Electric Charges And Fields Class 12 NCERT Solutions: Exercise Questions
Electric Charges and Fields Class 12 NCERT Solutions offer a solution to each question in the exercises of the chapter and allow students to gain a robust knowledge of electrostatics. Important concepts such as Coulomb law, electric field, electric flux, and Gauss theorem are explained with the right words in these solutions, hence preparing to undertake exams easy.
Q 1.3 Check that the ratio $\frac{ke^{2}}{Gm_{e}m_{p}}$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Answer:
Electrostatic force
$F=\frac{KQ^2}{r^2}$
So the dimension of
$[Ke^2]=[Fr^2]$ ..................(1)
The gravitational force between two bodies of mass M and m is
$F=\frac{GMm}{r^2}$
So, the dimension of
$[Gm_em_p]=[Fr^2]$ .............(2)
Therefore, from (1) and (2)
$[\frac{ke^{2}}{Gm_{e}m_{p}} ]$ is dimensionless
or
Here,
K = $1/4\pi \epsilon _{0}$ , where $\epsilon _{0}$ is the permittivity of space.
$[1/ \epsilon _{0}] =[C/V.m] = [Nm^2 C^{-2}]$
e = Electric charge ([e] = [C])
G = Gravitational constant. ([G]= $[Nm^2kg^{-2}]$ )
$m_{e}$ and $m_{p}$ are mass of electron and proton ([ $m_{e}$ ] = [ $m_{p}$ ] = [Kg])
Substituting these units, we get
$[\frac{ke^{2}}{Gm_{e}m_{p}} ] = [\frac{C^2 \times Nm^2C^{-2}}{Nm^2kg^{-2}\times kg\times kg} ] = M^{0}L^{0}T^{0}$
Hence, this ratio is dimensionless.
Putting the value of the constants
$\frac{ke^{2}}{Gm_{e}m_{p}} = \frac{9\times10^9\times(1.6\times10^{-19})^{2}}{6.67\times10^{11}\times9.1\times10^{-31}\times1.67\times10^{-27}}=2.3 \times 10^{40}$
The given ratio is the ratio of the electric force $\frac{ke^{2}}{R^2}$ to the gravitational force between an electron and a proton $\frac{Gm_{e}m_{p}}{R^2}$, considering the distance between them is constant!
Q 1.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
Answer:
The given statement "electric charge of a body is quantised" implies that the charge on a body can take only integral values. In other words, only an integral number of electrons can be transferred from one body to another and not in fractions.
Therefore, a charged body can only have an integral multiple of the electric charge of an electron.
Q 1.7 (b) Explain why two field lines never cross each other at any point.
Answer:
A tangent drawn at any point on a field line gives the direction of force experienced by a unit positive charge due to the electric field at that point. If two lines intersect at a point, then the tangent drawn there will give two directions of force, which is not possible. Hence, two field lines cannot cross each other at any point.
Q 1.8 (a) Two point charges $q_{A}=3 \mu C$ and $q_{B}=-3 \mu C$ are located 20 cm apart in vacuum. What is the electric field at the midpoint O of the line AB joining the two charges?
Answer:
Given, AB = 20 cm
Since, O is the midpoint of the line AB.
AO = OB = 10 cm = 0.1m
The electric field at a point caused by charge q, is given as,
$E = \frac{kq}{r^2}$
Where q is the charge, r is the distance between the charges and the point O
k = 9x109 N m2 C -2
Now,
Due to the charge at A, the electric field at O will be $E_{A}$ and in the direction AO.
$E_{A} =\frac{ 9\times10^9 \times 3\times10^{-6}}{0.1^2}$
Similarly, the electric field at O due to the charge at B, also in the direction AO
$E_{B} =\frac{ 9\times10^9 \times (-3\times10^{-6})}{0.1^2}$
Since both the forces are acting in the same direction, we can add their magnitudes to get the net electric field at O:
E' = $E_{A}$ + $E_{B}$ = 2E (Since their magnitudes are same)
$E' =2\times \frac{ 9\times10^9 \times 3\times10^{-6}}{0.1^2} = 5.4\times 10^4 N/C$ along the direction AO.
