Question

Seawater at a frequency \[f = 9 \times {10^2}Hz\] , has permittivity \[ \in = 80{ \in _0}\] and resistivity \[\rho = 0.25\,\Omega m\] . Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source \[V\left( t \right){\text{ = }}{V_0}sin\left( {2\pi ft} \right)\] . Then the conduction current density becomes ${10^x}$ times the displacement current density after time \[t = {\text{ }}\dfrac{1}{{800\,}}\] . Find the value of $x$. (Given : \[\dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}\] )

Answer 1

Hint:In the given question frequency, permittivity and resistivity of seawater are given. Also we are given the alternating voltage source given to the parallel plate capacitor. We first need to find displacement current using its differential formula. Then we find the conduction current. Finally we use the given relationship between ${I_d}$ and ${I_c}$ to find the value of $x$ .

Formula Used:
Displacement current, ${I_d} = \dfrac{{dq}}{{dt}}$
where $q = CV$ given that C is the capacitance of the given parallel plate capacitor
Capacitance, $C = \dfrac{{ \in d}}{A}$
where d is the distance between the plates of the capacitor and A is the area of the plate of parallel plate capacitor.
Conduction current, ${I_c} = \dfrac{V}{R}$
where R is the resistance of the capacitor.

Complete step by step solution:

Given: Frequency of seawater, \[f = 9 \times {10^2}Hz\]
Permittivity of seawater, \[ \in = 80{ \in _0}\]
Resistivity of seawater, \[\rho = 0.25\,\Omega m\]
Alternating Voltage given to the capacitor: \[V\left( t \right){\text{ = }}{V_0}sin\left( {2\pi ft} \right)\] ...(1)
Also, ${I_c} = {10^x} \times {I_d}$ ...(2)
Since, \[ \in = 80{ \in _0}\] therefore, from the relation \[ \in = { \in _r}{ \in _0}\] we get that \[{ \in _r} = 80\] .

Now, we know that the displacement current is, ${I_d} = \dfrac{{dq}}{{dt}}$ where $q = CV$ and $C = \dfrac{{ \in A}}{d}$ ...(3)
Thus, we get, ${I_d} = C\dfrac{{dV}}{{dt}}$ ...(4)
This is because $V$ depends on $t$ whereas $C$ does not.
Differentiating $V$ with respect to $t$ we get,
$\dfrac{{dV}}{{dt}} = {V_0}2\pi f\cos \left( {2\pi ft} \right)$ ...(5)

Combining equations (1), (3), (4) and (5), we get
\[{I_d} = \dfrac{{ \in A}}{d}\left( {{V_0}2\pi f\cos \left( {2\pi ft} \right)} \right)\]
Thus, we get final equation as: \[{I_d} = \dfrac{{{ \in _r}{ \in _0}A}}{d}{V_0}(2\pi f)\cos \left( {2\pi ft} \right)\] ...(6)
Similarly, we find an equation for conduction current which is given by the formula:
 ${I_c} = \dfrac{V}{R}$ ...(7)
 where R is the resistance of the capacitor and is given by $R = \rho \dfrac{d}{A}$ . ...(8)
Now combining the equations (1), (7) and (8), we get
${I_c} = \dfrac{{{V_0}sin\left( {2\pi ft} \right)}}{{\rho \dfrac{d}{A}}}$
This implies that ${I_c} = \dfrac{{{V_0}Asin\left( {2\pi ft} \right)}}{{\rho d}}$ ...(9)

\Now we will divide equation (6) by equation (9), we get
 \[\dfrac{{{I_d}}}{{{I_c}}} = \dfrac{{\dfrac{{{ \in _r}{ \in _0}A}}{d}{V_0}(2\pi f)\cos \left( {2\pi ft} \right)}}{{\dfrac{{{V_0}Asin\left( {2\pi ft} \right)}}{{\rho d}}}}\]
  \[\dfrac{{{I_d}}}{{{I_c}}} = { \in _r}{ \in _0}\rho (2\pi f)\cot \left( {2\pi ft} \right)\] ...(10)
Since, \[\dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}\]
Therefore, we can say that \[{ \in _0} = \dfrac{1}{{4\pi \times 9 \times {{10}^9}}}\]

Now, we put the given values in the equation (10), we get
\[\dfrac{{{I_d}}}{{{I_c}}} = \dfrac{1}{{4\pi \times 9 \times {{10}^9}}} \times 80 \times 0.25 \times (2\pi \times 9 \times {10^2})\cot \left( {2\pi \times 9 \times {{10}^2} \times \dfrac{1}{{800\,}}} \right)\]

After doing calculations, we get
\[\dfrac{{{I_d}}}{{{I_c}}} = \dfrac{{{{10}^3}}}{{{{10}^9}}}\cot \left( {\dfrac{{9\pi }}{{4\,}}} \right)\]
As we know that $\cot \left( {\dfrac{{9\pi }}{4}} \right) = 1$
Therefore, we get \[\dfrac{{{I_d}}}{{{I_c}}} = \dfrac{1}{{{{10}^6}}}\]
This equation can be written as
 \[{I_c} = {10^6}{I_d}\] ...(11)
Comparing equations (2) and (11),
We get, $x = 6$ .

Hence, $x = 6$ is the answer.

Note:One can make mistakes while combining the different equations to achieve the final expressions of displacement current and conduction current. In such types of questions we can also be asked to draw plots between different quantities like $V$ and $I$ etc as an additional part of the question. Also frequency might not be given directly, rather angular frequency or time period may be given in which case we first find the value of frequency and then proceed as above.