Potential at P due to charge at A is
\(V_{P A}=\frac{k q}{r+l}\)
Potential at P due to charge at B is
\(V_{P B}=-\frac{k q}{r-l}\)
Net potential at P is
\( V_{P} =V_{P A}+V_{P B}=k q\left[\frac{1}{r+l}-\frac{1}{r-l}\right] \)
\(=k q\left[\frac{r-l-r-l}{r^{2}-l^{2}}\right]=-\frac{k q(2 l)}{r^{2}-l^{2}} \)
\(V_{P} =-\frac{k p}{r^{2}-l^{2}} \ [\therefore p=q(2 l)]\)
In case \(r \gg l, V_{P}=-\frac{k p}{r^{2}}, \text { i.e. } V_{P} \propto \frac{1}{r^{2}}\) whereas due to a single charge potential at a point is \(V \propto \frac{1}{r}\)

(a) All three charges Q₁, Q₂ and Q₃ will contribute to the electric field.
(b) Only the enclosed charges, i.e. Q₁ and Q₂.
(c) No, the electric field will exist on the surface.

(a) Since, the number of field lines entering the hemisphere is equal to number of field lines leaving. Hence, the net electric flux through it is zero.
(b) As no charge is enclosed, therefore net electric flux is given by

\(\phi=\phi_{1}+\phi_{2}=0\)
where \(\phi_{1}\) = Electric flux through the curved surface area
\(\phi_{2}\) = Electric flux through the plane surface area
\(\therefore \ \phi_{1}=-\phi_{2}\)
\(\Rightarrow \phi_{1}=-E \cdot \pi R^{2} \cos 180^{\circ}=E \cdot \pi R^{2}\)

According to the question,
Capacitors of 20μFand C are connected in series.
(i) The equivalent capacitance
\((4\mu F)=\frac { (20\mu F)\times C }{ 20+C } \)
(20 + C) = 5C 4C = 20
∴ C = 5μF
(ii) Charge on capacitor (equivalent)
q = (4μF) x 12 = 48μ​​​​​​​C
same charge 48μ​​​​​​​C lies on both the capacitors
(iii) Potential drop across 20 μFcapacitor
\({ V }_{ 1 }=\frac { q }{ { C }_{ 1 } } =\frac { 48\mu C }{ 20\mu F } =2.4V\)
Potential drop across 5μFcapacitor
\({ V }_{ 2 }=\frac { q }{ { C }_{ 2 } } =\frac { 48\mu C }{ 5\mu F } =9.6V\)​​​​​

After disconnection from battery and doubling the separation between two plates
(i) Charge on capacitor remains same.
i.e, CV = C'V'
\(\Rightarrow CV=\left( \frac { C }{ 2 } \right) V'\Rightarrow V'=2V\)
∵ Electric field between the plates
\(E'=\frac { V' }{ d' } =\frac { 2V }{ 2d } \)
\(E'=\frac { V }{ d } =E\)
⇒ Electric field between the two plates remains same.
(ii) Capacitance reduces to half of original value as
C ∝ \(\frac { 1 }{ d } \Rightarrow C'=\frac { C }{ 2 } \)
(iii) Energy stored in the capacitor before disconnection from battery
\({ U }_{ 1 }=\frac { { q }^{ 2 } }{ 2C } \)
Now, energy stored in the capacitor after disconnection from battery
\({ U }_{ 2 }=\frac { { q }^{ 2 } }{ 2(C') } =\frac { { q }^{ 2 } }{ 2\times \left( \frac { C }{ 2 } \right) } =\frac { { q }^{ 2 } }{ C } \)
\(\Rightarrow { U }_{ 2 }=2\left( \frac { { q }^{ 2 } }{ 2C } \right) =2{ U }_{ 1 } \ { U }_{ 2 }=2{ U }_{ 1 }\)
Energy stored in capacitor gets doubled to its initial value.

