Thinking Process

Find the nature of force between $q_{1}-q_{2}$ and $a_{1}-q_{3}$. Nature of force will give the type of charge $a_{1}$. Find the nature of force between newly introduced charge and charge $q_{1}$.

Answer

(a) The net force on $q_{1}$ by $q_{2}$ and $q_{3}$ is along the $+x$-direction, so nature of force between $q_{1}, q_{2}$ and $q_{1}, q_{3}$ is attractive. This can be represent by the figure given below

The attractive force between these charges states that $q_{1}$ is a negative charge (since, $q_{2}$ and $q_{3}$ are positive).

Thus, nature of force between $q_{1}$ and newly introduced charge $Q$ (positive) is attractive and net force on $q_{1}$ by $q_{2}, q_{3}$ and $Q$ are along the same direction as given in the diagram below

The figure given above clearly shows that the force on $q_{1}$ shall increase along the positive $x$-axis due to the positive charge $Q$.

Note Unlike charges repel each other and like charges attract each other.

Thinking Process

Bringing the point positive charge towards the conducting sphere, charges the sphere by induction process. Electric field lines passes through a charged body following some rules.

Answer

(a) When a positive point charge is brought near an isolated conducting sphere without touching the sphere, then the free electrons in the sphere are attracted towards the positive charge. This leaves an excess of positive charge on the rear (right) surface of sphere.

Both kinds of charges are bound in the metal sphere and cannot escape. They, therefore, reside on the surface.

Thus, the left surface of sphere has an excess of negative charge and the right surface of sphere has an excess of positive charge as given in the figure below

attracted negative

An electric field lines start from positive charge and ends at negative charge (in this case from point positive charge to negative charge created inside the sphere).

Also, electric field line emerges from a positive charge, in case of single charge and ends at infinity.

Here, all these conditions are fulfilled in Fig. (a).

Answer

(d) Gauss’ law of electrostatics state that the total of the electric flux out of a closed surface is equal to the charge enclosed decided by the permittivity i.e., $Q$ electric $=\frac{Q}{\varepsilon_{0}}$.

Thus, electric flux through a surface doesn’t depend on the shape, size or area of a surface but it depends on the number of charges enclosed by the surface.

So, here in this question, all the figures same electric flux as all of them has single positive charge.

Answer

(b) According to Gauss’ law, the term $q$ on the right side of the equation $\int_{S} E \cdot d S=\frac{q}{\varepsilon_{0}}$ includes the sum of all charges enclosed by the surface.

The charges may be located anywhere inside the surface, if the surface is so chosen that there are some charges inside and some outside, the electric field on the left side of equation is due to all the charges, both inside and outside $S$.

So, $\mathrm{E}$ on LHS of the above equation will have a contribution from all charges while $q$ on the RHS will have a contribution from $q_{2}$ and $q_{4}$ only.

Thinking Process

Find the electric field strength on the charges of dipole.

Force varies directly with electric field strength i.e., higher the electric field strength greater the force and vice-versa.

Answer

(c) The space between the electric field lines is increasing, here from left to right and its characteristics states that, strength of electric field decreases with the increase in the space between electric field lines. As a result force on charges also decreases from left to right.

Thus, the force on charge $-q$ is greater than force on charge $+q$ in turn dipole will experience a force towards left.

Answer

(a) When a point positive charge brought near an isolated conducting plane, some negative charge developes on the surface of the plane towards the charge and an equal positive charge developes on opposite side of the plane. This process is called charging by induction.

Answer

(a) When the point is situated at a point on diameter away from the centre of hemisphere charged uniformly positively, the electric field is perpendicular to the diameter. The component of electric intensity parallel to the diameter cancel out.

Answer

(c, $d)$

$\oint_{S} E . d S=0$ represents electric flux over the closed surface.

In general, $\oint_{S}$ E.dS means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface.

When $g_{S} E . d S=0$, it means that the number of flux lines entering the surface must be equal to the number of flux lines leaving it.

Now, from Gauss’ law, we know that $g_{S} E . d S=\frac{q}{\varepsilon_{0}}$ where $q$ is charge enclosed by the surface. When $\int_{S} E \cdot d S=0, q=0$ i.e., net charge enclosed by the surface must be zero. Therefore, all other charges must necessarily be outside the surface. This is because charges outside because of the fact that charges outside the surface do not contribute to the electric flux.

