"The electric field between the two parallel plates of a capacitor of $1.5 \mu \mathrm{F}$ capacitance drops to one third of its initial value in $6.6 \mu \mathrm{s}$ when the plates are connected by a thin wire. The resistance of this wire is ________ $\Omega$. (Given, $\log 3=1.1$)"

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To find the resistance of the wire connecting the two plates of the capacitor, we need to apply the formula that relates the time constant $\tau$ (in seconds) of a capacitor-resistor (CR) circuit to the capacitance C (in Farads) and resistance R (in Ohms). The time constant $\tau$ is given by:

$\tau = R \times C$

The time constant also defines the time it takes for the voltage across the capacitor (and therefore the electric field between the plates, since they are directly related) to drop to approximately $\frac{1}{e}$ (where $e$ is the base of natural logarithms, approximately equal to 2.718) of its initial value. However, the question states that the electric field drops to one-third of its initial value. Using the natural logarithm properties, we can relate this decay process to the concept of the time constant.

The formula for the voltage (or electric field) across a discharging capacitor as a function of time $t$ is:

$V(t) = V_0 \times e^{-\frac{t}{RC}}$

where:

$V(t)$ is the voltage across the capacitor at time $t$,
$V_0$ is the initial voltage,
$R$ is the resistance,
$C$ is the capacitance, and
$t$ is the time.

Given that the electric field drops to one-third of its initial value in $6.6 \mu \mathrm{s}$, we can set $V(t)$ to $\frac{1}{3}V_0$ and solve for $R$:

$\frac{1}{3}V_0 = V_0 \times e^{-\frac{6.6 \mu s}{R \times 1.5 \mu F}}$

By dividing both sides by $V_0$, we simplify to:

$\frac{1}{3} = e^{-\frac{6.6}{R \times 1.5}}$

Taking the natural logarithm of both sides to solve for $R$:

$\ln\left(\frac{1}{3}\right) = -\frac{6.6}{R \times 1.5}$

Given that $\ln\left(\frac{1}{3}\right) = \ln(3^{-1}) = -\ln(3) = -1.1$ (since $\log 3 = 1.1$ and using the natural log instead of common log), we have:

$-1.1 = -\frac{6.6}{R \times 1.5}$

Solving for $R$:

$R = \frac{6.6}{1.5 \times 1.1}$

Now, calculate the value of $R$:

$R = \frac{6.6}{1.65} = 4\, \Omega$

Therefore, the resistance of the wire connecting the two plates of the capacitor is $4 \, \Omega$.

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The electric field between the two parallel plates of a capacitor of $1.5 \mu \mathrm{F}$ capacitance drops to one third of its initial value in $6.6 \mu \mathrm{s}$ when the plates are connected by a thin wire. The resistance of this wire is ________ $\Omega$. (Given, $\log 3=1.1$)

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The electric field between the two parallel plates of a capacitor of $1.5 \mu \mathrm{F}$ capacitance drops to one third of its initial value in $6.6 \mu \mathrm{s}$ when the plates are connected by a thin wire. The resistance of this wire is ________ $\Omega$. (Given, $\log 3=1.1$)

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