Inside
Outside                                                  

Electric field inside the hollow spherical charged conductor is zero. So, no work is done in moving a charge inside the shell. Thus, potential is constant and therefore, equal to its value at the surface, i.e. \(V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R}\)

P.E. of a pair of charges = work done to assemble the charges = \(\frac{1}{4\pi\epsilon_0r}.\frac{q_1q_2}{r}\)
qA = 6 x 10^-6 C = qB, qC = -6 x 10^-6 C and r = 0.1 m
then work done on the system
= W_AB+ W_BC + W_AC
= \(\frac{1}{4\pi\epsilon_0r}\)[ q_Aq_B + q_Bq_C + q_Aq_C ]
= 9 x 10^-9 x \(\frac{1}{0.1}\)[36 - 36 - 36] x 10^-12
= 9 x 10^-9 x (-36) x 10^-11
= -3.24 J

Consider the figure, |P_A| = P_C = P

The magnitude of resultant P_R ,
\({ P }_{ R }=\sqrt { { { p }_{ 1 } }^{ 2 }+{ { p }_{ 2 } }^{ 2 }+2{ p }_{ 1 }{ p }_{ 2 }\cos { \theta } } \)
\(=\sqrt { { p }_{ 2 }+{ p }_{ 2 }+2{ p }^{ 2 }\cos { \theta } } \)
\(=\sqrt { { 2p }^{ 2 }(1+\cos { \theta } } \)
\(=\sqrt { 2{ p }^{ 2 }\times 2{ cos }^{ 2 }\frac { \theta }{ 2 } } =2p \ cos\frac { \theta }{ 2 } \)
\(\tan\alpha =\frac { { p }_{ 2 }sin\theta }{ { p }_{ 1 }+{ p }_{ 2 }cos\theta } =\frac { psin{ 120 }^{ 0 } }{ p+pcos{ 120 }^{ 0 } } \)
\(=\frac { p\sqrt { 3/2 } }{ p } =\sqrt { 3 }\)
\(p-\frac { p }{ 2 } \)
\(|{ P }_{ R }|=2pcos\frac { \theta }{ 2 } =2pcos\frac { { 120 }^{ 0 } }{ 2 } =2p\times \frac { 1 }{ 2 } =p\)
\({ P }_{ R } \ will \ subtend \ an \ angle \ of \ { 30 }^{ 0 } \ with \ X \ axis.\)
Now torque acting on the system.
\(\tau ={ P }_{ R }\times E={ p }_{ R }Esin\theta =\frac { 1 }{ 2 } { p }E\)
The torque will work to align the dipole in the direction of electric field E.

(i)   \(\because \)  Electric field intensity and potential difference are related as,
\(E=-\frac { \Delta V }{ \Delta r } \ \Rightarrow \ \Delta V \ = \ -E\Delta r\)
\(\\ { V }_{ A } \ - \ { V }_{ C } \ =-4E \ or \ { V }_{ C } \ - \ { V }_{ A } \ = \ 4E\)
[\(\because \) By Pythagoras law, AC² = AB²+ BC²]
\(\Rightarrow \) 5³-3²  =  AB²  \(\Rightarrow \) AB = \(\Delta \)r = 4
(ii) As, V_C- V_A = 4E, is positive.
\(\therefore \)  V_C > VA
Potential is greater at point C than at point A, as potential decreases along the direction of electric field.

As, the positively charged particle experiences electrostatic force in the direction of electric field. Some work is done by the electric field on the positive charge, hence electrostatic potential energy decreases.

For the protons, the charge on a proton is + e.
If the number of up quarks is a, then the number of down quarks are (3-a) as the total number of quarks are 3. So, \(a\times \)up quark charge + (3-a) down quark charge) = +e
\(a\times \left( \frac { 2 }{ 3 } e \right) +\left( 3-a \right) \left( -\frac { e }{ 3 } \right) =e\)
\(\Rightarrow \frac { 2ae }{ 3 } -\frac { \left( 3-a \right) e }{ 3 } =e\)
\(\Rightarrow 2a-3+a=3\\ \Rightarrow 3a=6\\ \Rightarrow a=2\)
Thus, in the proton, there are two up quarks and one down quark.
\(\therefore \)  Possible quark composition for proton = uud
For the neutron, the charge on neutron is 0.
Let the number of up quarks is b and the number of down quarks be 3-b.
so, \(b\times \) up quark charge + (3-b) down quark charge = 0
\(b\times \left( \frac { 2e }{ 3 } \right) +\left( 3-b \right) \left( -\frac { e }{ 3 } \right) =0\)
\(\Rightarrow 2b-3+b=0\)
\(\Rightarrow 3b=3\\ \Rightarrow b=1\)
Thus, in neutron, there is one up quark and two down quarks.
\(\therefore \)   Possible quark composition for neutron = udd.

