Physics Ncert Solutions Class 12th

2.9 Dielectric constant of the mica sheet, k =6

While the voltage supply remained connected :

V = 100 V

Initial capacitance, C = 17.708 $*{10}^{-12}$ F

New capacitance,  $C_1$ = kC = 6 $*$ 17.708 $*{10}^{-12}F$ = 106.25 $*{10}^{-12}$ F

= 106.25 pF

New charge,  $q_1$ = $C_{1}V$ = 106.25 $*{10}^{-12}*100$ C = 10.62 $*{10}^{-9}$ C

If the supply voltage is removed, then there will be constant amount of charge in the plates.

Charge, q = CV = 17.708 $*{10}^{-12}*100$ C = 1.7708 $*{10}^{-9}$ C

Potential across plates,  $V_1$ = $\frac{q}{C_1}$ = $\frac{1.7708\mathrm{}*{10}^{-9}}{106.25\mathrm{}*{10}^{-12}}$ = 16.66 V

2.8 Area of each plate, A = 6 $*{10}^{-3}$ $m^2$

Distance between plates, d = 3 mm = 3 $*{10}^{-3}$ m

Supply voltage, V = 100 V

Capacitance C of the parallel plate is given by C = $\frac{k{\epsilon }_{0}A}{d}$

In case of air, dielectric constant k = 1 and

${\epsilon }_0$ = permittivity of free space = 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

Hence C = $\frac{8.854\mathrm{}*{10}^{-12}*6\mathrm{}*{10}^{-3}}{3\mathrm{}*{10}^{-3}}$ F = 17.708 $*{10}^{-12}$ F = 17.71 pF

Charge on each plate of the capacitor is given by q = VC = 100 $*17.71*{10}^{-12}$ C

= 1.771 $*{10}^{-9}$ C

Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.77 $*{10}^{-9}$ C

2.7 Let the three capacitors be

$C_1$ = 2 pF, $C_2$ = 3 pF and $C_3$ = 4 pF

The equivalent capacitance, $C_{eq}$ is given by

$C_{eq}=C_1+C_2+C_3$ = 2 + 3 + 4 = 9 pF

$\left(b\right)$ When the combination is connected to a 100 V supply, the voltage in all three capacitors = 100 V

$\mathrm{T}\mathrm{h}\mathrm{e}$ charge in each capacitor is given by the relation, q = VC

Hence for $C_1$ , $q_1$ = 100 $*2pC$ = 200 pC = 2 $*{10}^{-10}$ C,

for $C_2$ , $q_2$ = 100 $*3pC$ = 300 pC = 3 $*{10}^{-10}$ C,

for $C_3$ , $q_3$ = 100 $*4pC$ = 400 pC = 4 $*{10}^{-10}$ C

2.6 Capacitance of each of the three capacitors, C = 9 pF

The equivalent capacitance $C_{eq}$ when the capacitors are connected in series is given by

$\frac{1}{C_{eq}}$ = $\frac{1}{C}$ + $\frac{1}{C}+\frac{1}{C}$ = $\frac{3}{C}$ = $\frac{3}{9}$ = $\frac{1}{3}$

Hence,  $C_{eq}$ = 3 pF

Supply voltage, V = 120 V

Potential difference ( $V_1$ ) across each capacitor is given by $V_1$ = $\frac{V}{3}$ = $\frac{120}{3}$ = 40 V

2.4 Radius of the spherical conductor, r = 12 cm = 0.12 m

Uniformly distributed charge, q = 1.6 $*{10}^{-7}$ C

The electrical field inside the conductor is zero.

Electric field E just outside the conductor is given by

E = $\frac{1}{4\pi {\epsilon }_0}*\frac{q}{r^2}$ where ${\epsilon }_0$ = permittivity of free space = 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

E = $\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}}*\frac{1.6\mathrm{}*{10}^{-7}}{{0.12\mathrm{}}^2}$ = 99863.8 N $C^{-1}$ $\cong {10}^5$ N $C^{-1}$

At a point 18 cm from the centre of the sphere. Hence r = 18 cm = 0.18 m

From the above equation we get

E at 18 cm from the centre = $\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}}*\frac{1.6\mathrm{}*{10}^{-7}}{{0.18\mathrm{}}^2}$ = 4.438 $*{10}^4$ N $C^{-1}$

2.3 The arrangement is represented in the adjoining figure.

An equipotential surface is the plane on which total potential is zero everywhere. The plane is normal to line AB. The plane is located at the midpoint of the line AB because the magnitude of the charge is same.

The direction of the electric field at every point on that surface is normal to the plane in the direction of AB.

2.2

The charge at each corner, $q_1$ = 5 $\mu C=5*{10}^{-6}$ C

The sides of the hexagon, AB = BC = CD = DE = EF =FA = 10 cm = 0.1 m

Let O be the centre. The electric potential at O is given by

V = $\frac{1}{4\pi {\epsilon }_0}*\frac{6q_1}{d}$ , where ${\epsilon }_0$ = permittivity of free space = 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

Hence V = $\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}}*\frac{6*5*{10}^{-6}}{0.1}$ = 2.696 $*{10}^6$ V

Therefore, the potential at the centre is 2.7 $*{10}^6$ V

 A p-type semiconductor is electrically neutral despite having more holes, because the number of positively charged holes is exactly balanced/equal by the number of negatively charged acceptor ions introduced during doping. so practically untill any volatage is applied the semiconductor remains chargeless in other words doesn't produce any current even after doping.

 As per the NCERT Textbooks information"Although the number of holes is more than the number of electrons in a p-type semiconductor, the material as a whole is electrically neutral because the charge of holes is balanced by the negatively charged acceptor ions.”