Electric field due to a uniformly charged thin spherical shell :


(i) When point P lies outside the spherical shell :Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.
Let \(\overrightarrow{E}\) be the electric field at point P, then the electric flux through area element is \(\overrightarrow{ds}\) given by
\(d\phi = \overrightarrow{E}.\overrightarrow{\Delta S}\)
 Since \(\Delta\)S is also along normal to the surface,
\(d\phi = E ds\)
\(\therefore\) Total electric flux through the Gaussian surface is given by,
\(\phi = \oint_s Eds = E\oint_s ds\)
 Now  \(\oint_s ds = 4\pi r^2\)
\(\phi = E \times 4\pi r^2\)................(i)
Since the charge enclosed by the Gaussian surface is q₁ according to Gauss theorem,
\(\phi = \frac{q}{\epsilon_0}\)   .......................(ii)
From equations (i) and (ii),we obtain
\(E\times4\pi r^2 = \frac{q}{\epsilon_0}\)
\(E = \frac{1}{4\pi\epsilon_0}\frac{q}{r_2}\) ( for ( r > R ) )
(ii) When point P lies inside the spherical shell : In such a case, the Gaussian surface encloses no charge. According to Gauss law,
 \(E\times4\pi r^2 = 0\)
i.e., E = 0 (r < R)
Graph showing the variation of electric field as a function of r :

(i) Fot region II,
\(E_{II} = \frac{1}{2\pi\epsilon_0}(\sigma_1-\sigma_2)\)
towards tight side/from sheet A to sheet B
(ii) For region III,
\(E_{III} = \frac{1}{2\pi\epsilon_0}(\sigma_1+\sigma_2)\)
towards tight side/away from the two sheets

(i) Given, C₁ = 1\(\mu F\)  C₂ = 2\(\mu F\) C₃ = 3\(\mu F\)
The combined capacity (C_s) in series combination is given by
 \(\frac { 1 }{ { C }_{ s } } =\frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } =\frac { 1 }{ 1 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } =\frac { 11 }{ 6 } \)
\(\Rightarrow { C }_{ s }=\frac { 6 }{ 11 } \mu F\)
The combined capacity (C_p) in parallel combination is given by 
C_p = C₁ + C₂ +C₃ = 1 + 2 + 3 = 6\(\mu F\) 
\(\Rightarrow { C }_{ p }=11{ C }_{ s }\) 
(ii)  As, \({ \ C }_{ p }>{ C }_{ s }\) 
\(\therefore \) The charge supplied by 100 V battery
\({ q }_{ p }={ C }_{ p }V=6\mu F\times 100=6\times { 10 }^{ -6 }\times 100\)   
\(\\ { q }_{ p }=6\times { 10 }^{ -4 }C\)​​​​= 600 \(\mu\)C

Surface density near the hole = \(\sigma \) 
Unit vector = \(\hat { n } \) (normal directed outwards)

Let P be the point on the hole. The electric field at point P closed to the surface to conductor, according to Gauss' theorem.
\(\oint { E.dS=\frac { q }{ { \varepsilon }_{ 0 } } } \)
where, q is the charge near the hole.
\(EdS\cos { \theta } =\frac { \sigma dS }{ { \varepsilon }_{ 0 } } (\because \sigma =q|dS\therefore q=\sigma dS,where\quad dS=area)\)     
\(\therefore\)  Angle between electric field and area vector is 0⁰
\(\therefore \ E \ dS=\frac { \sigma \ dS }{ { \varepsilon }_{ 0 } } \)   
 \(\Rightarrow \ E=\frac { \sigma }{ { \varepsilon }_{ 0 } } \Rightarrow E=\frac { \sigma }{ { \varepsilon }_{ 0 } } \hat { n } \)  
This electric field due to the filled up hole and the field due to the rest of the charged conductor. The two fields inside the conductor are equal and opposite.
So, there is no electric field inside the conductor. Outside the conductor, the electric fields are equal in the same direction.
So, the electric field at point P due to each part
 \(=\frac { 1 }{ 2 } =E=\frac { \sigma }{ { 2\varepsilon }_{ 0 } } \hat { n } \)