Answer

$(b, d)$

The electric field due to a charge $Q$ at a point in space may be defined as the force that a unit positive charge would experience if placed at that point. Thus, electric field due to the charge $Q$ will be continuous, if there is no charge at that point. It will be discontinuous if there is a charge at that point.

Answer

(c, $d)$

Gauss’ law states that $\oint_{s} E . d S=\frac{q}{\varepsilon_{0}}$, where $q$ is the charge enclosed by the surface. If the charge is outside the surface, then charge enclosed by the surface is $q=0$ and thus, $\oint_{s}$ E.dS $=0$. Here, electric flux doesn’t depend on the type or nature of charge.

Answer

(c, $d)$

When there are various types of charges in a region, but the total charge is zero, the region can be supposed to contain a number of electric dipoles.

Therefore, at points outside the region (may be anywhere w.r.t. electric dipoles), the dominant electric field $\propto \frac{1}{r_{3}}$ for large $r$.

Further, as electric field is conservative, work done to move a charged particle along a closed path, away from the region will be zero.

Answer

$(a, c)$

Gauss’ law states that total electric flux of an enclosed surface is given by $\frac{q}{\varepsilon_{0}}$ where $q$ is the charge enclosed by the surface. Thus, from figure,

Total charge inside the surface is $=Q-2 Q=-Q$

$\therefore$ Total flux through the surface of the sphere $=\frac{-Q}{\varepsilon_{0}}$

Now, considering charge $5 Q$. Charge $5 Q$ lies outside the surface, thus it makes no contribution to electric flux through the given surface.

Answer

$(a, b, c)$

The positive charge $Q$ is uniformly distributed at the outer surface of the enclosed sphere. Thus, electric field inside the sphere is zero.

So, the effect of electric field on charge $q$ due to the positive charge $Q$ is zero.

Now, the only governing factor is the attractive and repulsive forces between charges

$(Q$ and $q)$ there are two cases arise.

Case I When charge $q>0$ i.e., $q$ is a positive charge, there creates a repulsive force between charge $q$ and $Q$.

The repulsive forces of charge $Q$ from all around the charge $q$ will push it towards the centre if it is displaced from the centre of the ring.

Case II When charge $q<0$ i.e., $q$ is a negative charge then there is an attractive force between charge $Q$ and $q$

If $q$ is shifted from the centre, then the positive charges nearer to this charge will attract it towards itself and charge $q$ will never return to the centre.

Answer

From Gauss’ law, the electric flux through an enclosed surface is given by $\oint_{s} E \cdot d S=\frac{q}{\varepsilon_{0}}$. Here, $q$ is the net charge inside that enclosed surface.

Now, the net charge on a dipole is given by $-q+q=0$

$\therefore \quad$ Electric flux through a surface enclosing a dipole $=\frac{-q+q}{\varepsilon_{0}}=\frac{0}{\varepsilon_{0}}=0$

Thinking Process

Let us draw the diagram as per the given situation. Using the induction process of charging distribute the charge on whole spherical shell. Now, find the required surface charge density.

Answer

Here, the charge placed at the centre of the spherical cavity is positively charged. So, the charge created at the inner surface of the sphere, due to induction will be $-Q$ and due to this charge created at outer surface of the sphere is $+Q$.

Now, surface charge density on the inner surface $=\frac{-Q}{4 \pi R_{1}^{2}}$ and

Surface charge density on the inner surface $=\frac{+Q}{4 \pi R_{2}^{2}}$

Answer

The protons and electrons are bound into a atom with distinct and independent existence and neutral in charge.

Electrostatic fields are caused by the presence of excess charges.

But there can be no excess charge on the inter surface of an isolated conductor. So, the electrostatic fields inside a conductor is zero despite the fact that the dimensions of an atom are of the order of an Angstrom.

Answer

Gauss’ law also implices that when the surface is so chosen that there are some chargas inside and some outside.

The flux in such situation is given by $\oint$ E.dS $=\frac{q}{\varepsilon_{0}}$.

In such situations, the electric field in the LHS is due to all the charges both inside and outside the surface. The term $q$ on the right side of the equation given by Gauss’ law represent only the total charge inside the surface.

Thus, despite being total charge enclosed by a surface zero, it doesn’t imply that the electric field everywhere on the surface is zero, the field may be normal to the surface.

Also, conversely if the electric field everywhere on a surface is zero, it doesn’t imply that net charge inside it is zero.

i.e.,

we get

$$ \text { Putting } E=0 \text { in } g \text { E.dS }=\frac{q}{\varepsilon_{0}} $$

$$ q=0 . $$

Thinking Process

According to general properties, electric field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity.

