
A $1.2\,m$ radius cylindrical region containing a uniform electric field that is increasing uniformly with time. At $t = 0$ the field is $0$ and at $t = 5.0\,s$ the field is $200\,\dfrac{V}{m}$. The total displacement current through a cross- section of the region is
A. $4.5 \times {10^{ - 16}}\,A$
B. $2.0 \times {10^{ - 15}}\,A$ )
C. $3.5 \times {10^{ - 10}}\,A$
D. $1.6 \times {10^{ - 9}}\,A$
Answer
548.1k+ views
Hint: In this question we need to determine the total displacement current through a cross section of the region. Here we will use the formula of the rate of change of electric field. And rearrange the formula. As there is a rate of change in time from $t = 0$ to $t = 5.0\,s$, there also occurs a small change. Then we will apply the values and evaluate to determine the required solution.
Complete step by step answer:
Now, it is given that the radius of the cylindrical region contains a uniform electric field along the cylindrical axis, $r = 1.2\,m$.
We need to find the total displacement current through a cross-section of the region.
Also, it is given that at $t = 0$ the field is $0$ and at $t = 5.0\,s$ the field is $200\,\dfrac{V}{m}$ .
Therefore, the rate of change of electric field is given by,
$\dfrac{{dE}}{{dt}} = \dfrac{{200 - 0}}{{5 - 0}}$
Now, we know that,
$\dfrac{{dE}}{{dt}} = \dfrac{I}{{A{\varepsilon _0}}}$
By rearranging,
$I = A{\varepsilon _0}\dfrac{{dE}}{{dt}}$
Then,
$I - \dfrac{{dq}}{{dt}} = A{\varepsilon _0}\dfrac{{dE}}{{dt}}$
As there is a rate of change in time from $t = 0$ to $t = 5.0\,s$, there occurs a small change which is $\dfrac{{dq}}{{dt}}$ .
$I = {\varepsilon _0} \times \pi {\left( {1.2} \right)^2} \times \dfrac{{200 - 0}}{{5 - 0}}$
We know that,
${\varepsilon _0} = 8.8 \times {10^{ - 12}}$ and $\pi = \dfrac{{22}}{7}$
By applying the values, we have,
$ I = 8.8 \times {10^{ - 12}} \times \dfrac{{22}}{7} \times 1.44 \times 40$
$\therefore I = 1.6 \times {10^{ - 9}}\,A$
Hence option D is the correct answer.
Note: In this question it is important to note that the displacement current is defined as the rate of change of electric displacement field. However, we can also find the total displacement current through a cross-section of the region by the formula $i = \dfrac{E}{r}$ where $r$ is the radius and $E$ electric field intensity. Here, $r = 1.2\,m$ and $E = 200\,\dfrac{V}{m}$. By applying the values and evaluating it we can get the required solution.
Complete step by step answer:
Now, it is given that the radius of the cylindrical region contains a uniform electric field along the cylindrical axis, $r = 1.2\,m$.
We need to find the total displacement current through a cross-section of the region.
Also, it is given that at $t = 0$ the field is $0$ and at $t = 5.0\,s$ the field is $200\,\dfrac{V}{m}$ .
Therefore, the rate of change of electric field is given by,
$\dfrac{{dE}}{{dt}} = \dfrac{{200 - 0}}{{5 - 0}}$
Now, we know that,
$\dfrac{{dE}}{{dt}} = \dfrac{I}{{A{\varepsilon _0}}}$
By rearranging,
$I = A{\varepsilon _0}\dfrac{{dE}}{{dt}}$
Then,
$I - \dfrac{{dq}}{{dt}} = A{\varepsilon _0}\dfrac{{dE}}{{dt}}$
As there is a rate of change in time from $t = 0$ to $t = 5.0\,s$, there occurs a small change which is $\dfrac{{dq}}{{dt}}$ .
$I = {\varepsilon _0} \times \pi {\left( {1.2} \right)^2} \times \dfrac{{200 - 0}}{{5 - 0}}$
We know that,
${\varepsilon _0} = 8.8 \times {10^{ - 12}}$ and $\pi = \dfrac{{22}}{7}$
By applying the values, we have,
$ I = 8.8 \times {10^{ - 12}} \times \dfrac{{22}}{7} \times 1.44 \times 40$
$\therefore I = 1.6 \times {10^{ - 9}}\,A$
Hence option D is the correct answer.
Note: In this question it is important to note that the displacement current is defined as the rate of change of electric displacement field. However, we can also find the total displacement current through a cross-section of the region by the formula $i = \dfrac{E}{r}$ where $r$ is the radius and $E$ electric field intensity. Here, $r = 1.2\,m$ and $E = 200\,\dfrac{V}{m}$. By applying the values and evaluating it we can get the required solution.
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