Q27P - Fundamentals Of Physics

Question

In Fig, a uniform electric field $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}$ collapses. The vertical axis scale is set by $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{ }\mathbf{N}\mathbf{/}\mathbf{C}$, and the horizontal axis scale is set by  ${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{\mu }\mathbf{s}$. Calculate the magnitude of the displacement current through a $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{ }{\mathbf{m}}^{\mathbf{2}}$ area perpendicular to the field during each of the time intervals a,b, and cisshown on the graph.

Step-by-Step Solution

Answer

The magnitude of the displacement current through a  $1.6 {\text{m}}^{\text{2}}$ area perpendicular to the field during each time interval a, b and c is  0.71 A, 0 A  and 2.8 A  respectively.

Step 1: Identification of the given data

The vertical scale is, $E_s=6\times {10}^5 \mathrm{N}/\mathrm{C}$

The horizontal scale is, $t_s=12 \mu \text{s}$

The perpendicular area to the field is, $A=1.6 {\text{m}}^{\text{2}}$

Step 2: Determining the concept

Substitute the expression for electric flux in the formula for displacement current. Find the rate of change of electric field for three different regions, and substituting these values and the given area in the displacement current formula, find the displacement current for three different regions.

Formulae are as follows:

 $i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$

${\varphi }_E=EA$

Where, $i_d$ is the displacement current, $\varphi$ is the flux.

Step 3: Determining the magnitude of the displacement current through a 1 . 6     m 2 area perpendicular to the field during each time interval a, b and c .

For region a:

The displacement current is given by,

$i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$

 

Its magnitude is,

$\left|i_d\right|={\epsilon }_0\left|\frac{d{\varphi }_E}{dt} \right|$

 

The electric flux is given as,

${\varphi }_E=EA$

 

Therefore,

 $\left|i_d\right|={\epsilon }_{0}A\left|\frac{dE}{dt} \right|$

From the graph for region ,

When $t=0, E=6\times {10}^5 \mathrm{N}/\mathrm{C}$

When $t=4 \text{s}, E=4\times {10}^5 \mathrm{N}/\mathrm{C}$

Thus,

$\left|\frac{dE}{dt}\right|=\left|\frac{4\times {10}^5 -6\times {10}^5 }{4 \mu \text{s}-0 }\right|$

 $\begin{array}{rcl}\left|\frac{dE}{dt}\right|& =& \left|\frac{4\times {10}^5 -6\times {10}^5 }{4 \mu \text{s}-0 }\right|\\ \left|i_d\right|& =& \left(8.85\times {10}^{-12} \mathrm{F}/\mathrm{m}\right) \left(1.6 {\text{m}}^{\text{2}}\right)\left|\frac{4\times {10}^5 -6\times {10}^5 }{4\times {10}^{-6} \text{s}-0 }\right|\\ \left|i_d\right|& =& 0.71 \text{A}\end{array}$

For region b:

From the graph, for region  b, there is no change in the electric field, that is, the electric field is constant over the time interval 4 s  to 10 s.

Hence the change  $\frac{dE}{dt}=0$ and no displacement current.

$\left|i_d\right|=0$

 

For region c:

From the graph,

When $t=10 \text{s}, E=4\times {10}^5 \mathrm{N}/\mathrm{C}$

When $t=12 \text{s}, E= 0 \mathrm{N}/\mathrm{C}$

$\left|\frac{dE}{dt}\right|=\left|\frac{0-4\times {10}^5 }{12 -10 }\right|$

$\left|\frac{dE}{dt}\right|=\left|\frac{0-4\times {10}^5 }{12 -10 }\right|$

 $\left|i_d\right|={\epsilon }_{0}A\left|\frac{dE}{dt} \right|$

$\begin{array}{rcl}\left|i_d\right|& =& \left(8.85\times {10}^{-12} \right) \left(1.6 {\text{m}}^{\text{2}}\right)\left|\frac{0-4\times {10}^5 }{12\times {10}^{-6} \text{s }-10\times {10}^{-6} \text{s}}\right|\\ \left|i_d\right|& =& \left(8.85\times {10}^{-12} \right) \left(1.6 {\text{m}}^{\text{2}}\right)\left|\frac{-4\times {10}^5 }{2\times {10}^{-6} \text{s}}\right|\\ \left|i_d\right|& =& 2.8 \text{A}\end{array}$

Therefore, the magnitude of the displacement current through a $1.6 {\text{m}}^{\text{2}}$ area perpendicular to the field during each time interval a, b and c is 0.71 A , 0 A and 2.8 A  respectively.

Using the displacement current formula, find the displacement current for the given regions.