NCERT Solutions for Class 12 Science Physics Chapter 1 - Electric Charges And Fields

Answer:

q₁  is a negative charge as the net force acting on it is along the positive x-axis.
When the Positive Charge Q is added at the point shown in the figure the force of attraction on the charge q₁ will increase and it will get attracted more towards the charge Q or we can say that the force on it shall increase along the positive x-axis.

Hence, the correct answer is  option (a).

Answer:

Positive charge acts as the source of electric field lines.  So the field lines will originate from the charge q.
Also, the field lines are always perpendicular to the surface of the conductor.
Tangent to the field line at any point gives the direction of the field at that point.
​Due to the presence of the positive charge q near the conducting sphere, negative charge will be induced on the left surface of the sphere and positive charge on the right surface of the sphere. 
Only figure 1 follows all of the above points. 
​Hence, the correct answer is option (a).

Answer:

According to the Gauss's Law electric flux out of a closed surface is equal to the net charge enclosed within the gaussian surface divided by the permittivity,
$\varphi \mathit{ }\mathit{=}\mathit{ }\frac{q_{net}}{{\epsilon }_{\mathit{0}}}$
Since the charge enclosed by each of the figures is equal to q so the net electric flux through each sphere will be the same.

Hence, the correct answer is option (d).

Answer:

According to the gauss law, the electric flux coming out of a closed surface depends only on the charge enclosed within that closed surface.
$\oint \overrightarrow{E}.\overrightarrow{dS}= \frac{q_{net}}{{\epsilon }_0}$
The electric field also depends on the charge enclosed within the closed gaussian surface.
The net charge enclosed within the gaussian surface is q₂  & q₄  only.
Hence, the correct answer is option (b).

Answer:

It can be seen from the figure that the field lines are not uniformly spaced, So it can be concluded that the field is not uniform.
Electric field strength decreases from left to right as the crowding of the field lines decreases, So the resultant force on charges also decreases from left to right.
Force on a charge -q is greater than the force exerted by the field on the charge +q.
The dipole will experience a net force towards the left.

Hence, the correct answer is option (c).

Answer:

When a small point charge is placed near a plane sheet of charge negative charge is induced on that side of the sheet which faces the +q charge whereas the charge induced on the other side of the sheet is opposite in nature i.e. +q 
 

Answer:

The component of the electric field parallel to the diameter cancels out and we are only left with the perpendicular components.

The resultant electric field is perpendicular to the diameter.

Hence, the correct answer is option (a).

 

Answer:

Answer:

The electric field in general is continuous but the field at a point cannot be continuous always. 

The electric field at a point is continuous if there is no charge at that point.

It will be discontinuous if there is a charge at that point.

Hence, the correct answer is option (b) and option (d).

Answer:

Gauss law states that the electric flux ${\oint }_s\mathbf{E}.\mathrm{d}\mathbf{S}$  through a closed surface depends on the charge enclosed within that closed surface.

So, If there is no charge inside the closed surface then the flux ${\oint }_s\mathbf{E}.\mathrm{d}\mathbf{S}$ will be equal to zero. 

The flux will be non-zero if there is a charge inside the Gaussian surface. ${\oint }_s\mathbf{E}.\mathrm{d}\mathbf{S}=\frac{q}{{\epsilon }_0}$ 


Hence, the correct answer is option (b) and (d).


 

Answer:

In the region the net charge is zero, So it can be said that inside the region there is a presence of equal and opposite charges inside the region which is basically a dipole.

So it can be said that the dipole is present inside the region and now we need to calculate the Electric Field at any Point outside this charge configuration.

The electric field due to the dipole at a large distance varies as $\frac{1}{r^3}$

Since the electric field is a conservative field so the work done in moving a charged particle in closed loop in the Electric field region is
zero.

So, the correct answers are option (c) and (d).
 

Answer:

The flux through the closed gaussian surface is given by the Gauss law:

$\varphi =\frac{q_{net}}{{\epsilon }_o}$

The net charge enclosed within the gaussian surface is given by 

$Q-2Q =-Q$

So the electric flux through the sphere is $\frac{-Q}{{\epsilon }_o}$

5Q charge lies outside the sphere, So the electric flux due to the 5Q charge is zero.

Hence, The correct answer is option (a) and (c).

Answer:

At the center of the ring, the net electric field is zero. So the net force experienced by the charge q is zero. If the charge is displaced from the center of the ring then there will be an electric field in the opposite direction. So the charge will be pushed back to its original position. 

A similar thing will happen for -q charge and it will also be pushed back to its original position by the above criteria.

The charge -q will perform SHM for small displacement.

The point at the center is a point of unstable equilibrium.
 

