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Question: A \(1.2\,m\) radius cylindrical region containing a uniform electric field that is increasing unifor...

A 1.2 m1.2\,m radius cylindrical region containing a uniform electric field that is increasing uniformly with time. At t=0t = 0 the field is 00 and at t=5.0 st = 5.0\,s the field is 200 Vm200\,\dfrac{V}{m}. The total displacement current through a cross- section of the region is
A. 4.5×10−16 A4.5 \times {10^{ - 16}}\,A
B. 2.0×10−15 A2.0 \times {10^{ - 15}}\,A )
C. 3.5×10−10 A3.5 \times {10^{ - 10}}\,A
D. 1.6×10−9 A1.6 \times {10^{ - 9}}\,A

Explanation

Solution

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=0t = 0 to t=5.0 st = 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 mr = 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=0t = 0 the field is 00 and at t=5.0 st = 5.0\,s the field is 200 Vm200\,\dfrac{V}{m} .
Therefore, the rate of change of electric field is given by,
dEdt=200−05−0\dfrac{{dE}}{{dt}} = \dfrac{{200 - 0}}{{5 - 0}}
Now, we know that,
dEdt=IAε0\dfrac{{dE}}{{dt}} = \dfrac{I}{{A{\varepsilon _0}}}
By rearranging,
I=Aε0dEdtI = A{\varepsilon _0}\dfrac{{dE}}{{dt}}
Then,
I−dqdt=Aε0dEdtI - \dfrac{{dq}}{{dt}} = A{\varepsilon _0}\dfrac{{dE}}{{dt}}
As there is a rate of change in time from t=0t = 0 to t=5.0 st = 5.0\,s, there occurs a small change which is dqdt\dfrac{{dq}}{{dt}} .
I=ε0×π(1.2)2×200−05−0I = {\varepsilon _0} \times \pi {\left( {1.2} \right)^2} \times \dfrac{{200 - 0}}{{5 - 0}}
We know that,
ε0=8.8×10−12{\varepsilon _0} = 8.8 \times {10^{ - 12}} and π=227\pi = \dfrac{{22}}{7}
By applying the values, we have,
I=8.8×10−12×227×1.44×40I = 8.8 \times {10^{ - 12}} \times \dfrac{{22}}{7} \times 1.44 \times 40
∴I=1.6×10−9 A\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=Eri = \dfrac{E}{r} where rr is the radius and EE electric field intensity. Here, r=1.2 mr = 1.2\,m and E=200 VmE = 200\,\dfrac{V}{m}. By applying the values and evaluating it we can get the required solution.