A) Step 1: Find capacitance between the two plates. Formula used: \(C=\frac{\varepsilon_0 A}{d}\)
Given, Distance between the plates, d=5 cm=0.05m
Radius of each circular plate,
r = 12cm = 0.12m
Now, area of each plate, A=πr²
A = 3.14 (0.12)² = 0.045216 m²
Capacitance between two plates,
\(C=\frac{\varepsilon_0 A}{d}\)
Here, ϵ₀ = permittivity of free space =8.85 × 10⁻¹²C²/Nm²
C=8.85×10⁻¹²× 0.045216/0.05
C = 8.0032×10⁻¹²
Step 2: Find change of potential difference between the two plates. Formula used: q = CV
Given, charging current, I=0.15A
Charge on each plate, q=CV
Where, V = Potential difference across the plates
Differentiating both sides with respect to time (t), we get,
\(\begin{array}{rlr} \Rightarrow \quad \frac{d q}{d t} & =C \cdot \frac{d V}{d t} \end{array}\)
\(\begin{array}{rlr} \Rightarrow \quad I & =C \cdot \frac{d V}{d t} \end{array}\)      \(\left[\because \frac{d q}{d t}=I\right]\)      
\(\begin{array}{rlr} \Rightarrow \quad \frac{d V}{d t} & =\frac{I}{C}=\frac{0.15}{8.0032 \times 10^{-12}} \end{array}\)
Step 2 : Find change of potential difference between teh two plates.
Formula used: q = CV
dV/dt= 1.87 × 1010V/s
The rate of change of potential difference between the plates is 1.87×10¹⁰V/s
Final answer : C=8.0032×10⁻¹²,1.87×10¹⁰V/s
B) Formula used: id=ϵ0(dϕEdt)
Given, charging current or conduction current, I=0.15 A
The displacement current across the plates, id=ϵ₀(dϕE/dt)
Using Gauss's law, electric flux ϕE=qϵ₀
id=ϵ₀(1dq/ϵ₀dt)=dq/dt
id=dqdt = conduction current =0.15A
Hence, the displacement current, id is 0.15 A.
Final answer : 0.15 A.
C) Kirchhoff’s first rule (junction rule):
It states that at a junction in an electrical circuit, the sum of currents flowing into the junction is equal to the sum of currents flowing out of the junction.
Kirchhoff’s first rule is valid at each plate of the capacitor provided that we consider the current to be the sum of both conduction and displacement currents.

(i) This wavelength (21 cm) corresponds to the radio waves.
(ii) This frequency (1057 MHz) also corresponds to the radio waves (short wavelength).
(iii) As, T = 2.7 K
Using the formula,
\(\lambda\)_m T = b = 0.29 cm-K
\(\lambda_m=\frac{0.29}{2.7}=0.11 \mathrm{~cm}\)
This wavelength corresponds to the microwaves region of the electromagnetic spectrum.
(iv) This wavelength lies in the visible region of the electromagnetic spectrum.
(v) Energy, E = 14.4 keV
= 14.4 \(\times\) 10³ \(\times\) 1.6 \(\times\) 10^-19 J
Frequency of wave,
\(\begin{aligned} v & =\frac{E}{h}=\frac{14.4 \times 1.6 \times 10^{-16}}{6.6 \times 10^{-34}} \end{aligned}\)
= 3.5 \(\times\) 10¹⁸ Hz
This frequency lies in the X-ray region of the electromagnetic spectrum.