Answer:

The net charge enclosed within the closed surface is given by the sum of the two charges of the dipole i.e. -q+q =0 

So the net electric flux is given by the formula $\varphi = \frac{0}{{\epsilon }_o}=0$

So, the net electric flux through this closed surface is zero.

Answer:

The charge induced on the inner surface is -Q

The charge on the outer surface of the sphere is +Q.

The surface charge density is given by charge divided by the Area of the surface.

 $$\begin{gathered} \left(i\right) {\sigma }_{inner} = \frac{=Q}{4\pi R_1^2} \\ \left(ii\right) {\sigma }_{outer} = \frac{Q}{4\pi R_2^2} \end{gathered}$$

Answer:

There is an equal number of protons and neutrons present in a neutral atom, So the net charge on an atom is zero.
 

Answer:

According to the gauss law, the electric field through a closed Gaussian surface depends only on the charge enclosed within that gaussian surface.

Mathematically, $\oint E.ds = \frac{q_{net}}{{\epsilon }_o}$

But the Electric field at the surface could also be due to a charge present outside the closed gaussian surface.

So, the net charge inside the closed gaussian surface decides the field coming out of the gaussian surface only, It does tell us about the field due to any external charges, which could be present.

So it would be wrong to say that the electric field at the surface is zero is the net charge enclosed within the surface is zero.

Now if the Electric field on the surface is zero, then the charge enclosed within the gaussian surface will necessarily be zero. This is the correct statement.

 

Answer:

ans

Answer:

(a) Using the symmetry we can draw the Gaussian surface such that it encloses the corner A.

Total flux through one of the cube is given by $\frac{1}{8}\times \frac{q}{{\epsilon }_o}= \frac{q}{8{\epsilon }_o}$

(b) Again we can draw the Gaussian surface such that it enclose the charge completely. So the flix is now given by $\frac{1}{4}\times \frac{q}{{\epsilon }_o}= \frac{q}{4{\epsilon }_o}$

(c) The flux, in this case, is given by $\frac{1}{2}\times \frac{q}{{\epsilon }_o}= \frac{q}{2{\epsilon }_o}$

(d) The flux, in this case, is also given by $\frac{1}{2}\times \frac{q}{{\epsilon }_o}=\frac{q}{2{\epsilon }_o}$

Answer:

Mass of paisa coin = 0.75 g (made up Al only)
Atomic mass of Al = 27u

Number of moles of 0.75 g of $\mathrm{Al}=\frac{0.75}{27}$
Number of Aluminium atoms = number of moles × N_A
Where, N_A is the Avogadro number.
Thus, number of Al atom $=\frac{0.75}{27}\times 6.022\times {10}^{23}$
But, atomic number of Al = 13
That means the atom of Aluminium has 13e^– and 13p+ in neutral state.
So, number of either electron and proton:
Total no. of e^– = no. of atoms × no. of e^– in one atom

                      $=\frac{0.75}{27}\times 6.022\times {10}^{23}\times 13$
$\mathrm{Total} \mathrm{charge}=\frac{0.75}{27}\times 6.022\times {10}^{23}\times 13\times 1.6\times {10}^{-19}\mathrm{C}$

Total Charge = 3.48 × 10⁴ C

The coin has equal amount of positive and negative charge $\pm$3.48 × 10⁴C.

Answer:

Given:
q₁ = 3.48 × 10⁴ C
q₂ = –3.48 × 10⁴ C
(i) r = 1 cm = 10^–2 m

$F=\left|K \frac{q_1{q}_2}{r^2}\right|=\frac{9\times {10}^9\times {\left(3.48\times {10}^4\right)}^2}{{\left({10}^{-2}\right)}^2}=1.09\times {10}^{23} \mathrm{N}.$

(ii)Similarly for, r = 100 m = 10² m

Answer:

(i) Due to symmetry of figure, some of all attractive force acting on the chlorine atom is equal to zero.
 

Answer:

Let the charge is placed at point 'P', at a distance 'x' on left of charge 'q' as shown below,

At P, the charge '2q' experience force towards left due to 'q' and towards right due to '–3q'
$$\begin{gathered} \therefore \frac{2q^2}{4\mathrm{\pi }{\epsilon }_{\mathrm{o}}{x}^2}=\frac{6q^2}{4\mathrm{\pi }{\epsilon }_{\mathrm{o}}{\left(x+d\right)}^2} \\ \therefore {\left(d+x\right)}^2=3x^2 \\ \therefore 2x^2-2dx-d^2=0 \end{gathered}$$
Solving the quadratic equation we get:
$x=\frac{d}{2}\pm \frac{\sqrt{3}}{2}d$
Negative sign would be between 'q' and '–3q' and hence it is not acceptable.
Thus, $x=\frac{d}{2}\left(1+\sqrt{3}\right)$ to left of q.

