When we rub two bodies, due to friction, some electrons are transferred from one body to another. The body which gains electrons becomes negatively charged and which loses electrons becomes positively charged by equal amount.

\(3\times { 10 }^{ -6 }J\)

0.026 mA
\(I_{ c }=\frac { V_{ L } }{ R_{ L } } =\frac { 0.5 }{ 800 } =\frac { 5 }{ 8 } mA\)
\( I_{ b }=I_{ e }-I_{ c }=\frac { I_{ c } }{ \alpha } -I_{ c }=I_{ c }\left( \frac { 1 }{ \alpha } -1 \right) \)
\( =\frac { 5 }{ 8 } \left( \frac { 1 }{ 0.96 } -1 \right) =0.026mA\)

As the field is directed outwards, charge on the wire must be positive.
\(V_B-V_A=-\int _{ A }^{ B }{ \overrightarrow { E } .\overrightarrow { dt } } =-\int _{ r=0.2m }^{ r=0.4m }{ \frac { 1000 }{ r } } dr\)
\(V_B-V_A=-1000[log_er]^{r=0.4m}_{r=0.2m}\)
= - 1000\([log_e0.4-log_e0.2]\)
= - 1000 log_e2
\(V_B-V_A=-1000\times 0.6931\)
\(=-693.1 volt\) 

Given, charge, q = 10C (q₁ = q₂ = q)
Each side of equilateral triangle, r = 2 m
Potential energy (PE) = ?
Potential energy between two charges is given by
\(PE=\frac { k{ q }_{ 1 }{ q }_{ 2 } }{ r } \) [\(\because \) r = distance between q₁ and q₂]
\(\therefore \) PE of system will be three times the potential energy between the two charges as the equal charge is placed at the vertices of equilateral triangle.
So, PE_net  = \(\frac { 3\times kqq }{ r } =\frac { 3{ kq }^{ 2 } }{ r } =\frac { 3\times 9\times { 10 }^{ 9 } \times 10\times 10 }{ 2 } \)
= 1.35 x 10¹² J

Electric field intensity due to a long wire : A line charge is in the form of a thin charged rod with uniform linear charge density A (charge per unit length),

To determine the electric field intensity \(\overrightarrow{E}\) at any point P at a perpendicular distance r from the rod, let a right circular closed cylinder of radius r and length I with the infinitely long line of charge as its  axis (as shown in figure). The magnitude of \(\overrightarrow{E}\) at every point on the curved surface of the cylinder is the same as all such points are at the same distance from the line charge. Also \(\overrightarrow{E}\) and unit vector \(\hat{n}\) it normal to curved surface are in the same direction so, \(\theta\)= 0°.
\(\therefore\) Contribution of curved surface of cylinder towards electric flux,
\(\oint_s \overrightarrow{E} \overrightarrow{ds} = \oint_s \overrightarrow{E} \hat{n}{ds}\)
= \(E \oint_s ds = E(2\pi r l)\)
where ( \(2\pi r l\) ) is area of the curved surface of the cylinder.
On the ends of cylinder, the angle between electric field \(\overrightarrow{E}\) and  \(\hat{n}\) is 90°. So it will not contribute to electric flux on cylinder.
\(\Longrightarrow\) \(E \oint_s ds = E(2\pi r l)\)
Charge enclosed in the cylinder linear charge density length
q = \(\lambda \)I
According to Gauss's theorem \(\oint_s \overrightarrow{E} \overrightarrow{ds} = \frac{q}{\epsilon_0}\)
\(\Longrightarrow\)   \((2\pi r l)\times E = \frac{\lambda l}{\epsilon_0}\)
\( E = \frac{\lambda l}{2\pi\epsilon_0rl}\)
\( E = \frac{l}{r}\)
(As \(\frac{\lambda}{2\pi\epsilon_0r} \) = constant)

Any surface that has same electric potential at every point on it is called equipotential surface.
(i) Equipotential surface in case of single point charge

(ii) Equipotential surfaces when the electric field is in Z-direction.

The equipotential surfaces due to a single point charge is represented by concentric spherical shells of increasing radius, so they are not equidistant.
(iii) No, the electric field does not exist tangentially to an equipotential surface because no work done in moving a charge from one point to other on equipotential surface. This indicates that the component of electric field along the equipotential surface is zero. Hence, the equipotential surface is perpendicular to field line.

v = 2.2 x 10⁸ ms^-1
\(\lambda\) = ?
\(\lambda =\frac { h }{ mv } =\frac { 6.63\times { 10 }^{ -34 } }{ 9.1\times { 10 }^{ -31 }\times 22\times { 10 }^{ 8 } } \)
\(=0.331\times { 10 }^{ -11 }=3.3\times { 10 }^{ -12 }\)

When a message is received, it gets attenuated through the channel. Therefore, the receiving antenna is to be followed by an amplifier and a detector. The carrier frequency is.usually changed to a lower frequency (IF) stage. The detected signal may not be strong enough to be use and hence, is required to be amplified. (1) In order to obtain the original message signal m(t) of angular frequency, a simple method is used which is shown below in the form of a block diagram:

When the received modulated signal is passed through a rectifier, an envelope signal is produced. This envelope signal is the message signal. In order to retrieve the message, the signal is passed through an envelope detector

(i) It is because reverse break down voltage of LED is very low i.e., nearly 5 V.
(ii) Because solar cell can work with any light whose photon energy is more than the band gap energy.
(ill) The heavy doping of p and n sides of p-n junction, makes the deplection region very thin, hence for a small reverse bias voltage, electric field is very high.

The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R₁ = +10 cm, R₂ = -15 cm.
Refractive index of air is taken as unity.
We use the lens formula. The sign convention has to be applied for f, R₁ and R₂.
Substituting the values, we have
\(\frac { 1 }{ 12 } =(n-1)\left( \frac { 1 }{ 10 } -\frac { 1 }{ 15 } \right) \)
This gives n = 1.5.
(iii) For a glass lens in air, n₂ = 1.5, n₁ = 1, f = +20 cm. Hence, the lens formula gives
\(\frac { 1 }{ 20 } =0.5\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
For the same glass lens in water, n₂ = 1.5, n₁ = 1.33. Therefore \(\frac { 1.33 }{ f } =(1.5-1.33)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
Combining these two equations, we find f = + 78.2 cm.