Here, R = 10 cm = 0.1 cm;
C = 200 pF = 200 x 10^-12 F = 2 x 10^-10 F;
E_rms = 200 V; \(\omega\) = 200 rad s^-1 ;
r = 2.0 x 10^-2 m.
(a) \({ I }_{ rms }=\frac { { E }_{ rms } }{ 1/\omega C } =\omega C{ E }_{ rms }\)
= 200 x (2 x 10^-10) x 200
(b) Yes, because I_D = 1
(c) \({ I }_{ 0 }=\sqrt { 2 } { I }_{ rms }=\sqrt { 2 } \times 8\times { 10 }^{ -6 }\)
= 11.312 x 10^-6 A
(d) Consider a loop of radius r between two circular plates of parallel plate capacitor placed coaxially with them. The area of this loop \({ A }^{ \prime }=\pi { r }^{ 2 }\)
By symmetry, the magnetic field \(\overrightarrow { B } \) is equal in magnitude and is tangentially to the circle at every point. In this case, only a part of displacement current I_D will cross the loop of area \({ A }^{ \prime }\) . Therefore, the current passing through the area \({ A }^{ \prime }\)
\({ I }^{ \prime }=\frac { { I }_{ D } }{ \pi { R }^{ 2 } } \times \pi { r }^{ 2 }=\frac { { I }_{ D } }{ { R }^{ 2 } } { r }^{ 2 }\)
Using Ampere's Maxwell law we have, \(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }\times \)(total current through the area \({ A }^{ \prime }\)) or \(B=\frac { { \mu }_{ 0 }{ I }_{ 0 }r }{ 2\pi { R }^{ 2 } } =\frac { 4\pi \times { 10 }^{ -7 }\times 11.312\times { 10 }^{ -6 }\times 2\times { 10 }^{ -2 } }{ 2\pi \times { \left( 0.1 \right) }^{ 2 } } \)
or \(2\pi rB={ \mu }_{ 0 }\frac { { I }_{ 0 } }{ { R }^{ 2 } } { r }^{ 2 }\)
= 4.525 x 10^-12 T

(i) We know that Electric energy density \({ U }_{ E }=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 }\). and magnetic energy desity \({ U }_{ B }=\frac { 1 }{ 2 } \frac { { B }^{ 2 } }{ \mu _{ 0 } } \) 
Total average energy density \({ U }_{ av }={ U }_{ E }+{ U }_{ B }=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 }+\frac { 1 }{ 2 } \frac { { B }^{ 2 } }{ { \mu }_{ 0 } } \) 
Given \(E={ E }_{ 0 } \ sin \ (kz-\omega t)\) 
\(B={ B }_{ 0 } \ sin \ (kz-\omega t)\)
Time average value of \({ E }^{ 2 }\) over complete cycle \(=\frac { { E }_{ 0 }^{ 2 } }{ 2 } \)  and time average value of \(=\frac { B_{ 0 }^{ 2 } }{ 2 } \) 
\(\therefore \ { U }_{ av }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }\frac { { E }_{ 0 }^{ 2 } }{ 2 } +\frac { 1 }{ 2 } { \mu }_{ 0 }\frac { { B }_{ 0 }^{ 2 } }{ 2 } =\frac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }+\frac { { B }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 } } \) 
(ii) Average intensity \({ I }_{ av }={ U }_{ av }c=\left[ \frac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }+\frac { { B }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 } } \right] c\) 
\(=\left[ \frac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }+\frac { { E }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 }^{ c2 } } \right] c\quad \left[ \therefore { B }_{ 0 }=\frac { E_{ c } }{ c } \right] \) 
\(\frac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }\quad c\left[ 1+\frac { 1 }{ { \varepsilon }_{ 0 }{ \mu }_{ 0 }^{ c2 } } \right] \)  
\(\frac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 } \ c\left[ 1+1 \right] \ \left[ \therefore \frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } =c^{ 2 } \right] \) 
or \({ I }_{ av }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }c{ E }_{ 0 }^{ 2 }\) 

Speed (or Velocity).

Radio waves < Microwaves < X-rays < Gamma rays 
Uses of Microwaves :
(a) Aircraft navigation,
(b) Ovens

The direction of propagation of the electromagnetic wave is perpendicular to both electric field vector E and magnetic field vector B, i.e. In the direction of E x B.
This can be seen by the diagram given below.

