Macroscopic form of Ohm's law is V = IR
Where V - Potential difference
I - current 
R - resistance of a conductor.

By using superposition principle, the net electric field at point A is
\(\vec { E_{ A } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1A }^{ 2 } } \hat { { r }_{ 1A } } +\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 2A }^{ 2 } } \hat { { r }_{ 2A } } \)
where r_1A and r_2A are the distances of point A from the two charges respectively
\(\vec { E_{ A } } =\frac { 9\times 10^{ 4 }\times 1\times { 10 }^{ -6 } }{ (2\times { 10 }^{ -3 })^{ 2 } } (\hat { j } )+\frac { 9\times 10^{ 9 }\times 1\times { 10 }^{ -6 } }{ (2\times 10^{ -3 })^{ 2 } } (\hat { i } )\)\(\)
= 2.25 x 10⁹\(\hat { j } \) + 2.25 x 10⁹\(\hat { i } \) = 2.25 x 10⁹ (\((\hat { i } +\hat { j } )\)
The magnitude of electric field
\(|\vec { { E }_{ A } } |=\sqrt { (2.25\times { 10 }^{ 9 })^{ 2 }+(2.25\times 10^{ 9 })^{ 2 } } \)
= 2.25 x \(\sqrt { 2 } \) x 10⁹ NC^-1
The direction of \(\vec { { E }_{ A } } \) is given by \(\frac { \vec { { E }_{ A } } }{ |\vec { { E }_{ A } } | } =\frac { 2.25\times 10^{ 9 }(\hat { i } +\hat { j } ) }{ 2.25\times \sqrt { 2 } \times { 10 }^{ 9 } } =\frac { (\hat { i } +\hat { j } ) }{ \sqrt { 2 } } \), which is the unit vector along OA as shown in the figure.

The acceleration experienced by an electron placed at point A is
\(\vec { a_{ A } } =\frac { \vec { F } }{ m } =\frac { q\vec { { E }_{ A } } }{ m } \)
=\(\frac { (-1.6\times 10^{ -19 })\times (2.25\times 10^{ 9 })(\hat { i } +\hat { j } ) }{ 9.1\times { 10 }^{ -31 } } \)
= -3.95 x 10²⁰ \((\hat { i } +\hat { j } )\)Nkg^-1
The electron is accelerated in a direction exactly opposite to \(\vec { { E }_{ A } } \).

(i) The electric field lines start at q₂ and end at q₁. In figure (a), q₂ is positive and q₁ is negative. The number of lines starting from q₂ is 18 and number of the lines ending at q₁ is 6. So q₂ has greater magnitude. The ratio of \(\left| \frac { { q }_{ 1 } }{ { q }_{ 2 } } \right| =\frac { { N }_{ 1 } }{ { N }_{ 2 } } =\frac { 6 }{ 18 } =\frac { 1 }{ 3 } \). It implies that |q₂| = 3|q₁|.
(ii) In figure (b), the number of field lines emanating from both positive charges are equal (N = 18). So the charges are equal. At point A, the electric field lines are denser compared to the lines at point B. So the electric field at point A is greater in magnitude compared to the field at point B. Further, no electric field line passes through C, which implies that the resultant electric field at C due to these two charges is zero.
(iii) In the figure (c), the electric field lines start at q₁ and q₃ and end at q₂. This implies that q₁ and q₃ are positive charges. The ratio of the number of field lines is \(\left| \frac { { q }_{ 1 } }{ { q }_{ 2 } } \right| =\frac { 8 }{ 16 } =\left| \frac { { q }_{ 3 } }{ { q }_{ 2 } } \right| =\frac { 1 }{ 2 } \), implying that q₁ and q₃ are half of the magnitude of q₂. So q₁ = q₃ = +10 nc.