Electric flux through the closed surface S is 
\( { \phi }_{ S }=\frac { \Sigma q }{ { \varepsilon }_{ 0 } } =\frac { +2q-q }{ { \varepsilon }_{ 0 } } =\frac { q }{ { \varepsilon }_{ 0 } }\)
\( \\ \Longrightarrow { \ \ \phi }_{ S }=\frac { q }{ { \varepsilon }_{ 0 } } \)
Charge +3q is outside the closed surface S, therefore, it would not be taken into consideration in applying Gauss' theorem

As capacitors are connected in series, charge on each capacitor must be same.
Charge on \(6\mu F\) capacitor = charge on 12\(\mu C\) capacitor
\(6\times 10^{-6}\times 2=12\times 10^{-6}\times V_2\)
\(V_2={6\times 2\over 12}=1 volt\)
Total battery voltage = V₁ + V₂ = 2 + 1
= 3V

\((i)\phi_E={q\over \epsilon_o}={8.85\times 10^{-8}\over 8.85\times 10^{-12}}=10^4Nm^2C^{-1}\)
(ii) flux will remain the same as charge enclosed by the surface is same as in case (i)

As r >> a, so E  = \(2\overrightarrow { P } \over 4\pi\epsilon_or^3\)
= \(9\times10^9\times 2\times 30\times 10^{-6}\times 10^{-2}\over (20\times10^{-2})^3\)
= \(6.25\times 10^5 NC\) 

Length of each side a = 0.1 m = 10 cm
distance of the centre of the cube from each corner,
\(r={a\sqrt3\over2}={10\sqrt3\over 2}=5\sqrt3\)
= \(5\sqrt3\times10^{-2}m\)
When identical point charges +q each are at the eight corners of the cube, net field intensity at the centre is zero (cancelling out in four pairs). When one of the corner charges is removed, three pairs of E values cancel. Due to charge on the seventh corner, 
\(E={1\over 4\pi\epsilon_o}{q\times 1\over r^2}=9\times 10^9{q\over (5\sqrt3\times10^{-2})^2}\)
= \((1.2\times 10^{12}q)NC^{-1}\) 

Zero, because equal forces are inclined at angle of \(120°\) so resultant is Zero.

(i) (a) When a capacitor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that electric field within the conductor. This happens until a static situation is achieved,i.e. when the two fields cancel each other and the net electrostatic field in the conductor becomes zero.

(b) In contrast to conductors, dielectrics are non-conducting substance, i.e. they have no charge carriers.Thus, in a dielectric, free movement of charges in not possible.It turns out that the external field induces dipole moment by stretching molecules of the dielectric.
The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produces a field that opposes the external field. However, the opposing field is so induced, that does not exactly cancel the extent of the effect depends on the nature of dielectric.

Both polar and non-polar dielectrics develop net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by P for linear isotropic dielectrics.
P = X^E
Where, X is constant of proportionality and is called electric susceptibility of the electric slab.
(ii) (a) At point C, inside the shell. Electric field inside a spherical shell is zero.
Thus, the force experienced by charge at centre C will also be zero.
\(\because \)  F_c = qE (E_inside  the shell = 0)
\(\therefore \) F_c = 0
At point A, | F_A | = 2Q \(\left[ \frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } .\frac { 3Q/2 }{ { x }^{ 2 } } \right] \\.\)
\( F \ = \ \frac { { 3Q }^{ 2 } }{ { { 4\pi \varepsilon }_{ 0 } }{ x }^{ 2 } } ,\) away from shell
Electric flux through the shell,
\(\Phi =\frac { 1 }{ { \varepsilon }_{ 0 } } \) x magnitude of charge enclosed by shell
\(=\frac { 1 }{ { \varepsilon }_{ 0 } } \times \frac { Q }{ 2 } \Rightarrow \Phi =\frac { Q }{ { 2\varepsilon }_{ 0 } } \)