Answer

Thus, the electric field lines will start from positive charges and move towards infinity as given in the figure below

Side view

Top view

Answer

(a) There are eight corners in a cube so, total charge for the cube is $\frac{q}{8}$.

Thus, electric flux at $A=\frac{q}{8 \varepsilon_{0}}$.

(b) When the charge $q$ is place at $B$, middle point of an edge of the cube, it is being shared equally by 4 cubes. Therefore, total flux through the faces of the given cube $=q / 4 \varepsilon_{0}$.

(c) When the charge $q$ is placed at $C$, the centre of a face of the cube, it is being shared equally by 2 cubes. Therefore, total flux through the faces of the given cube $=q / 2 \varepsilon_{0}$.

(d) Similarly, when charge $q$ is placed at $Q$, the mid-point of $B$ and $C$, it is being shared equally by 2 cubes. Therefore, total flux through the faces of the given cube $=q / 2 \varepsilon_{0}$.

Thinking Process

Treating the paisa coins made up of only Al, find the magnitude of equal number of positive and negative charges. What conclusion do you draw from this magnitude?

Answer

Here, given quantities are

Mass of a paisa coin $=0.75 \mathrm{~g}$

Atomic mass of aluminium $=26.9815 \mathrm{~g}$

Avogadro’s number $=6.023 \times 10^{23}$

$\therefore \quad$ Number of aluminium atoms in one paisa coin,

$$ N=\frac{6.023 \times 10^{23}}{26.9815} \times 0.75=1.6742 \times 10^{22} $$

As charge number of $\mathrm{Al}$ is 13 , each atom of $\mathrm{Al}$ contains 13 protons and 13 electrons.

$\therefore \quad$ Magnitude of positive and negative charges in one paisa coin $=\mathrm{N}$ ze

$$ \begin{aligned} & =1.6742 \times 10^{22} \times 13 \times 1.60 \times 10^{-19} \mathrm{C} \ & =3.48 \times 10^{4} \mathrm{C}=34.8 \mathrm{kC} \end{aligned} $$

This is a very large amount of charge. Thus, we can conclude that ordinary neutral matter contains enormous amount of $\pm$ charges.

Thinking Process

Force on a point charge $=\frac{|q|^{2}}{4 \pi \varepsilon_{0} r^{2}}$. Here, $q=$ magnitude of one charge, $r=$ distance between two charges.

Answer

Here,

$$ \begin{aligned} q= \pm 34.8 \mathrm{RC}= \pm 3.48 \times 10^{4} \mathrm{C} \newline\ r_{1}=1 \mathrm{~cm}=10^{-2} \mathrm{~m}, r_{2}=100 \mathrm{~m}, r_{3}=10^{6} \mathrm{~m} \newline \quad \quad \quad \quad \quad \quad \text { and } \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \newline \ F_{1}=\frac{|q|^{2}}{4 \pi \varepsilon_{0} r_{1}^{2}}=\frac{9 \times 10^{9}\left(3.48 \times 10^{4}\right)^{2}}{\left(10^{-2}\right)^{2}}=1.09 \times 10^{23} \mathrm{~N} \newline \ F_{2}=\frac{|q|^{2}}{4 \pi \varepsilon_{0} r_{2}^{2}}=\frac{9 \times 10^{9}\left(3.48 \times 10^{4}\right)^{2}}{(100)^{2}}=1.09 \times 10^{15} \mathrm{~N} \newline \ F_{3}=\frac{|q|^{2}}{4 \pi \varepsilon_{0} r_{3}^{2}}=\frac{9 \times 10^{9}\left(3.48 \times 10^{4}\right)^{2}}{\left(10^{6}\right)^{2}}=1.09 \times 10^{7} \mathrm{~N} \end{aligned} \newline $$

Conclusion from this result We observe that when $\pm$ charges in ordinary neutral matter are separated as point charges, they exert an enormous force. Hence, it is very difficult to disturb electrical neutrality of matter.

Thinking Process

(i) Net force on a charge due to two equal and opposite charges will be zero. Also electric field on a charge is given by $E=\frac{F}{q}$ where $E=$ electric field, $F=$ force on charge q due to electric field, $q=$ magnitude of charge $q$

(ii) If a Cs atom is removed from the corner A then a singly charged negative Cs ion at A will appear.