Q 1.9 A system has two charges $q_{A}=2.5\times 10^{-7}C$ and $q_{B}=-2.5\times 10^{-7}$ 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?
Answer:
Given,
$q_{A}=2.5\times 10^{-7}C$ and $q_{B}=-2.5\times 10^{-7}$
The total charge of the system = $q_{A} + q_{B} = 0$
$\therefore$ The system is electrically neutral. (All dipole systems have net charge zero!)
Now, distance between the two charges, d = 15 + 15 = 30 cm = 0.3 m
We know, the electric dipole moment of the system, p = $q_{A}$ x d = $q_{B}$ x d (i.e, the magnitude of charge x distance between the two charges)
$\therefore p = 2.5\times10^{-7} C \times 0.3 m = 7.5\times10^{-8} Cm$
The direction of a dipole is towards the positive charge. Hence, in the positive z-direction.
Q 1.18 A point charge of $2.0\mu C$ is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Answer:
Given,
q = net charge inside the cube = $2.0\mu C$
Using Gauss’s law, we know that the flux of the electric field through any closed surface S is $1/\epsilon _{0}$ times the total charge enclosed by S.
i.e. $\phi = q/\epsilon _{0}$
where, q = net charge enclosed and $\epsilon _{0}$ = permittivity of free space (constant)
$\therefore \phi = 2.0\times10^{-6}/8.85\times10^{-12}\ Nm^2C^{-1} = \boldsymbol{2.2\times10^5\ Nm^2C^{-1}}$
(Note: Using Gauss's formula, we see that the electric flux through the cube is independent of the position of the charge and the dimension of the cube! )
Q 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 $1.5 \times 10^{3}\frac{N}{C}$ and points radially inward, what is the net charge on the sphere?
Answer:
We know that, for determining the electric field at r>R for a conducting sphere, the sphere can be considered as a point charge located at its centre.
Also, the electric field intensity at a point P, located at a distance r, due to the net charge q is given by,
$E = k\frac{q}{r^2}$
Given, r = 20 cm = 0.2 m (From the centre, not from the surface!)
$\\ \therefore 1.5\times10^3 = 9\times10^9\times \frac{q}{0.2^2} \\ \implies q = \frac{1.5\times0.04}{9}\times10^{-6} = 6.67\times10^{-9}\ C$
Therefore, charge on the conducting sphere is $- 6.67\ nC$ (since flux is inwards)
NCERT Solution for Class 12 Chapter 1: Additional Questions
Q 1. 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?
Answer:
When two spheres of the same size are touched, on attaining the equipotential state, the total charge of the system is equally distributed on both of them.
Therefore, (i) When uncharged third sphere C is touched with A, charge left on A = $0.5\times 6.5\times 10^{-7}C$
and charge attained by C = $0.5\times 6.5\times 10^{-7}C$
(ii) Now, charge on B + charge on C = $6.5\times 10^{-7}C$ + $0.5\times 6.5\times 10^{-7}C$ = $1.5\times 6.5\times 10^{-7}C$
When touched, charge left on B = $0.5\times1.5\times 6.5\times 10^{-7}C$
Therefore $q_{A}\rightarrow 0.5\times q_{A}\ and\ q_{B}\rightarrow 0.75\times q_{B}$
Therefore,
$F' = 0.5\times0.75 \times F = 0.375 \times 1.5\times10^{-2} N = \boldsymbol{5.7\times10^{-3}\ N}$
Q 2. An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 \times 10^{4}NC^{-1}$ (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm-3. Estimate the radius of the drop.
