Electromagnetic Waves Physics - Electromagnetic Waves

1. Capacitance between the two plates is given by the relation,

$\text{C}=\frac{\varepsilon_0\text{A}}{\text{d}}$

Where,

A = Area of each plate $=\pi\text{r}^2$

$\text{C}=\frac{\varepsilon_0\pi\text{r}^2}{\text{d}}$

$=\frac{8.85\times10^{-12}\times\pi\times12^2}{0.05}$

$=8.0032\times10^{-12}\text{F}=80.032\text{pF}$

Charge on each plate, q = CV

Where,

V = Potential difference across the plates

Differentiation on both sides with respect to time (t) gives:

$\frac{\text{dq}}{\text{dt}}=\text{C}\frac{\text{dV}}{\text{dt}}$

But, $\frac{\text{dq}}{\text{dt}}=\text{current }(I)$

$\therefore\frac{\text{dV}}{\text{dt}}=\frac{I}{\text{C}}$

$\Rightarrow\frac{0.15}{80.032\times10^{-12}}=1.87\times10^9\text{V}/\text{s}$

Therefore, the change in potential difference between the plates is $1.87\times10^9\text{V}/\text{s}$.

2. The displacement current across the plates is the same as the conduction current. Hence, the displacement current, i_d is 0.15 A.
3. Yes

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