Class 12th

2.13

Length of the side of a cube = b

Charge at each vertices= q

d = diagonal of each face, d = $\sqrt{b^2+b^2}$ = b $\surd 2$

l = length of the solid diagonal, l = $\sqrt{d^2+b^2}$ = b $\surd 3$

If r is the distance from the centre of the cube to its corner, then r = l/2 = $\frac{\surd 3}{2}b$

The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.

V = $\frac{8q}{4\pi {\epsilon }_{0}r}$ = $\frac{8q}{4\pi {\epsilon }_0\frac{\surd 3}{2}b}$ = $\frac{4q}{\surd 3\pi {\epsilon }_{0}b}$

Therefore, the potential at the centre of the cube is $\frac{4q}{\surd 3\pi {\epsilon }_{0}b}$

The electric field at the centre of the cube, due to eight charges, gets cancelled, since the charges are distributed symmetrically. The electric field is zero at the centre.

2.12

Charge located at the origin O, q=8 mC= 8 $*{10}^{-3}$ C

Magnitude of a small charge, which is taken from point P to Q, $q_1$ = -2 $*{10}^{-9}$ C

Pont P is at a distance, $d_1$ = 3 cm = 0.03 m from origin, along Z axis

Point Q is at a distance, $d_2$ = 4 cm = 0.04 m from origin, along y axis

Potential at point P, $V_1$ = $\frac{q}{4\pi {\epsilon }_0{d}_1}$

Potential at point Q, $V_2$ = $\frac{q}{4\pi {\epsilon }_0{d}_2}$

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

Work done, W = $q_1$ $(V_2-V_1)$ = $\frac{q{q}_1}{4\pi {\epsilon }_0}(\frac{1}{d_2}\mathrm{}-\mathrm{}\frac{1}{d_1})\mathrm{}$ = $\frac{8\mathrm{}*{10}^{-3}*-2\mathrm{}*{10}^{-9}}{4*\pi *8.854\mathrm{}*{10}^{-12}}(\frac{1}{0.04}\mathrm{}-\mathrm{}\frac{1}{0.03})\mathrm{}$

= (-0.1438) $*(-8.333)$ = 1.198 J

2.11 Capacitance of the first capacitor, C = 600 pF = 600 $*{10}^{-12}$ F

Potential difference, V = 200V

Electrostatic energy of the first capacitor, $E_1$ = $\frac{1}{2}$ C $V^2$ = $\frac{1}{2}*$ 600 $*{10}^{-12}*{200}^2$ J

= 1.2 $*{10}^{-5}$ J

After the supply is disconnected and the first capacitor is connected to another capacitor of 600 pF capacitance in series, the equivalent capacitance $C_{eq}$ is given by

$\frac{1}{C_{eq}}$ = $\frac{1}{c}$ + $\frac{1}{c}$ = $\frac{1}{600}$ + $\frac{1}{600}$ = $\frac{1}{300}$

$C_{eq}=300pF$

New electrostatic energy, $E_2$ = $\frac{1}{2}{C}_{eq}{V}^2$ = $\frac{1}{2}*300*{10}^{-12}*{\left(200\right)}^2$ = 6 $*{10}^{-6}$ J

Electrostatic energy lost = $E_1-E_2$ = 1.2 $*{10}^{-5}$ $-$ 6 $*{10}^{-6}$ = 6 $*{10}^{-6}$ J

2.10 Capacitance C = 12 pF = 12 $*{10}^{-12}$ F

Potential difference, V = 50V

Electrostatic energy stored in the capacitor is given by the relation,

E = $\frac{1}{2}$ C $V^2$ = $\frac{1}{2}*$ 12 $*{10}^{-12}*{\left(50\right)}^2$ J = 1.5 $*{10}^{-8}$ J

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