Here, electromagnetic wave is along the Z-direction which is given by the cross product of E and B

We know that , I _d = \(\varepsilon _{ 0 }A\frac { dE }{ dt } \)
Where, \(\varepsilon _{ 0 }\) = 8.85 x 10^-12 C² N^-1m^-2
Area, A = 1 cm² = 10^-4 m^-2
∴ \(\frac { dE }{ dt } \) = 3 x 10⁶ Vm^-1 s^-1
∵  I _d = 8.85 x 10^-12 x 10^-4 x 3 x 10⁶= 2.7 x 10^-9A

If an ideal capacitor is charged by DC battery, current flows momentarily till capacitor gets fully charged after that no current flows. however, when an AC source is connected, the conduction current, \(I_c=\frac{d q}{d t}\) starts flowing in the connecting wire. As charging polarity of AC current changes, the capacitor is alternatively charged and discharged with time. this causes change in electric field between plates of the capacitor which causes change in electric flux and gives rise to displacement current in the region between plates of cpacitor, as displacement current, \(I_d=\varepsilon_0 \frac{d {\phi}_E}{d t}\) and I_d = I_e at all instants.

While dealing with the charging of a parallel plate capacitor with varying current, it was found that Ampere's circuital law is not logically consistent, because \(\oint { \overset { \rightarrow }{ B } ,\overset { \rightarrow }{ dl } } \)has not same value on the two sides of a plate of charged capacitor. The inconsistency of Ampere's circuital law was removed by maxwell by predicting the presence of displacement current in the region between the plates of the capacitor. when the charge on the capacitor is changing with time. Maxwell also predicted that the sum of conduction current and displacement current has the property of continuity.
The generalised form of Ampere's circuital law, modified by Maxwell states that
\(\oint { \overset { \rightarrow }{ B } ,\overset { \rightarrow }{ dl } = } { \mu }_{ 0 }(I+{ I }_{ D })={ \mu }_{ 0 }\left( (I+{ \epsilon }_{ 0 }\frac { { d\phi }_{ E } }{ dt } ) \right) \)

Ratio = 1; because of the velocity of both the wave. length in a vacuum is same \(\left(=3 \times 10^8 \mathrm{~ms}^{-1}\right)\)

Not necessarily, A displacement current such as that between the plates of a charging capacitor) can also produce loops of \(\vec B\)

As, we know that
Electric field intensity, 
\( E= \ \sqrt { \frac { { Pc\mu }_{ 0 } }{ { 4\pi r }^{ 2 } } }\)
\( i.e.\ E\alpha \ \sqrt { \frac { P }{ { r }^{ 2 } } } \)
As, r is constant in this question.
So,  \(E\alpha \sqrt { P } \)
For two different situations,
\({ \ E }_{ 1 }\alpha \sqrt { { P }_{ 1 } } and \ { E }_{ 2 } \ \alpha \sqrt { { P }_{ 2 } } \)
\(frac { { E }_{ 1 } }{ { E }_{ 2 } } =\left( \frac { { P }_{ 1 } }{ { P }_{ 2 } } \right) ^{ { 1 }/{ 2 } }\Rightarrow \frac { { E }_{ 1 } }{ { E }_{ 2 } } =\left( \frac { 100 }{ 50 } \right) ^{ { 1 }/{ 2 } }=\sqrt { 2 }\)
\(\Rightarrow \ { E }_{ 2 }=\frac { { E }_{ 1 } }{ \sqrt { 2 } }\)
\(As,d { E }_{ 1 }=E\Rightarrow { E }_{ 2 }=\frac { E }{ \sqrt { 2 } } \)

The generalised form of Ampere's circuital law, modified by Maxwell, states that
\(\oint B.dI={ \mu }_{ 0 }(I+{ I }_{ d })={ \mu }_{ 0 }\left( I+{ \varepsilon }_{ 0 }\quad \frac { { d\phi }_{ E } }{ dt } \right) \)
While dealing with the charging of a parallel plate capacitor with varying current, it was found that Ampere's circuital law is not logically consistent, because \(\oint B.dI\) has not the same value on the two sides of a plate of charged capacitor. The inconsistency of Ampere's circuital law was removed by Maxwell by predicating the presence of displacement current in the region between the plates of capacitor, when the charge on the capacitor is changing with time. So, displacement current is the missing term (which is related with the changing electric field which passes through the surface between the plates of capacitor). In this way, Maxwell pointed out that for consistency of Ampere's circuital law, there must be displacement current I_d along with the conduction current in the closed loop as (I+I_d)  has the property of continuity.