Electrostatic Potential and Capacitance

2.21 Charge –q is located at (0,0,-a) and charge +q is located at (0,0,a). Hence, they form a dipole. Point (0,0,z) is on the axis of this dipole and point (x,y,0) is normal to the axis of the dipole. Hence electrostatic potential at point (x,y,0) is zero. Electrostatic potential at point (0,0,z) is given by,

V = $\frac{1}{4\pi {\epsilon }_0}$ ( $\frac{q}{z-a})+\frac{1}{4\pi {\epsilon }_0}$ ( $-\frac{q}{z+a})$ = $\frac{q(z+a-z+a)}{4\pi {\epsilon }_0(z^2-a^2)}$ = $\frac{2qa}{4\pi {\epsilon }_0(z^2-a^2)}$ = $\frac{p}{4\pi {\epsilon }_0(z^2-a^2)}$

Where ${\epsilon }_0$ = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to the distance. i.e. V $\propto$ $\frac{1}{r^2}$

Electrostatic potential ( $V_1)$ at point (5,0,0) is given by

$V_1$ = $\frac{-q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(5-0)}^2+{(-a)}^2}}+\frac{q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(5-0)}^2+{\left(a\right)}^2}}$

= 0

Electrostatic potential ( $V_2)$ at point (-7,0,0) is given by

$V_2$ = $\frac{-q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(-7)}^2+{(-a)}^2}}+\frac{q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(-7)}^2+{\left(a\right)}^2}}$

= 0

Hence, no work is done in moving a small test charge from point (5,0,0) to point (-7,0,0) along the x-axis.

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

2.19

Let the charge in Proton 1 be $q_1$ = 1.6 $*{10}^{-19}$ C, in Proton 2, $q_2$ = 1.6 $*{10}^{-19}$ C and the charge in the electron, $q_3$ = $-$ 1.6 $*{10}^{-19}$ C

Distance between Proton 1 & Proton 2, $d_1$ = 1.5 Å = 1.5 $*{10}^{-10}$ m

Distance between Proton 1 and Electron, $d_2$ = 1.0 Å = 1.0 $*{10}^{-10}$ m

Distance between Proton 2 and Electron, $d_3$ = 1.0 Å = 1.0 $*{10}^{-10}$ m

The potential energy of the system,

V = $\frac{q_1{q}_2}{4\pi {\epsilon }_0{d}_1}$ + $\frac{q_2{q}_3}{4\pi {\epsilon }_0{d}_3}$ + $\frac{q_3{q}_1}{4\pi {\epsilon }_0{d}_2}$ ,

Where ${\mathrm{\epsilon }}_0=\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}$ = 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

V = $\frac{1}{4\pi {\epsilon }_0}$ ( $\frac{q_1{q}_2}{d_1}$ + $\frac{q_2{q}_3}{d_2}$ + $\frac{q_3{q}_1}{d_2}$ )

= $\frac{1}{4\pi *8.854\mathrm{}*{10}^{-12}}*$ ( $\frac{1.6\mathrm{}*{10}^{-19}*1.6\mathrm{}*{10}^{-19}}{1.5\mathrm{}*{10}^{-10}}$ + $\frac{1.6\mathrm{}*{10}^{-19}*-1.6\mathrm{}*{10}^{-19}}{1.0\mathrm{}*{10}^{-10}}$ + $\frac{-1.6\mathrm{}*{10}^{-19}*1.6\mathrm{}*{10}^{-19}}{1.0\mathrm{}*{10}^{-10}}$ )

= 8.988 $*{10}^9*(1.707*{10}^{-28}$ $-2.56*{10}^{-28}-2.56*{10}^{-28})$

= $-3.067*{10}^{-18}$ J

= $-3.067*{10}^{-18}*\frac{1}{1.6\mathrm{}*{10}^{-19}}$ eV

= - 19.17 eV

2.18 The distance between electron – proton of a hydrogen atom, d = 0.53 Å = 0.53 $*{10}^{-10}$ m

Charge of an electron, $q_1$ = $-$ 1.6 $*{10}^{-19}$ C

Charge of a proton, $q_2$ = 1.6 $*{10}^{-19}$ C

Potential at infinity = 0

Potential energy of the system = Potential energy at infinity – Potential energy at a distance d

= 0 - $\frac{q_1{q}_2}{4\pi {\epsilon }_{0}d}$ , where ${\mathrm{\epsilon }}_0=\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}$ == 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

= 0 - $\frac{1.6\mathrm{}*{10}^{-19}*1.6\mathrm{}*{10}^{-19}}{4\pi *8.854\mathrm{}*{10}^{-12}*0.53\mathrm{}*{10}^{-10}}$ = - 4.34 $*{10}^{-18}$ J = $\frac{4.34\mathrm{}*{10}^{-18}}{1.6\mathrm{}*{10}^{-19}}$ = - 27.13 eV

(Since 1 eV = 1.6 $*{10}^{-19}$ J)

Kinetic energy = $\frac{1}{2}$ of potential energy = $\frac{1}{2}$ $*-$ 27.13 eV = -13.57 eV

Total energy = - 13.57 – (-27.13) = 13.57 eV

Therefore, the minimum work required to free the electron is 13.57 eV

In case of zero potential energy, $d_1$ = 1.06 Å = 1.06 $*{10}^{-10}$ m

Potential energy of the system = Potential energy at $d_1$ - Potential energy at d

= $\frac{q_1{q}_2}{4\pi {\epsilon }_{0}d}$ $*\frac{1}{1.6\mathrm{}*{10}^{-19}}$ - 27.13 eV = $\frac{{(1.6\mathrm{}*{10}^{-19})}^2}{4\pi *8.854\mathrm{}*{10}^{-12}*1.06\mathrm{}*{10}^{-10}}*\frac{1}{1.6\mathrm{}*{10}^{-19}}-$ 27.13 = 13.56 -27.13 eV= -13.56 eV

2.15 Charge placed at the centre of the shell is +q. Hence a charge of magnitude of –q will be induced to the inner surface of the shell. Therefore, the total charge on the inner surface of the shell is –q.

Surface charge density at the inner surface of the shell is given by the relation,

${\sigma }_1=\frac{Totalcharge}{Innersurfacearea}=\frac{-q}{4\pi r_1^2}$ …………(1)

When a charge of magnitude of Q is placed on the outer shell and the radius of outer shell is $r_2$ . The total charge experienced by the outer shell will be Q + q. The surface charge density at the outer surface of the shell is given by

${\sigma }_2=\frac{Totalcharge}{Outersurfacearea}=\frac{Q+q}{4\pi r_2^2}$ …………(2)

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape.