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Step-by-Step Solution
Step 1: Identify the Relevant Formula
A parallel-plate capacitor with a dielectric of finite resistivity behaves as if there were a resistor in parallel with the capacitor. The resistance R of this dielectric (assuming uniform thickness and uniform cross-sectional area the same as the capacitor plates) can be written as:
$R = \rho \dfrac{d}{A}$
Meanwhile, the capacitance C is given by:
$C = \kappa \varepsilon_0 \dfrac{A}{d}$
Here,
$\rho$ is the resistivity of the dielectric,
$\kappa$ is the relative permittivity of the material,
$\varepsilon_0$ is the permittivity of free space,
A is the area of the plates, and
d is the distance between the plates.
Combining these, you can eliminate $d$ and $A$ to obtain the effective resistance in terms of the capacitance:
$R = \rho \dfrac{\kappa \varepsilon_0}{C}
$
The leakage current I through this resistance under an applied potential difference V is given by Ohm’s law:
$I = \dfrac{V}{R} = \dfrac{VC}{\rho\,\kappa\,\varepsilon_0}$
Step 2: Substitute the Given Values
From the question, we have:
$\rho = 200\:\Omega\cdot\text{m}$
$C = 2\times 10^{-12}\:\text{F}$
$V = 40\:\text{V}$
$\kappa = 50$
$\varepsilon_0 = 8.85\times 10^{-12}\:\text{F/m}$
Step 3: Plug the Values into the Formula
Using
$I = \dfrac{VC}{\rho\,\kappa\,\varepsilon_0}$,
we get:
$I = \dfrac{(40)\,\bigl(2\times 10^{-12}\bigr)}{(200)\,(50)\,\bigl(8.85\times 10^{-12}\bigr)}
$
Step 4: Perform the Calculation
1. Multiply the numerator:
$40 \times 2\times 10^{-12} = 80\times 10^{-12} \\[6pt]$
2. Multiply the denominator:
$200 \times 50 = 10,000$, and
$10,000 \times 8.85\times 10^{-12} = 8.85\times 10^{-8} \\[6pt]$
3. So,
$I = \dfrac{80\times 10^{-12}}{8.85\times 10^{-8}}
$
Divide:
$I = \bigl(\dfrac{80}{8.85}\bigr)\times 10^{-4} \approx 9.0\times 10^{-4}\:\text{A} = 0.9\times 10^{-3}\:\text{A} = 0.9\:\text{mA}
Step 5: State the Final Answer
Therefore, the leakage current flowing out of the capacitor is
$0.9\:\text{mA}.$