As, P(1, 0, 4) and (2,-1, 5)
\(\therefore \ 2l=PQ=[(2-1)\hat { i } +(-1-0)\hat { j } +(5-4)\hat { k] }\\ =(\hat { i } -\hat { j } +\hat { k } )\)
and \(\ q=\pm 9\times { 10 }^{ -6 }C,E=0.20\hat { i } V/cm.\tau =?\)
Since, \(\tau =\rho \times E=q(2l)\times E\)
\(\therefore \ \tau =9\times { 10 }^{ -6 }(\hat { i } -\hat { j } +\hat { k } )\times 0.20=18\times { 10 }^{ -7 }(\hat { k } -\hat { j) } \)
Magnitude of torque,
\(\tau =18\times { 10 }^{ -7 }[\sqrt { { (1) }^{ 2 }+{ (1) }^{ 2 } } ]=25.45\times { 10 }^{ -7 }N-m\)

No, it is not correct to write the unit of electric dipole moment as mC. The symbol mC represents milli-coulomb, i.e. unit of electric charges. IN SI system unit symbols are written in alphabetical order.
∴ Unit of dipole moment is C-m.

Let the two capacitors be C₁ and C_2, capacitance will be maximum when connected in parallel.
i.e.   C₁ + C₂ = 25
Capacitance will be minimum when connected in series.
i.e.  \(\frac { { C }_{ 1 }{ C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } =4\)
Since, we are left with only two values 5 \(\mu \) F and 20\(\mu \) F.
So, the value of capacitances will be  5 \(\mu \) F and 20\(\mu \) F.

Electric field lines do not pass through a conductor.Hence,  the interior of the conductor is free from the influence of the electric field

Total charge q = CV and total energy \(U={1\over 2}CV^2\)
Now, V is constant and \(C_p>C_s\)
therefore, parallel combination is required or storing greater charge and greater energy.

(i) Capacitors are used in radio circuits for tuning purposes.
(ii) Capacitors are used in power supplies for smoothing the rectified current.

If Changed particle was Initially at rest, it will move along the direction of electric field. If initial velocity of the charged particle makes a certain angle with electric field, then the charged particle will no move along the line of force.

Electric field intensity at the centre of the square will be zero.

In a satellite, there is a condition of weightlessness. Therefore, mg = 0. On account of electrostatic force of repulsion between the balls, the strings would become horizontal. Therefore, angle between the strings = 180^o.
Also, tension in each string = force of repulsion 
\(T={1\over 4\pi\epsilon_o}{q^2\over(2l)^2}N\)

Conservation of electric charge means that the total charge on an isolated system remains unchanged with time.

Potential difference, V = 15 x 10⁶ V
Dielectric strength of the surrounding gas = 5 x 10⁷ V/m
Electric field intensity, E = Dielectric strength = 5 x 10⁷ V/m
Minimum radius of the spherical shell required for the purpose is given by,
\(r=\frac{V}{E}\)
\(=\frac{15 \times 10^{6}}{5 \times 10^{7}}=0.3 \mathrm{~m}=30 \mathrm{~cm}\)
Hence, the minimum radius of the spherical shell required is 30 cm.

\(1.88\times {{10}^{5}}\)Nm²C^-1

\(60\mu F\)

(i) The magnitude,
\(\left| { E }_{ AB } \right| =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \times \frac { q }{ { a }^{ 2 } } =E\)
\(\left| { E }_{ AC } \right| =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \times \frac { 2q }{ { a }^{ 2 } } =2E\)

\({ E }_{ net }=\sqrt { { E }_{ AB }^{ 2 }+{ E }_{ AC }^{ 2 }+{ 2E }_{ AB }{ E }_{ AC }cos\theta } \)
\(=\sqrt { { (2E) }^{ 2 }+{ E }^{ 2 }+2\times 2E\times E\times \left( -\frac { 1 }{ 2 } \right) } \)
\(=\sqrt { { 4E }^{ 2 }+E^{ 2 }-{ 2E }^{ 2 } } =E\sqrt { 3 } ....(i)\)
\(We\quad know\quad that,\quad E=\frac { q }{ 4\pi { \varepsilon }_{ 0 }{ a }^{ 2 } } \)
\(So,{ E }_{ net }=\frac { q\sqrt { 3 } }{ 4\pi { \varepsilon }_{ 0 }{ a }^{ 2 } } \)
(ii) Direction of resultant electric field at vertex.
\(tan\alpha =\frac { { E }_{ AB }{ sin120 }^{ 0 } }{ { E }_{ AC }+{ E }_{ AB }{ cos120 }^{ 0 } } \)
\(=\frac { E\times \frac { \sqrt { 3 } }{ 2 } }{ 2E+E\times \left( -\frac { 1 }{ 2 } \right) } \)
\(tan\alpha =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \alpha ={ tan }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) \)
α = 30⁰

