=2.26×10^{5} Nm^{2}C^{-1}\]
Question 1.19 : 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 :
(a) The Electric Flux depends only on the charge enclosed by the Gaussian surface and independent of the size of the Gaussian surface. The electric flux through new Gaussian surface. The electric flux through new Gaussian surface remains same i.e –1.0 × 103 Nm2 /C because the charge enclosed remains same in this case also.
(b) \[Using\;\phi = \frac{q}{\varepsilon_{0}}\]
\[we\; get \; q=\varepsilon_{0}\phi\]
\[=(8.85×10^{-12})(-1×10^{3})\]
\[i.e. \; q = -8.85×10^{-9}C\]
Question 1.20 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 :
$$E = \frac{9×10^{9}(q)}{r^{2}} we \ get$$
Since Electric field points radially inward, so E is taken as -ve
$$q =-\frac{Er^{2}}{9×10^{9}}$$
$$q= - \frac{(1.5×10^{3})[20×10^{-2}]^{2}}{9×10^{9}}$$
$$q=-\frac{(1.5×10^{3})(.2)^{2}}{9×10^{9}}$$
$$q=-6.67nC$$
Question 1.21 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 :
(a) Charge on the sphere is given by
\[q=\sigma×4\pi R^{2}\\Here \;\sigma= 80×10^{-6}Cm^{-2}\;and\\R=\frac{2.4}{2}=1.2m\\\therefore\;q=(80\times10^{-6})\times4\pi(1.2)^{2}\\=1.45\times10^{-3}C\]
(b) Total electric flux
$$ \phi_{total} = \frac{Q}{\epsilon_{0} }$$
$$\epsilon_{0} = 8.854 \times 10^{-12}C^{2}m^{-2}N^{-1}$$
$$Q=1.447 \times 10^{-3}C$$
$$\phi_{total} = \frac{1.447 \times 10^{-3}}{8.854 \times 10^{-12}}$$
$$\phi_{total} = 1.63 \times 10^{8}C^{-1}m^{2}N$$
Therefore, the total electric flux leaving the surface of the sphere is $1.63 \times 10^{8}C^{-1}m^{2}N$
Question 1.22 An infinite line charge produces a field of 9 ×104 N/C at a distance of 2 cm. Calculate the linear charge density.
Solution :
\[Using \;E=\frac{\lambda}{2\pi\varepsilon_{0}r},\;we\;get\\\lambda=2\pi\varepsilon_{0}Er=4\pi\varepsilon_{0}\frac{Er}{2}\\=\frac{1}{9\times10^{9}}\times\frac{9\times10^{4}\times0.02}{2}\\=10^{-7}Cm^{-1}\]
Question 1.23 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?