Answer:
The force due to the electric field is balancing the weight of the oil droplet.
weight of the oil drop = density x volume of the droplet x g = $\rho \times \frac{4}{3}\pi r^3 \times g$
Force due to the electric field = E x q
charge on the droplet, q = No. of excess electrons x charge of an electron = $12\times q_{e} = 12\times 1.6\times10^{-19} = 1.92\times10^{-18} C$
Balancing forces:
$\rho \times \frac{4}{3}\pi r^3 \times g = E\times q$
Putting known and calculated values:
$\\ r^3 =\frac{3}{4\pi} E\times q / \rho\times g $
$ \Rightarrow r^3 = \frac{3}{4}\times\frac{7}{22}\times(2.55\times10^4)\times 1.92\times10^{-18} / (1.26\times10^3 \times10 ) $
$ \Rightarrow r^3 = 0.927\times10^{-18} m^3$
$r = 0.975\times10^{-6} m = 9.75\times10^{-4} mm$
Q 6. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left (\frac{\sigma}{2 \epsilon_{0}} \right )$ $\widehat{n}$ , where $\widehat{n}$ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.
Answer:
Let the surface area of the sphere be S.
And assume that the hole is filled. For a point B, just above the hole, considering a Gaussian surface passing through B, we have
$\oint E.dS = q/\epsilon_{0}$
Now, since the electric field is always perpendicular to the surface of the conductor.
$\\ \therefore \oint E.dS = E.S= q/\epsilon_{0} = (\sigma.S)/\epsilon_{0} \\ \implies E = \sigma/\epsilon_{0}$
Using Superposition principle, $E =E_{1} + E_{2}$ ,
where $E_{1}$ is due to the hole and $E_{2}$ is due to the rest of the conductor. (both pointing outwards, i.e away from the centre)
Again, for a point A, just below the hole, the Electric field will be zero because of electrostatic shielding.
Using the superposition principle, this will be due to $E_{1}$ pointing inwards(towards the centre) and due to $E_{2}$ (Pointing away from the centre)
$0 =E_{1}-E_{2}$ $\implies E_{1}=E_{2}$
Using this relation, we get:
$E =E_{1} + E_{2} = 2E_{1} \implies E_{1} = E/2 = \sigma/2\epsilon_{0}$
Since this is pointing outwards,
$\boldsymbol{E_{1}} = \sigma/2\epsilon_{0}\ \boldsymbol{\widehat{n}}$ is the electric field in the hole.
(Trick:
1. Assume the hole to be filled.
2. Consider 2 points just above and below the hole.
3. Electric fields at these points will be due to the hole and the rest of the conductor. Use the superposition principle.)
Q 7. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law.
[Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Answer:
Let AB be a long, thin wire of uniform linear charge density λ.
Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.
The charge on a small length dx on the line AB is q, which is given as q = λdx.
So, according to Coulomb’s law, the electric field at P due to this length dx is
$dE'=\frac{1}{4\pi \epsilon_o}\frac{\lambda dx}{(PC)^2}$
But $P C=\sqrt{h^2+x^2}$
$
\Rightarrow d E=\frac{1}{4 \pi \epsilon_o} \frac{\lambda d x}{\left(h^2+x^2\right)}
$
This electric field at P can be resolved into two components as dEcosθ and dEsinθ. When the entire length AB is considered, then the dEsinθ components add up to zero due to symmetry. Hence, there is only the dEcosθ component.