(a) Charge placed at the centre of a shell is +q. Hence, a charge of magnitude - q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is - q.
Surface charge density at the inner surface of the shell is given by the relation,
\(\sigma_{1}=\frac{\text { Total charge }}{\text { Inner surface area }}=\frac{-q}{4 \pi r_{1}^{2}}\)       .........(i)
A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell
\(\sigma_{2}=\frac{\text { Total charge }}{\text { Outer surface area }}=\frac{Q+q}{4 \pi r_{2}^{2}}\)        ..........(ii)
(b) Yes
The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.

(i) Charge of proton q = 1.6 x 10^-19C
Force on proton at A is F = qE_A
= (1.6 x 10^-19C) (40 N / C)
= 6.4 x 10^-18N
(ii) Since, electric field,
 E\(\propto \) Number of electric field lines/Area
 Hence, E_B = 1/2 E_A
 = 1/2(40 N / c)
 = 20 N / C.

(a)Electrostatic field lines cannot start from a negative charge.
(b)Electrostatic field lines cannot end at positive charge
(c)Electrostatic field lines cannot form closed loops.

Here, q = Ze = \(78\times1.6\times10^{-19}C\)
Potential gradient at a point is numerically equal to electric field at that point, i.e., 
\({dV\over dr}=E={q\over 4\pi\epsilon_or^2}\)
At \(r=10^{-12}m\)
\({dV\over dr}=E={9\times10^9\times78\times1.6\times 10^{-19}\over(10^{-12})^2}\)
\(=1.123\times 10^{17}Vm^{-1}\)
At r = \(5\times 10^{-15}m\)
\({dV\over dr}=E={q\over 4\pi\epsilon_or^2}={9\times 10^9\times78\times1.6\times10^{-19}\over(5\times10^{-15})^2}\)
\(=4.5\times 10^{21}Vm^{-1}\) 

Here q₁ = q₂ = q = ?
F = 3.7 x 10^-9N, r = 5\(\mathring { A } \) = 5 x 10^-10m
n = ?
As F=\({kq_1q_2\over r^2}={kq^2\over r^2}\)
\(q=\sqrt{r^2F/K}\)
\(=\sqrt{{1\over 9\times10^9}(5\times10^{-10})^2\times3.7\times10^{-9}}\)
\(={\sqrt{ 25\times3.7}\over 9}\)=3.2 x 10^-19C
From q = ne
\(n={q\over e}={3.2\times 10^{-19}\over 1.6\times 10^{-19}}=2\)

Since \({ F }_{ e }=-\frac { { ke }^{ 2 } }{ { r }^{ 2 } } \)
So     \({ ke }^{ 2 }={ -F }_{ e }{ r }^{ 2 }\)
And    \({ F }_{ G }=-G\frac { { m }_{ p }{ m }_{ e } }{ { r }^{ 2 } } \)
So \({ Gm }_{ p }{ m }_{ e }={ -F }_{ G }{ r }^{ 2 }\)
So dimensions of \(\frac { { ke }^{ 2 } }{ { Gm }_{ e }{ m }_{ p } } =\frac { { F }_{ e }{ r }^{ 2 } }{ { F }_{ G }{ r }^{ 2 } } =\frac { { F }_{ e } }{ { F }_{ G } } \)
 \(=\frac { { [MLT }^{ -2 }] }{ { [MLT }^{ -2 }] } =[{ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }]\)
 = No dimensions
The value of \(\frac { { ke }^{ 2 } }{ { Gm }_{ e }{ m }_{ p } } \)
  \(=\frac { 9\times { 10 }^{ 9 }\times (1.6\times { 10 }^{ -19 })^{ 2 } }{ 6.67\times { 10 }^{ -11 }(1.67\times { 10 }^{ -27 })(9.1\times { 10 }^{ -31 }) } \)
  \(=2.9\times { 10 }^{ 39 }\)
From Eq. (1) and (2), we have
  \(\left| \frac { { F }_{ e } }{ { F }_{ G } } \right| =\frac { { ke }^{ 2 } }{ { Gm }_{ p }{ m }_{ e } } =2.9\times { 10 }^{ 39 }\)
The ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