R₀ = 3 Ω, T = 100^oC, T₀ = 20^oC
α = 0.004/^oC, R_T = ?
R_T = R₀(1 + α(T - T₀))
R₁₀₀ = 3(1 + 0.004 x 80)
R₁₀₀ = 3.96 Ω

Resistance of the conductor, \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}\)
From Graph - I,
Resistance of the conductor A is, \(\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{4}=0.5 \Omega\)
Resistance of the conductor B is, \(\mathrm{R}_{\mathrm{B}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{4}{3}=1.33 \Omega\)
Resistance of the conductor C is, \(\mathrm{R}_{\mathrm{c}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{5}{2}=2.5 \Omega\)
From Graph - II,
Resistance of the conductor D is, \(\mathrm{R}_{\mathrm{D}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{4}{2}=2 \Omega\)
Resistance of the conductor E is, \(\mathrm{R}_{\mathrm{E}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{3}{4}=0.75 \Omega\)
Resistance of the conductor F is, \(\mathrm{R}_{\mathrm{F}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{5}=0.4 \Omega\)
The conductor F has least resistance, \(R_F=\mathbf{0. 4} \Omega\)
The conductor C has maximum resistance, \(\mathbf{R}_{\mathrm{c}}=2.5 \Omega\)

(a) In the figure

1 - Electrostatic force F_e = QE
2 - Weight W = mg
3 - Elastic force F = -kx
4 - Upward force = Normal reaction = N
b) In this figure

1- Electrostatic force F = qE
2 - Weight W = mg
3 - Tension acting along the string is T
(c) The charge is attracted towards the positively charged plate because it is a negative charge.

1 - Force = qE
2 - Downward force F = mg

Hysteresis is the phenomenon of lagging of magnetic induction behind the magnetising field.

When an external electric field is applied, the centers of positive and negative charges are separated by a small distance which induces dipole moment in the direction of the external electric field. Then the dielectric is said to be polarized by an external electric field.

House hold appliances are always connected in parallel so that even if one is switched off, the other devices could function properly.

There are three quantities required to specify the magnetic field of the earth on its surface, which are often called as the elements of the earth's magnetic field. they are
(a) Magnetic declination (D)
(b) Magnetic dip or inclination (I)
(c) the horizontal component of the Earth's magnetic field (B_H)

According to the superposition principle, the total electric potential at a point is equal to the sum of the potentials due to each charge at that point.
Consider the point at which the total potential zero is located at a distance x from the charge +q as shown in the figure.

The total electric potential at P is zero.
V_tot = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ x } -\frac { 2q }{ (9-x) } \right) \)=0
Which gives \(\frac { q }{ x } -\frac { 2q }{ (9-x) } \)
or \(\frac { 1 }{ x } =\frac { 2 }{ (9-x) } \)
Hence, x = 3m

In a right angle triangle OPN let the angle \(\angle\)OPN = \(\theta \)₁ which implies, \({ \varphi }_{ 1 }=\frac { \pi }{ 2 } -{ \theta }_{ 1 }\) and also in a right angle triangle OPM, 
\(\angle\)OPN = \(\theta \)₂ which implies, \({ \varphi }_{ 2 }=\frac { \pi }{ 2 } +{ \theta }_{ 2 }\)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } \left( cos\left( \frac { \pi }{ 2 } -{ \theta }_{ 1 } \right) -cos\left( \frac { \pi }{ 2 } +{ \theta }_{ 2 } \right) \right) \hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (si{ n }_{ 1 }+{ sin }_{ 2 })\hat { n } \)

(i) Suppose n cells, each of emf volts and internal resistance r ohms are connected in series with an external resistance R as shown in Figure

(ii) The total emf of the battery = nr
The total resistance in the circuit = nr + R
By Ohm's law, the current in the circuit is
\(I=\cfrac { total\ emf }{ total\ resistance } =\cfrac { n\xi }{ nr+5 } \)
Case (a) If r << R, then
\(I=\cfrac { n\xi }{ R } ={ nl }_{ 1 }\)
where II is the current due to a single cell
\(\left( { I }_{ 1 }=\cfrac { \xi }{ R } \right) \)
(iii) Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell
Case (b) If >> R, \(I=\cfrac { n\xi }{ nr } =\cfrac { \xi }{ r } \)
(iv) It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells.
(v) Thus series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R. Cells in parallel
(i) In parallel connection all the positive terminals of the cells are connected to one point and all the negative terminals to a second point. These two points form the positive and negative terminals of the battery.
(ii) Let n cells be connected in parallel between the points A and B and a resistance R is connected between the points A and B as shown in Figure. Let be the emf and r the internal resistance of each cell.

