Physics 1 Marks Question

1. From the given electric field vector, it can be inferred that the electric field is directed along the negative x direction. Hence, the direction of motion is along the negative y

diraction i.e., $-\hat{\text{j}}$.

2. It is given that,

$\vec{\text{E}}=3.1 \ \text{N}/\text{C}\cos\Big[(1.8 \ \text{rad}/ \text{m})\text{y}+(5.4\times10^8 \ \text{rad}/\text{s})\text{t}\Big]\hat{\text{i}}\dots(1)$

$\vec{\text{E}}=\text{E}_0\sin(\text{kx}-\omega\text{t})\dots(2)$

Electric field amplitude, E₀ = 3.1 N/C

$\omega$ = 5.4 x 10⁸ rad/s

$\lambda=\frac{2\pi}{1.8}=3.490 \ \text{m}$

3. Frequency of wave is given as:

$\text{v}=\frac{\omega}{2\pi}$

$=\frac{5.4\times10^8}{2\pi}=8.6\times10^7 \ \text{Hz}$

4. Magnetic field strength is given as:

$\text{B}_0=\frac{\text{E}_0}{\text{c}}$

Where,

$\therefore\ \text{B}_0=\frac{3.1}{3\times10^8}=1.03\times10^{-7} \ \text{T}$

5. On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:

$\vec{\text{B}}=\text{B}_0\cos(\text{ky}+\omega\text{t})\hat{\text{k}}$

$=\Big\{(1.03\times10^{-7}\text{T})\cos\Big[(1.8 \ \text{rad}/\text{m})\text{y}+(5.4\times10^6\text{rad}/\text{s})\text{t}\Big]\Big\}\hat{\text{k}}$

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

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

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

V = Potential difference across the plates

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

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

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

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Alternate answer

Hint: $\text{W = K qV} = \frac{\text{P}^{2}}{2\text{m}}\text { or } \text{p}= \sqrt{2\text{mqV}}.$

Alternate Answer

Alternate Answer

Alternate Answer

1. $2,000\mathring{\text{A}}-\text{Ultraviolet radiation}.$

2. $5,000\mathring{\text{A}}-\text{Visible light}.$

3. $10,000\mathring{\text{A}}-\text{Infrared radiation}.$

4. $1.0\mathring{\text{A}}-\text{X-rays}.$