Answer:

(a) A and C are positive  charges. (Electric field lines are outwards).
(b) The electric field line density is maximum at point C. Hence, the charge 'C' has the largest magnitude.
(c) (i) near A

Answer:

(a)
(i) Due to symmetry of figure, the electric field at centre 'O' of the polygon is zero.
(ii) When the charge 'q' is removed then net electric field due to the other charges at O,
$E=\frac{q}{4\mathrm{\pi }{\epsilon }_{\mathrm{o}}{r}^2}$ (along OA)
(iii) If charge 'q' at A is replaced by '–q' then the value of net electric field at O,
$E=2\times \left(\frac{q}{4\mathrm{\pi }{\epsilon }_{\mathrm{o}}{r}^2}\right)=\frac{2q}{4\mathrm{\pi }{\epsilon }_{\mathrm{o}}{r}^2}$  (along OA.)

(b) Similar method can be applied as in part (a) to calculate the electric field for regular polygon of n sides (symmetry is used to solve).

Answer:

(a) Let us suppose that universe is a perfect sphere of radius R.

Answer:

(a) Let us consider a sphere 'S' of radius 'R' and two hypothetical sphere of radius r < R and r > R.
 

Charge on the sphere,

$$\begin{gathered} \Rightarrow q={\int }_0^R\rho dV={\int }_0^R\left(kr\right)4\pi r^2·dr \\ \Rightarrow q=\frac{4\pi k{R}^4}{4}=2e \\ \Rightarrow k=\frac{2e}{\pi R^4} \end{gathered}$$

If proton 1 and 2 are embedded at distance 'r' from the centre of the sphere as shown, then attractive force on proton 1 due to charge distribution is


$F_1=eE=\frac{-ek{r}^2}{4{\epsilon }_0}$

Repulsive force on proton 1 due to proton 2 is,

$F_2=\frac{e^2}{4\pi {\epsilon }_0{\left(2r\right)}^2}$

Net force on 1,
F = F₁ + F₂

So, $F=\left[\frac{-e{r}^2}{4{\epsilon }_0}·\frac{2e}{\pi R^4}+\frac{e^2}{16\pi {\epsilon }_0{r}^4}\right]$

Thus, net force on proton 1 will be zero, when

$$\begin{gathered} \frac{e{r}^{2}2e}{4\pi {\epsilon }_0\pi R^4}=\frac{e^2}{16\pi {\epsilon }_{0}r} \\ \Rightarrow r^4=\frac{R^4}{8} \\ \Rightarrow r=\frac{R}{{\left(8\right)}^{{\displaystyle \frac{1}{4}}}} \end{gathered}$$

Answer:

(a) Net electric field at plate γ before collision is equal to the sum of electric field at plate γ due to plate α and β.
 

Answer:

(i) $F=\frac{qQ}{r^2}=1 \mathrm{dyne}=\frac{{\left[1\mathrm{esu} \mathrm{of} \mathrm{charge}\right]}^2}{{\left[1 \mathrm{cm}\right]}^2}$

Answer:

Given, two charge '–q' at 'A' and 'B'.

AB = AO + OB = 2d

x = small distance perpendicular to O.



So, force of attraction at 'P' towards A and B are each

$F=\frac{q\left(q\right)}{4{\mathrm{\pi \epsilon }}_o{r}^2},$ where AP = BP = r.

Horizontal components of these forces F_n are cancel out. Vertical components along PO add.

If ∠APO = 0, the net force on 'q' along PO is F′

⇒ F′ = 2F cosθ

$$\begin{gathered} \Rightarrow \frac{2q^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{r}^2}·\left(\frac{x}{r}\right) \\ \Rightarrow \frac{2q^{2}x}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\left(d^2+x^2\right)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

When, $x<<d, F\text{'}=\frac{2q^{2}x}{4{\mathrm{\pi \epsilon }}_0{d}^3}=K^x$
Where, $K=\frac{2q^2}{4{\mathrm{\pi \epsilon }}_0{d}^3}$
⇒ F ∝ x

Force on charge q is proportional to its displacement from the centre O and it is directed towards O.

Hence, motion of charge 'q' would be the simple harmonic, where

$\omega =\sqrt{\frac{K}{m}}$

and $T=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=2\mathrm{\pi }\sqrt{\frac{\mathit{m}}{\mathit{K}}}$

$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{d}^3}{2q^2}}={\left[\frac{8{\mathrm{\pi }}^3{\mathrm{\epsilon }}_{\mathrm{o}} {\mathrm{md}}^3}{q^2}\right]}^{\frac{1}{2}}$

Answer:

(a) Let the charge 'q' is displaced towards the axis of ring.