Answer

(i) From the given figure, we can analyse that the chlorine atom is at the centre of the cube i.e., at equal distance from all the eight corners of cube where cesium atoms are placed. Thus, due to symmetry the forces due to all Cs tons, on $\mathrm{Cl}$ atom will cancel out.

Hence,

$$ E=\frac{F}{q} \text { where } F=0 $$

$$ \therefore \quad E=0 $$

(ii) Thus, net force on $\mathrm{Cl}$ atom at $A$ would be,

$$ F=\frac{e^{2}}{4 \pi \varepsilon_{0} r^{2}} $$

where, $\quad r=$ distance between $\mathrm{Cl}$ ion and $\mathrm{Cs}$ ion.

Applying Pythagorous theorem, we get

Now,

$$ \begin{aligned} r & =\sqrt{(0.20)^{2}+(0.20)^{2}+(0.20)^{2}} \times 10^{-9} \mathrm{~m} \newline \ & =0.346 \times 10^{-9} \mathrm{~m} \newline \ F & =\frac{q^{2}}{4 \pi \varepsilon_{0} r^{2}}=\frac{e^{2}}{4 \pi \varepsilon_{0} r_{2}} \newline \ & =\frac{9 \times 10^{9}\left(1.6 \times 10^{-19}\right)^{2}}{\left(0.346 \times 10^{-9}\right)^{2}}=1.92 \times 10^{-9} \mathrm{~N} \newline \end{aligned} $$

Thinking Process

The force on any charge will be zero only if all forces are balanced i.e., force of attraction is balanced by force of repulsion.

Answer

Here, let us keep the charge $2 q$ at a distance $r$ from $A$.

Thus, charge $2 q$ will not experience any force.

When, force of repulsion on it due to $q$ is balanced by force of attraction on it due to $-3 q$, at $B$, where $A B=d$.

Thus, force of attraction by $-3 q=$ Force of repulsion by $q$

$$ \begin{array}{rlrl} \Rightarrow & & \frac{2 q \times q}{4 \pi \varepsilon_{0} x^{2}} & =\frac{2 q \times 3 q}{4 \pi \varepsilon_{0}(x+d)^{2}} \ \Rightarrow & & (x+d)^{2} & =3 x^{2} \ \Rightarrow & x^{2}+d^{2}+2 x d & =3 x^{2} \ \Rightarrow & & =2 x^{2}-d^{2} \end{array} $$

$$ \begin{aligned} \therefore \quad 2 x^{2}-2 d x-d^{2} & =0 \ x & =\frac{d}{2} \pm \frac{\sqrt{3} d}{2} \end{aligned} $$

(Negative sign be between $q$ and $-3 q$ and hence is unadaptable.)

$$ \begin{aligned} x & =-\frac{d}{2}+\frac{\sqrt{3} d}{2} \ & =\frac{d}{2}(1+\sqrt{3}) \text { to the left of } q . \end{aligned} $$

Thinking Process

(i) Electric lines of forces always starts from a positive charge and ends at a negative charge. In case of a single charge, electric lines of force start from positive charge ends at infinity.

(ii) The magnitude of a charge depends on the number of lines of force enamating from a charge i.e., higher the number of lines of forces, higher the magnitude of charge and vice-versa.

Answer

(i) Here, in the figure, the electric lines of force emanate from $A$ and $C$. Therefore, charges $A$ and $C$ must be positive.

(ii) The number of electric lines of forces enamating is maximum for charge $C$ here, so $C$ must have the largest magnitude.

(iii) Point between two like charges where electrostatic force is zero is called netural point. So, the neutral point lies between $A$ and $C$ only.

Now the position of neutral point depends on the strength of the forces of charges. Here, more number of electric lines of forces shows higher strength of charge $C$ than $A$. So, neutral point lies near $A$.

Thinking Process

Due to symmetry forces by all the charges are cancelled out.

Answer

(a) (i) The point $O$ is equidistant from all the charges at the end point of pentagon. Thus, due to symmetry, the forces due to all the charges are cancelled out. As a result electric field at $O$ is zero.

(ii) When charge $q$ is removed a negative charge will develop at $A$ giving electric field $E=\frac{q \times 1}{4 \pi \varepsilon_{0} r^{2}}$ along $O A$.

(iii) If charge $q$ at $A$ is replaced by $-q$, then two negative charges $-2 q$ will develop there. Thus, the value of electric field $E=\frac{2 q}{4 \pi \varepsilon_{0} r^{2}}$ along $O A$.