SI units of electric flux is N-m²C^-1.
Dimensional formula of electric flux is [ M¹L¹T^-3A^-1].
\(\phi={{\sum {q}_{inside}}\over{{\epsilon}_{0}}}={{-q+3q+5q}\over{{\epsilon}_{0}}}={{7q}\over{{\epsilon}_{0}}}\)
As \(\phi={{q}\over{{\epsilon}_{0}}},\) electric flux is not affected by area/shape of the surface. So, the electric flux remains unaffected.
This theorem emphasis that total normal flux from the surface depends only on algebraic sum of charges enclosed by the surface, irrespective of their location. The charges outside the surface do not affect the electric flux. In day-to-day life, the theorem implies that the knowledge you can import depends only on what you have stored within you. You
cannot emanate what you have not absorbed or what lies outside your reach?
In the examination, you can write what you have learnt by heart? Extraneous help from outside by unfair means will never serve your purpose.

(a) Consider an electric dipole of charges +q and - q separated by 2x distance being placed in free space. Let P be the point at which the electric field is to be determined due to the electric dipole.

The electric field \(\overrightarrow{E}\) at P is given by
\(\overrightarrow{E}\) = \(\overrightarrow{E_A}\)+ \(\overrightarrow{E_B}\)
\(\left| \vec { { E }_{ A } } \right| \)= \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (d+x) }^{ 2 } } \)
\(\left| \vec { { E }_{ B } } \right| \)=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (d-x) }^{ 2 } } \)
From diagram, it is clear that \(\overrightarrow{E_A}\)< \(\overrightarrow{E_B}\)and they are opposite in direction. Hence,
E = \(\left| \vec { E } \right| =\left| \vec { { E }_{ B } } -\vec { { E }_{ A } } \right| \)
\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \frac { q }{ { (d-x) }^{ 2 } } -\frac { q }{ { (d+x) }^{ 2 } } \right] \)
\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q\times 4dx }{ { { (d }^{ 2 }-{ x }^{ 2 }) }^{ 2 } } \)
\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \)\(\frac { 2Pd }{ { { (d }^{ 2 } }-{ x }^{ 2 })^{ 2 } } \)
When x <  \({ E }_{ axial }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2P }{ { d }^{ 3 } } alongPx\)
\(Potential(V)=\frac { { E }_{ axial } }{ d } .d\)
\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2P }{ { d }^{ 2 } } \)
(b) The capacitor can be considered as split into two capacitor connected in parallel.
Here \({ C }_{ 1 }=\frac { { K }_{ 1 }{ \varepsilon }_{ 0 }l }{ 2d } and{ C }_{ 2 }=\frac { { K }_{ 1 }{ \varepsilon }_{ 0 }l }{ 2d } \)
In parallel combination,
\({ C }_{ eq }={ C }_{ 1 }+{ C }_{ 2 }\)
\(\frac { { K }_{ 1 }{ \varepsilon }_{ 0 }l }{ 2d } +\frac { { K }_{ 2 }{ \varepsilon }_{ 0 }l }{ 2d } \)
\(\frac { { \varepsilon }_{ 0 }l }{ 2d } \left( { K }_{ 1 }+{ K }_{ 2 } \right) \)
From this figure
\({ C }_{ eq }=\frac { { K }_{ 2 }{ \varepsilon }_{ o }l }{ 2d } \)
From eqns. (i) & (ii) 
\(\frac { { K }_{ 1 }{ \varepsilon }_{ 0 }l }{ 2d } =\frac { { \varepsilon }_{ o }l }{ 2d } \left( { K }_{ 1 }+{ K }_{ 2 } \right) \)
\(K={ K }_{ 1 }+{ K }_{ 2 }\)

(a) (i) Presence of mind.
(ii) Ability to take prompt decisions.
(iii) Sense of responsibility
(b) Given, q₁ = 5\(\mu\)C = 5 x 10^-9C
q₂ = -2\(\mu\)C = -2 x 10^-9C.
The distance between the charges
= x = x₂ - x₁
= 23 cm - 5cm
= 18 cm
= 0.18 m
Thus, electrostatic potential energy of charges
 \(U = \frac{1}{4\pi\epsilon_0}.\frac{q_1q_2}{x}\)
= \(\frac{9\times10^9\times5\times10^{-9}\times-2\times10^{-9}}{0.18}\)
= \(\frac{-9\times10^{-8}}{18\times10^{-2}}\)
= - 5 x 10^-7 J

