Electrostatic Potential and Capacitance

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 $*{10}^{-12}*100$ 

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

Distance between plates, d = 3 mm = 3 $\times {10}^{-3}$ 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 $\times {10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

Hence C = $\frac{8.854\mathrm{}\times {10}^{-12}\times 6\mathrm{}\times {10}^{-3}}{3\mathrm{}\times {10}^{-3}}$ F

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

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 ( $V_1$ )

2.5 Capacitance between the parallel plates of the capacitor, C = 8 pF

Let us assume, initially, distance between the parallel plates was d and it was filled with air.

Dielectric constant of air, k = 1

Capacitance C is given by the formula,

C = $\frac{k{\epsilon }_{0}A}{d}$ = $\frac{{\epsilon }_{0}A}{d}$ , since k = 1…………………….(1)

Where A =

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

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

The electrical field inside the conductor is zero.

Electric field E just outside the conductor is given by

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

E

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

2.2

The charge at each corner, $q_1$ = 5 $\mu C=5\times {10}^{-6}$ 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 = $\frac{1}{4\pi {\epsilon }_0}\times \frac{6q_1}{d}$ , where ${\epsilon }_0$ = permittivity of free space = 8.854 $\times {10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

Hence V = $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}\times \frac{6\times 5\times {10}^{-6}}{0.1}$ = 2.69

2.1

Let the charges be

$q_1$ = 5 $\times {10}^{-8}$ C and $q_2$ = -3 $\times {10}^{-8}$ C

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges

R = distance of the point P from the charge $q_1$

Let the electrical potential (V) at point P is zero

Potential at point