(b) When pentagon is replaced by $n$ sided regular polygon with charge $q$ at each of its corners, the electric field at $O$ would continue to be zero as symmetricity of the charges is due to the regularity of the polygon. It doesn’t depend on the number of sides or the number of charges.

Thinking Process

Expansion of the universe will start if the coulomb repulsion on a hydrogen atom, at $R$, is larger than the gravitational attraction.

Answer

(a) Let us suppose that universe is a perfect sphere of radius $R$ and its constituent hydrogen atoms are distributed uniformly in the sphere.

As hydrogen atom contains one proton and one electron, charge on each hydrogen atom.

$$ e_{H}=e_{P}+e=-(1+Y) e+e=-Y e=(Y e) $$

If $E$ is electric field intensity at distance $R$, on the surface of the sphere, then according to Gauss’ theorem,

$$ \begin{aligned} \text { g. E.ds } & =\frac{q}{\varepsilon_{0}} \text { i.e., } E\left(4 \pi R^{2}\right)=\frac{4}{3} \frac{\pi R^{3} N|Y e|}{\varepsilon_{0}} \newline\ E & =\frac{1}{3} \frac{N|Y e| R}{\varepsilon_{0}} \newline \end{aligned} $$

Now, suppose, mass of each hydrogen atom $\simeq m_{P}=$ Mass of a proton, $G_{R}=$ gravitational field at distance $R$ on the sphere.

Then

$$ -4 \pi R^{2} G_{R}=4 \pi G m_{P} \quad \frac{4}{3} \pi R^{3} N $$

$$ \Rightarrow \quad G_{R}=\frac{-4}{3} \pi G m_{P} N R $$

$\therefore$ Gravitational force on this atom is $F_{G}=m_{P} \times G_{R}=\frac{-4 \pi}{3} G m_{P}^{2} N R$

Coulomb force on hydrogen atom at $R$ is $F_{C}=(Y e) E=\frac{1}{3} \frac{N Y^{2} e^{2} R}{\varepsilon_{0}}$

Now, to start expansion $F_{C}>F_{G}$ and critical value of $Y$ to start expansion would be when

$$ \begin{aligned} F_{C} & =F_{G} \newline \ \Rightarrow \quad \frac{1}{3} \frac{N Y^{2} e^{2} R}{\varepsilon_{0}} & =\frac{4 \pi}{3} G m_{P}^{2} N R \newline \ \Rightarrow \quad Y^{2} & =\left(4 \pi \varepsilon_{0}\right) G \frac{m_{P}{ }^{2}}{e} \newline \ & =\frac{1}{9 \times 10^{9}} \times\left(6.67 \times 10^{-11}\right) \frac{\left(1.66 \times 10^{-27}\right)^{2}}{\left(1.6 \times 10^{-19}\right)^{2}}=79.8 \times 10^{-38} \newline \ \Rightarrow \quad Y & =\sqrt{79.8 \times 10^{-38}}=8.9 \times 10^{-19} \simeq 10^{-18} \end{aligned} $$

Thus, $10^{-18}$ is the required critical value of $Y$ corresponding to which expansion of universe would start. (b) Net force experience by the hydrogen atom is given by

$$ F=F_{C}-F_{G}=\frac{1}{3} \frac{N Y^{2} e^{2} R}{\varepsilon_{0}}-\frac{4 \pi}{3} G m_{P}^{2} N R $$

If acceleration of hydrogen atom is represent by $d^{2} R / d t^{2}$, then

$$ \begin{aligned} m_{p} \frac{d^{2} R}{d t^{2}} & =F=\frac{1}{3} \frac{N Y^{2} e^{2} R}{\varepsilon_{0}}-\frac{4 \pi}{3} G m_{p}^{2} N R \newline \ & =\frac{1}{3} \frac{N Y^{2} e^{2}}{\varepsilon_{0}}-\frac{4 \pi}{3} G m_{p}^{2} N R \newline \ \therefore \quad \frac{d^{2} R}{d t^{2}} & =\frac{1}{m_{p}} \frac{1}{3} \frac{N Y^{2} e^{2}}{\varepsilon_{0}}-\frac{4 \pi}{3} G m_{p}^{2} N R=\alpha^{2} R \end{aligned} $$

where, $\quad \alpha^{2}=\frac{1}{m_{p}} \frac{1}{3} \frac{N Y^{2} e^{2}}{\varepsilon_{0}}-\frac{4 \pi}{3} G m_{p}^{2} N$

The general solution of Eq (iv) is given by $R=A e^{\alpha t}+B e^{-\alpha t}$. We are looking for expansion, here, so $B=0$ and $R=A e^{\alpha t}$.