(i) Potential of three shells
At shell A
Potential,
V_A = \(\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \left[ \frac { { q }_{ a } }{ a } -\frac { { q }_{ b } }{ b } +\frac { { q }_{ c } }{ c } \right] \)
 \(=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \left[ \frac { { 4\pi a }^{ 2 }\sigma }{ a } -\frac { { 4\pi b }^{ 2 }\sigma }{ b } +\frac { { 4\pi c }^{ 2 }\sigma }{ c } \right] \left( \because \sigma =\frac { q }{ { 4\pi r }^{ 2 } } \right) \\ \)
\(=\frac { \sigma }{ { \varepsilon }_{ 0 } } \left( a-b+c \right) \)
At shell B
Potential, V_B = \(\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \left[ \frac { { q }_{ a } }{ a } -\frac { { q }_{ b } }{ b } +\frac { { q }_{ c } }{ c } \right] \)
= \(\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \left[ \frac { { 4\pi a }^{ 2 }\sigma }{ b } -\frac { { 4\pi b }^{ 2 }\sigma }{ b } +\frac { { 4\pi c }^{ 2 }\sigma }{ c } \right] \left[ \because \sigma =\frac { q }{ { 4\pi r }^{ 2 } } \right] \)
\( \\ =\frac { \sigma }{ { \varepsilon }_{ 0 } } \left[ \frac { { a }^{ 2 }-{ b }^{ 2 } }{ b } +c \right]\)
At shell C
Potential, V_c =  \(\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \left( \frac { { q }_{ a } }{ c } -\frac { { q }_{ b } }{ c } +\frac { { q }_{ c } }{ c } \right) \)
\(\\ =\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \left( \frac { { 4\pi a }^{ 2 }\sigma }{ c } -\frac { { 4\pi b }^{ 2 }\sigma }{ c } +\frac { { 4\pi c }^{ 2 }\sigma }{ c } \right) \left[ \because \sigma \frac { q }{ { 4\pi r }^{ 2 } } \right] t)\)
\(=\frac{\sigma}{\varepsilon_{0}}\left(\frac{a^{2}-b^{2}+c^{2}}{c}\right)\)
(ii) Relation between the radii Now,
Now,   V_A = V_C (given)
\(\frac{\sigma}{\varepsilon_{ }}(a-b+c)=\frac{\sigma}{\varepsilon} \frac{\left(a^{2}-b^{2}+c^{2}\right)}{c}\)
\(a-b+c=\frac{a^{2}-b^{2}+c^{2}}{c}=\frac{a^{2}-b^{2}}{c}+c\)
\(c(a-b)=a^{2}-b^{2}\)
\(\Rightarrow \ c=a+b\left[\because\left(a^{2}-b^{2}\right)=(a-b)(a+b)\right]\)

(i) Examples of paramagnetic material are aluminum; chromium; oxygen.
(ii) Iron and cobalt are two ferromagnetic materials. The ferromagnetic substances, but to a much larger degree. For example, relative magnetic permeability of paramagnetic substances is slightly greater than 1

Charge located at the origin, q = 8 mC = 8 x 10^-3 C
The magnitude of the charge taken from the point P to R and then to Q, q₁ = 2 x 10^-9 C
Here OP= d₁= 3 cm = 3 x 10^-2 m
OQ = d₂= 4 cm = 4 x 10^-2 m
Potential at the point P, V₁ \(=\frac{q}{4 \pi \epsilon_{0} d_{1}}\) 

Potential at the point Q, V₁ \(=\frac{q}{4 \pi \epsilon_{0} d_{1}}\)
The work done (W) is independent of the path
Therefore, W = q₁[V₁ – V₂]
\(=V_{2}=q_{1}\left[\frac{q}{4 \pi \epsilon_{0} d_{2}}-\frac{q}{4 \pi \epsilon_{0} d_{1}}\right]\)
\(=V_{2}=\frac{q q_{1}}{4 \pi \epsilon_{0}}\left[\frac{1}{d_{2}}-\frac{1}{d_{1}}\right]\)
Where, \(\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\)
Therefore,
\(W=9 \times 10^{9} \times 8 \times 10^{-3} \times\left(-2 \times 10^{-9}\right)\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{3 \times 10^{-2}}\right]\)
= -144 x 10^-3 x (-100/12)
= 1.2  Joule
Therefore, the work done during the process is 1.2 J