Class 12th

1.28 (a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.

Let q be the charge inside the conductor and ${?}_0$ the permittivity of free space.

According to Gauss's law, Flux,$\phi$  = E.ds = $\frac{q}{{?}_0}$

Here, E = 0, hence $\frac{q}{{?}_0}$ = 0, so q = 0 (as ${?}_0\ne 0)$

Therefore, the charge inside the conductor is zero. The entire charge Q appears on the outer surface of the conductor.

(b) The outer surface of the conductor A has a charge amount Q. Another conductor B, having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount –q will be induced in the inner surface of the conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of the conductor A is Q+q.

(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.

8.2 Radius of each circular plate, R = 6.0 cm = 0.06 m

Capacitance of parallel capacitor, C = 100 pF = 100 $*{10}^{-12}$ F

Supply voltage, V = 230 V

Angular frequency, $\omega$ = 300 rad/s

rms value of the conduction current, I = $\frac{V}{X_c}$ , where $X_c=\mathrm{c}\mathrm{a}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}=\mathrm{}\frac{1}{\mathrm{\omega }\mathrm{C}}$

Hence, I = V $*\omega *C$ = 230 $*300*$ 100 $*{10}^{-12}$ = 6.9 $*{10}^{-6}$ A = 6.9 $\mu A$

Yes, the conduction current is equal to the displacement current.

Magnetic field is given as, B = $\frac{{\mu }_{0}r}{2\pi R^2}{I}_0$ , where

${\mu }_0$ = Free space permeability = 4 $\pi *{10}^{-7}$ N $A^{-2}$

$I_0$ = Maximum value of current = $\surd 2I$

r = distance between two plates on the axis = 3.0 cm = 0.03 m

then, B = $\frac{4\pi *{10}^{-7}*0.03}{2\pi *{0.06}^2}$ $*\surd 2*$ 6.9 $*{10}^{-6}$ = 1.626 $*{10}^{-11}$ T

8.1 Radius of the each circular plate, r = 12 cm = 0.12

Distance between the plates, d = 5 cm = 0.05 m

Charging current, I = 0.15 A

Permittivity of free space, ${?}_0$ = 8.85 $*{10}^{-12}$ $C^2{N}^{-1}{m}^{-2}$

Capacitance between two plates is given by the relation,

C = $\frac{{?}_{0}A}{d}$  , where A = Area of each plate = $?r^2$ = $?*{0.12}^2$

C = $\frac{8.85\mathrm{}*{10}^{-12}*\mathrm{}?*{0.12}^2}{0.05}$  = 8.007 $*{10}^{-12}$

F

Charge on each plate, q = CV, where V = potential difference across plates

Differentiating both sides w.r.t. t, we get

$\frac{dq}{dt}$ = C  $\frac{dV}{dt}$

But $\frac{dq}{dt}$  = I, therefore

$\frac{dV}{dt}=\frac{I}{C}$  = $\frac{0.15}{8.007\mathrm{}*{10}^{-12}}$  = 1.87 $*{10}^{10}$  V/s$$

$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e},\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}$ 1.87 $*{10}^{10}$ V/s

The displacement current across the plates is the same as the conduction current. Hence the displacement current is 0.15 A

Kirchhoff's first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

Ans.1.26

(a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.

(b) The field lines shown in (b) do not represent electrostatic field lines because field lines can not emerge from a negative charge and cannot terminate at a positive charge.

(c) The field lines shown in (c) represent electrostatic field line. This is because the field lines emerge from the positive charge and repel each other.

(d) The field lines shown in (d) do not represent electrostatic field lines because the field lines should not intersect each other.

(e) The field lines shown in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.

1.24 The given conditions are explained in the adjacent diagram

Where A and B represent two large, thin metal plates, parallel and close to each other. The outer surface of A is shown as I, outer surface of B is shown as II and the surface in between A and B is shown as III.

Charge density of plate A, $\sigma$ = 17.0 $*{10}^{-22}$C/$m^2$

Charge density of plate B, $\sigma$ = $-$ 17.0 $*{10}^{-22}$
 C/ $m^2$$$

(a) & (b) In the region, I and III, electric field E is zero, because charge is not enclosed by the respective plates.

(c) Electric field, E in the region II is given by

E = $\frac{\sigma }{{?}_0}$
 , where

${?}_0$ = Permittivity of free space = 8.854  $*{10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$

E = $\frac{17.0\mathrm{}*{10}^{-22}}{8.854\mathrm{}*{10}^{-12}}$ N/C = 1.92$*{10}^{-10}$  N/C