$\Rightarrow \quad$ Velocity of expansion, $v=\frac{d R}{d t}=A e^{\alpha t}(\alpha)=\alpha A e^{\alpha t}=\alpha R$

Hence, $v \propto R$ i.e., velocity of expansion is proportional to the distance from the centre.

Thinking Process

According to the given charge density distribution of the sphere of radius $R$ i.e., $p(r)=K r$ for $r \leq R=0$ for $r>R$ it is obvious that the electric field is radial.

Answer

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

Now, for point $r<R$, electric field intensity will be given by,

$$\phi E.ds = \frac{1}{e_0} \int pdV $$

$ [for dV , V = \frac{4}{3} \Pi r^3 \Rightarrow dV = 3 $ $\times \frac{4}{3} \Pi r^3dr = 4 \pi r^2] \newline $ $$\phi E.ds = \frac{1}{e_0} 4 \pi K \int _0^r r^3dr $$ $$E = \frac{1}{4E_0}Kr^2$$

$(\because p(r)=K r)$

Here, charge density is positive.

So, direction of $\mathrm{E}$ is radially outwards.

For points $r>R$, electric field intensity will be given by

$$ \begin{array}{rlrl} \text { gE.dS } =\frac{1}{\varepsilon_{0}} \int \rho . d V \newline \ \Rightarrow E\left(4 \pi^{2}\right) =\frac{4 \pi K}{\varepsilon_{0}} \int_{0}^{R} r^{3} d r=\frac{4 \pi K}{\varepsilon_{0}} \frac{R^{4}}{4} \newline \ \Rightarrow E =\frac{K}{4 \varepsilon_{0}} \frac{R^{4}}{r^{2}} \newline \end{array} $$

Charge density is again positive. So, the direction of $\mathrm{E}$ is radially outward.

(b) The two protons must be on the opposite sides of the centre along a diameter following the rule of symmetry. This can be shown by the figure given below. Charge on the sphere,

$$ \begin{aligned} q & =\int_{0}^{R} \rho d V=\int_{0}^{R}(K r) 4 \pi r^{2} d r \newline \ q & =4 \pi K \frac{R^{4}}{4}=2 e \newline \ \therefore \quad K & =\frac{2 e}{\pi R^{4}} \end{aligned} $$

If protons 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

$$ F_{1}=e E=\frac{-e K r^{2}}{4 \varepsilon_{0}} $$

Repulsive force on proton 1 due to proton 2 is

Net force on proton 1 ,

$$ F_{2}=\frac{e^{2}}{4 \pi \varepsilon_{0}(2 r)^{2}} $$

So,

$$ \begin{aligned} & F=F_{1}+F_{2} \newline \ & F=\frac{-e K r^{2}}{4 \varepsilon_{0}}+\frac{e^{2}}{16 \pi \varepsilon_{0} r^{2}} \newline \ & F=\frac{-r^{2}}{4 \varepsilon_{0}} \frac{Z e}{\pi R^{4}}+\frac{e^{2}}{16 \pi \varepsilon_{0} r^{4}} \end{aligned} $$

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

$$ \begin{array}{rlrl} & & \frac{e r^{2} 2 e}{4 \varepsilon_{0} \pi R^{4}} & =\frac{e^{2}}{16 \pi \varepsilon_{0} r} \ \Rightarrow & r^{4} & =\frac{R^{4}}{8} \ \Rightarrow & r & =\frac{R}{(8)^{1 / 4}} \end{array} $$

This is the distance of each of the two protons from the centre of the sphere.

Answer

(a) Net electric field at plate $\gamma$ before collision is equal to the sum of electric field at plate $\gamma$ due to plate $\alpha$ and $\beta$.

The electric field at plate $\gamma$ due to plate $\alpha$ is $E_{1}=\frac{-Q}{S\left(2 \varepsilon_{0}\right)}$, to the left.

The electric field at plate $\gamma$ due to plate $\beta$ is $E_{2}=\frac{q}{S\left(2 \varepsilon_{0}\right)}$, to the right.