So, the net electric field at P due to dx is
dE' = dE cosθ
$
\Rightarrow d E^{\prime}=\frac{\lambda d x \cos \theta}{4 \pi \epsilon_o \left(h^2+x^2\right)}
$-----(1)
$\qquad$
$\ln \triangle \mathrm{POC}$
$
\tan \theta=\frac{x}{h}
$
$
\Rightarrow \mathrm{x}=\mathrm{h} \tan \theta
$
Differentiating both sides w.r.t. $\theta$,
$
\begin{aligned}
& \frac{d x}{d \theta}=h \sec ^2 \theta \\
& \Rightarrow \mathrm{dx}=\mathrm{h} \sec ^2 \theta \mathrm{~d} \theta .........(2)
\end{aligned}
$
Also, $h^2+x^2=h^2+h^2 \tan ^2 \theta$
$
\begin{aligned}
& \Rightarrow h^2+x^2=h^2\left(1+\tan ^2 \theta\right) \\
& \Rightarrow h^2+x^2=h^2 \sec ^2 \theta \ldots \ldots \ldots \ldots...........(3)
\end{aligned}
$
(Using the trigonometric identity, $1+\tan ^2 \theta=\sec ^2 \theta$ )
Using equations (2) and (3) in equation (1),
$
\begin{aligned}
& d E^{\prime}=\frac{\lambda \cos \theta \times h \sec ^2 \theta}{4 \pi \epsilon_o \times h^2 \sec ^2 \theta} d \theta \\
& d E^{\prime}=\frac{\lambda \cos \theta}{4 \pi \epsilon_o h} d \theta
\end{aligned}
$
The wire extends from $\theta=-\frac{\pi}{2}$ to $\theta=\frac{\pi}{2}$ since it is very long. Integrating both sides,
$
E^{\prime}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\lambda \cos \theta}{4 \pi \epsilon_o h} d \theta
$
$\begin{aligned} & E^{\prime}=\frac{\lambda}{4 \pi \epsilon_o h} \int_{\frac{\pi}{-2}}^{\frac{\pi}{2}} \cos \theta d \theta \\ & E^{\prime}=\frac{\lambda}{4 \pi \epsilon_o h}\left(\sin \frac{\pi}{2}-\sin \left(-\frac{\pi}{2}\right)\right) \\ & E^{\prime}=\frac{\lambda}{4 \pi \epsilon_o h} \times 2 \\ & E^{\prime}=\frac{\lambda}{2 \pi \epsilon_o h}\end{aligned}$
This is the net electric field due to a long wire with linear charge density λ at a distance h from it.
In the question linear charge density =E
Therefore
$E'=\frac{E}{2\pi \epsilon_o h}$
Q 8. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + $\left ( \frac{2}{3} \right )$ e, and the ‘down’ quark (denoted by d) of charge $\left (- \frac{1}{3} \right )$ e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
Answer:
Given, a proton and a neutron consist of three quarks each.
And, ‘up’ quark is of charge + $\left ( \frac{2}{3} \right )$ e, and the ‘down’ quark of charge $\left (- \frac{1}{3} \right )$ e
Let the number of 'up' quarks be n. Therefore, the number of 'down' quarks is (3-n).
$\therefore$ The net charge = $\\ n\times\frac{2}{3}e + (3-n)\frac{-1}{3}e = (n-1)e$
Now, a proton has a charge +1e
$\therefore$ $(n-1)e = +1e \implies n =2$
Proton will have 2 u and 1 d, i.e, uud
Similarly, the neutron has a charge 0
$\therefore$ $(n-1)e = 0 \implies n =1$
Neutron will have 1 u and 2 d, i.e, udd
Q 10. A particle of mass m and charge (–q) enters the region between the two charged plates, initially moving along the x-axis with speed vx . The length of the plate is L, and a uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $\frac{qEL^{2}}{2mv_{x}^{2}}$ .
Answer:
Let s be the vertical deflection, t be the time taken by the particle to travel between the plates
$\therefore$ $s = ut + \frac{1}{2}at^2$
Here, u =0 , since initially there was no vertical component of velocity.
The particle in the electric field will experience a constant force (Since the electric field is constant.)
F = ma = -qE (Using Newton's Second Law, F = ma)
$\therefore$ a = -qE/m (-ve sign implies here in downward direction)
Again, t = Distance covered/ Speed = $L/ v_{x}$
(In x-direction, since there is no force, hence component of velocity in x-direction remains constant = $v_{x}$ .
And, the distance covered in the x-direction = length of the plate = L)
Putting these values in our deflection equation,
$\\ s = ut + \frac{1}{2}at^2 = (0)(L/v_{x}) + \frac{1}{2}(-qE/m)(L/v_{x})^2 \\ \implies s =\frac{-qEL^2}{2mv_{x}^2}$
(S is -ve, which implies it deflects downwards.)
$\therefore$ The vertical deflection of the particle at the far edge of the plate is $\frac{qEL^{2}}{2mv_{x}^{2}}$ .
This motion is similar to the motion of a projectile in a gravitational field, which is also a constant force. The force acting on the particle in the gravitational field is mg while in this case, it is qE. The trajectory will be the same in both cases.