(iii) The equivalent internal resistance of the battery is \(\cfrac { 1 }{ { { r }_{ eq } } } =\cfrac { 1 }{ r } +\cfrac { 1 }{ r } +...\cfrac { 1 }{ r } (netrms)=\cfrac { n }{ r } \)
So \(\cfrac { 1 }{ { r }_{ eq } } =\cfrac { r }{ n } \) and the total resistance in the circuit = \(R+\cfrac { r }{ n } \) The total emf is the potential difference between the points A and B, which is equal to \(\xi \) The current in the circuit is given by
\(I=\cfrac { \xi }{ \frac { r }{ n } +R } \)
\(I=\cfrac { n\xi }{ r+nR } \)
Case (a) If >> R,\(I=\cfrac { n\xi }{ r } ={ nl }_{ 1 }\)
Case (b) If < \(I=\cfrac { \xi }{ R } \)
where II is the current due to a single cell and is equal to \(\cfrac { \xi }{ r } \) when R is negligible. Thus, the current through the external resistance due to the whole battery is n times the current due to a single cell.
  

r = 25cm = 25 x 10^-2m
\(\sigma =\frac { 3 }{ \pi } { cm }^{ -2 }\)
\(AS,\quad \sigma =\frac { q }{ A } =\frac { q }{ 4\pi { r }^{ 2 } } \) [A - surface Area of the sphere]
\(q=(4\pi { r }^{ 2 })\sigma \)
\(=4\pi (0.25{ ) }^{ 2 }\times \frac { 3 }{ \pi } \)
= 0.75C.

The total positive charge in the nucleus is,
q = +ze = 79 x 1.6 x 10^-19C.
Volume charge density, \(\sigma =\frac { q }{ \frac { 4 }{ 3 } \times \pi { R }^{ 3 } } \)
\(s=\frac { q }{ \frac { 4 }{ 3 } \times 3.14\times (7.5\times { 10 }^{ -15 }{ ) }^{ 3 } } \)
= 0.088 x 1026
= 8.8 x 10²⁴ Cm^-3

When a bar magnet (magnetic dipole) of dipole moment \(\vec { { p }_{ m } } \) is held at an angle θ with the direction of a uniform magnetic field \(\vec { { B } } \), as shown in Figure the magnitude of the torque acting on the dipole is

\(|\vec { \tau _{ B } } |=|\vec { { p }_{ m } } ||\vec { B } |sin\theta \)
If the dipole is rotated through a very small angular displacement dθ against the torque ፒ_B at constant angular velocity, then the work done by external torque \((\vec { { \tau }_{ ext } } )\) for this small angular displacement is given by
dW = \(|\vec { { { \tau }_{ ext } } } |\) dθ
Since the bar magnet to be moved at constant angular velocity, it implies \(|\vec { { \tau }_{ B } } |=|\vec { \tau _{ ext } } |\)
dW = P_mB sinθ dθ
Total work done in rotating the dipole from θ' to θ is
W =\(\int _{ \theta ' }^{ \theta }{ \tau d\theta } =\int _{ \theta ' }^{ \theta }{ p_{ m } } Bsin\theta d\theta ={ p }_{ m }B[-cos\theta d\theta ]_{ \theta ' }^{ \theta }\)
W = p_mB(cosθ - cosθ')
This work done is stored as potential energy in bar magnet at an angle θ when it is rotated from θ' to θ and it can be written as
U = p_mB(cosθ - cosθ')           .........(1)
In fact, equation (1) gives the difference in potential energy between the angular positions θ' and θ. We can choose the reference point θ' = 90°, so that second term in the equation becomes zero and the equation (1) can be written as
U = -p_mB(cosθ)              ............(2)
The potential energy stored in a bar magnet in a uniform magnetic field is given by
U = -\(\vec { { p }_{ m } } .\vec { B } \)                     ............(3)
Case 1
(i) If θ = 0°, then
U = P_mB (cos0⁰) = - P_mB
(ii) If θ = 180°, then
U = P_mB (cos 180°) = p_mB
We can infer from the above two results, the potential energy of the bar magnet is minimum when it is aligned along the external magnetic field and maximum when the bar magnet is aligned anti-parallel to an external magnetic field.