Hence, the net electric field at plate $\gamma$ before collision.

$$ E=E_{1}+E_{2}=\frac{q-Q}{S\left(2 \varepsilon_{0}\right)} \text {, to the left, if } Q>q $$

(b) During collision, plates $\beta$ and $\gamma$ are together. Their potentials become same.

Suppose charge on plate $\beta$ is $q_{1}$ and charge on plate $\gamma$ is $q_{2}$. At any point $O$, in between the two plates, the electric field must be zero.

Electric field at $O$ due to plate $\alpha=\frac{-Q}{S\left(2 \varepsilon_{0}\right)}$, to the left

Electric field at $O$ due to plate $\beta=\frac{q_{1}}{S\left(2 \varepsilon_{0}\right)}$, to the right

Electric field at $O$ due to plate $\gamma=\frac{q_{2}}{S\left(2 \varepsilon_{0}\right)}$, to the left

As the electric field at $O$ is zero, therefore

$$ \begin{aligned} \frac{Q+q_{2}}{S\left(2 \varepsilon_{0}\right)} & =\frac{q_{1}}{S\left(2 \varepsilon_{0}\right)} \ Q+q_{2} & =q_{1} \ Q & =q_{1}-q_{2} \end{aligned} $$

As there is no loss of charge on collision,

$$ Q+q=q_{1}+q_{2} $$

On solving Eqs. (i) and (ii), we get

$$ \begin{aligned} & q_{1}=(Q+q / 2)=\text { charge on plate } \beta \ & q_{2}=(q / 2)=\text { charge on plate } \gamma \end{aligned} $$

(c) After collision, at a distance $d$ from plate $\beta$,

Let the velocity of plate $\gamma$ be $v$. After the collision, electric field at plate $\gamma$ is

$$ E_{2}=\frac{-Q}{2 \varepsilon_{0} S}+\frac{(Q+q / 2)}{2 \varepsilon_{0} S}=\frac{q / 2}{2 \varepsilon_{0} S} \text { to the right. } $$

Just before collision, electric field at plate $\gamma$ is $E_{1}=\frac{Q-q}{2 \varepsilon_{0} S}$.

If $F_{1}$ is force on plate $\gamma$ before collision, then $F_{1}=E_{1} Q=\frac{(Q-q) Q}{2 \varepsilon_{0} S}$

Total work done by the electric field is round trip movement of plate $\gamma$

$$ \begin{aligned} W & =\left(F_{1}+F_{2}\right) d \ & =\frac{\left[(Q-q) Q+(q / 2)^{2}\right] d}{2 \varepsilon_{0} S}=\frac{(Q-q / 2)^{2} d}{2 \varepsilon_{0} S} \end{aligned} $$

If $m$ is mass of plate $\gamma$, the KE gained by plate $\gamma=\frac{1}{2} m v^{2}$

According to work-energy principle, $\frac{1}{2} m v^{2}=W=\frac{(Q-q / 2)^{2} d}{2 \varepsilon_{0} S}$

$$ Y=(Q-q / 2){\frac{d}{m \varepsilon_{0} S}}^{1 / 2} $$

Answer

(i) From the relation, $F=\frac{Q q}{r^{2}}=1$ dyne $=\frac{[1 \text { esu of charge }]^{2}}{[1 \mathrm{~cm}]^{2}}$

So, 1 esu of charge $=(1 \text { dyne })^{1 / 2} \times 1 \mathrm{~cm}=F^{1 / 2} \cdot L=\left[\mathrm{MLT}^{-2}\right]^{1 / 2} \mathrm{~L}$

$\Rightarrow 1$ esu of charge $=\mathrm{M}^{1 / 2} \mathrm{~L}^{3 / 2} \mathrm{~T}^{-1}$.

Thus, esu of charge is represented in terms of fractional powers $\frac{1}{2}$ of $M$ and $\frac{3}{2}$ of $L$.