(a) The capacitance of the capacitor in the presence of dielectric is
C = \(\frac { { \varepsilon }_{ r }{ \varepsilon }_{ 0 }A }{ d } =\frac { 5\times 8.85\times 10^{ -12 }\times 6 \times 10^{-4}}{ 6\times 10^{ -3 } } \)
= 44.25 x 10^-13F = 4.425 pF
The stored charge is
Q = CV = 44.25 x 10^-13 x 10
= 442.5 x 10^-13C = 44.25pC
The stored energy is
\(U=\frac { 1 }{ 2 } \) CV² = \(\frac { 1 }{ 2 } \) x 44.25 x 10^-13 x 100
= 2.21 x 10^-10 J
(b) After the removal of the dielectric, since the battery is already disconnected the total charge will not change. But the potential difference between the plates increases. As a result, the capacitance is decreased.
New capacitance is
C₀=\(\frac { C }{ { \varepsilon }_{ r } } =\frac { 44.25\times 10^{ -12 } }{ 5 } \)
= 0.885 x 10^-12 F = 0.885 pF
The stored charge remains same and 44.25 pC. Hence newly stored energy is
U₀ =\(\frac { { Q }^{ 2 } }{ 2{ C }_{ 0 } } =\frac { { Q }^{ 2 }{ \varepsilon }_{ r } }{ 2C } =\varepsilon _{ r }U\)
= 5 x 2.21 x 10^-10J = 11.05 x 10^-10J
The increased energy is ΔU = (11.05 - 2.21) x 10^-10 J  = 8.84 x 10^-10 J
When the dielectric is removed, it experiences an inward pulling force due to the plates. To remove the dielectric, an external agency has to do work on the dielectric which is stored as additional energy. This is the source for the extra energy 8.84 x 10^–10 J.

The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T
Deflection is 
\(\theta =60°\Rightarrow tan60°=\sqrt { 3 } =1.732\)
\(I=\frac { 2R{ B }_{ H } }{ { \mu }_{ ° }N } tan\theta \)
\(=\frac { 2\times 0.12\times 25\times 1{ 0 }^{ -6 } }{ 4\times 1{ 0 }^{ -7 }\times 3.14\times 100 } \times 1.732=0.82\times 1{ 0 }^{ -1 }A\)
I = 0.082 A

Energy stored in the capacitor
i) Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge - Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
ii) To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = V dQ
Where \(V=\frac { Q }{ C } \)  .....(1)
iii) The total work done to charge a capacitor is
\(W=\int _{ 0 }^{ Q }{ \frac { Q }{ C } } dQ=\frac { { Q }^{ 2 } }{ 2C } \quad \quad ....(2)\)
This work done is stored as electrostatic potential energy (U_E) in the capacitor.
\({ U }_{ E }=\frac { { Q }^{ 2 } }{ 2C } =\frac { 1 }{ 2 } { CV }^{ 2 },\quad (\therefore Q=CV)\quad ....(3)\)
(iv) This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor.Substituting  \(C=\frac { { \varepsilon }_{ 0 }A }{ d } \) and V = Ed. 
\(U=\frac { 1 }{ 2 } \left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) { (Ed) }^{ 2 }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }(Ad){ E }^{ 2 }\quad \quad \quad \quad \quad ...(4)\)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \({ u }_{ E }=\frac { U }{ Volume } \)       
Equation (4) ⇒ \({ u }_{ E }=\frac{1}{2}{ \varepsilon }_{ 0 }{ E }^{ 2 }\).....(5)
(v) From equation (5),
(a) We infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.
(b) The energy density depends only on the electric field and not on the size of the plates of the capacitor. 
(c) This is true for the electric field due to any type of charge configuration.

(i) This device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle as shown in Figure 1.64 (a) and (b).
(ii) The lower end of the rod is connected to the copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike.
(iii) Since the induced charge density on thin sharp spike is large, it results in a corona discharge.
(iv) This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud.

(v) The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the earth.
(vi) The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.

Energy stored in the capacitor
\(U=\frac { 1 }{ 2 } { Cv }^{ 2 }\quad \quad ...(1)\)
This is rewritten as using \(C=\frac { { \varepsilon }_{ 0 }A }{ d } \& Ed=V\)
\(U=\frac { 1 }{ 2 } \left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) { (Ed })^{ 2 }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }(Ad)\quad { E }^{ 2 }...(2)\)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \({ U }_{ E }=\frac { U }{ Volume } \) From equation (4),
We get
\({ u }_{ E }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }{ E }^{ 2 }\quad \quad \quad \quad \quad ...(3)\)
(iv) The energy density depends only on the electric field and not on the size of the plates of the capacitor.

The drift velocity is the average velocity acquired by the electrons inside the conductor when it is subjected to an electric field. The average time between successive collisions is called the mean free time denoted by \(\tau \). The acceleration \(\vec { a } \) experienced by the electron in an electric field \(\vec { E } \)  is given by
\(\vec { a } =\cfrac { -e\vec { E } }{ m } \left( since\vec { F } =-e\vec { E } \right) \)
The drift velocity is given by
\({ \vec { V } }_{ d }=\vec { a } \tau \)
\({ \vec { V } }_{ d }=\cfrac { e\tau }{ m } \vec { E } \)
\({ \vec { V } }_{ d }=-\mu \vec { E } \)
Here \(\mu =\cfrac { e\tau }{ m } \) is the mobility of the electron and it is defined as the magnitude of the drift velocity per unit electric field \(\mu =\cfrac { \left| { \vec { v } }_{ d } \right| }{ \left| \vec { E } \right| } \)

The charge + q will be stable between B₁ and B₂ with respect to the displacement.

Force ∝ charge 
After the separation, the magnitude of charge will be increased. So the force will be more than before.

\(\vec {E}\) = 10x\(\hat{i},\)  when x = 2 m
\(\vec {E}\) = 10 x 2 x \(\hat{i}\) = 20\(\hat{i}\)
Since, \(E=\frac{-dV}{dx}\therefore V=+20 V\)

\(\frac{1}{C_s}=\frac{1}{2}+\frac{1}{2}=1 \ μF\)
C_p = 1 + 1 = 2 μF

Electric field between the plates \(= \frac{σ}{ε_0}\)
Electric force on an electron \(= e\frac{σ}{ε_0}\)
Magnetic force on an electron, F = BIl
But, \(I= \frac{e}{t}\)
∵Electron moves in a straight line. So,
E_F = M_F
\(e\frac{σ}{ε_0}=B(\frac{e}{t})l\)
\(\therefore t = ε_0\frac{lB}{σ}\)

\(\vec { B } =\frac { \beta }{ \sqrt { 3 } } =(\hat { i } +\hat { j } +\hat { k } )T\)
Using an equation,
Lorentz force, \(\vec{F}=Il\hat{j}\times\vec B\)
We can get,
Lorentz force \(F=\sqrt { \frac { 2 }{ { 3 } } } \beta Il\)

(d) Electrostatic shielding

Magnetic dipole moment

Capacitor

b) Gauss - electrical battery

Kirchhoff's I law - conservation of energy

Aurora borealis - Southern hemisphere

(a) Assertion and, Reason are correct and Reason is the correct explanation of Assertion.

(a) Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(b) Assertion and Reason are true but Reason is the false explanation of the Assertion.