(ii) Let 1 esu of charge $=x C$, where $x$ is a dimensionless number. Coulomb force on two charges, each of magnitude 1 esu separated by $1 \mathrm{~cm}$ is dyne $=10^{-5} \mathrm{~N}$. This situation is equivalent to two charges of magnitude $x C$ separated by $10^{-2} \mathrm{~m}$.

$$ \begin{array}{ll} \therefore F=\frac{1}{4 \pi \varepsilon_{0}} \frac{x^{2}}{\left(10^{-2}\right)^{2}}=1 \text { dyne }=10^{-5} \mathrm{~N} \newline \newline \ \therefore \frac{1}{4 \pi \varepsilon_{0}}=\frac{10^{-9}}{x^{2}} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}} \newline \newline \ \text { Taking, } x=\frac{1}{|3| \times 10^{9}}, \newline \newline \ \text { we get, } \frac{1}{4 \pi \varepsilon_{0}}=10^{-9} \times|3|^{2} \times 10^{18} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}=9 \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}} \newline \newline \ \text { If }|3| \rightarrow 2.99792458 \text {, we get } \frac{1}{4 \pi \varepsilon_{0}}=8.98755 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2} . \end{array} $$

Thinking Process

For motion of charge a to be simple harmonic, force on charge q must be proportional to its distance from the centre $\mathrm{O}$ and is directed towards $\mathrm{O}$.

Answer

Let us elaborate the figure first.

Given, two charge $-q$ at $A$ and $B$

$$ A B=A O+O B=2 d $$

$x=$ small distance perpendicular to $O$.

i.e., $x<d$ mass of charge $q$ is. So, force of attraction at $P$ towards $A$ and $B$ are each $F=\frac{q(q)}{4 \pi \varepsilon_{0} r^{2}}$, where $A P=B P=r$

Horizontal components of these forces $F_{n}$ are cancel out. Vertical components along $P O$ add.

If $\angle A P O=O$, the net force on $q$ along $P O$ is $F^{\prime}=2 F \cos Q$

$$ \begin{aligned} & =\frac{2 q^{2}}{4 \pi \varepsilon_{0} r^{2}} \frac{x}{r} \newline \ & =\frac{2 q^{2} x}{4 \pi \varepsilon_{0}\left(d^{2}+x^{2}\right)^{3 / 2}} \end{aligned} $$

When,

$$ x«d, F^{\prime}=\frac{2 q^{2} x}{4 \pi \varepsilon_{0} d^{3}}=K x $$

where,

$$ K=\frac{2 q^{2}}{4 \pi \varepsilon_{0} d^{3}} $$

$\Rightarrow$

$$ F \propto x $$

i.e., force on charge $q$ is proportional to its displacement from the centre $\mathrm{O}$ and it is directed towards $\mathrm{O}$.

Hence, motion of charge $q$ would be simple harmonic, where

and

$$ \begin{aligned} \omega & =\sqrt{\frac{K}{m}} \newline \ T & =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{K}} \newline \ & =2 \pi \sqrt{\frac{m \cdot 4 \pi \varepsilon_{0} d^{3}}{2 q^{2}}}=\frac{8 \pi^{3} \varepsilon_{0} m d^{3}}{q^{2}} \end{aligned} $$

Answer

Let us draw the figure according to question,

A gentle push on $q$ along the axis of the ring gives rise to the situation shown in the figure below.

Taking line elements of charge at $A$ and $B$, having unit length, then charge on each elements.

$$ d F=2-\frac{Q}{2 \pi R} q \times \frac{1}{4 \pi \varepsilon_{0}} \frac{1}{r^{2}} \cos \theta $$

Total force on the charge $q$, due to entire ring

$$ \begin{aligned} F & =-\frac{Q q}{\pi R}(\pi R) \cdot \frac{1}{4 \pi \varepsilon_{0}} \frac{1}{r^{2}} \cdot \frac{2}{r} \ F & =-\frac{Q q z}{4 \pi \varepsilon_{0}\left(Z^{2}+R^{2}\right)^{3 / 2}} \end{aligned} $$

Here, $Z<R, \quad F=-\frac{Q q z}{4 \pi \varepsilon_{0} R^{3}}=-K z$

where

$$ \frac{Q q}{4 \pi \varepsilon_{0} R^{3}}=\text { constant } $$

$F \propto-Z$

$\Rightarrow$

Clearly, force on $q$ is proportional to negative of its displacement. Therefore, motion of $q$ is simple harmonic.

$\Rightarrow \quad T=2 \pi \sqrt{\frac{4 \pi \varepsilon_{0} m R^{3}}{Q q}}$

$$ \begin{aligned} \omega=\sqrt{\frac{K}{m}} \text { and } T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{K}} \newline \ T=2 \pi \sqrt{\frac{m 4 \pi \varepsilon_{0} R^{3}}{Q q}} \newline \ T=2 \pi \sqrt{\frac{4 \pi \varepsilon_{0} m R^{3}}{Q q}} \end{aligned} $$