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Chapter 17: Problem 2

The charge on a parallel plate capacitor is varying as \(q=q_{0} \sin 2 \pi \cdot n t\) The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is (A) \(\frac{q}{\varepsilon_{0} A}\) (B) \(\frac{q_{0}}{\varepsilon_{0}} \sin 2 \pi n t\) (C) \(2 \pi n q_{0} \cos \pi n t\) (D) \(\frac{2 \pi n q_{0}}{\varepsilon_{0}} \cos 2 \pi n t\)

Short Answer

Expert verified
The displacement current through the parallel plate capacitor is given by \(I_d = \frac{2\pi n q_{0}}{\varepsilon_{0} A} \cos(2\pi nt)\), which corresponds to option (D).

Step by step solution

01

Find the time derivative of the charge

We are given the charge as a function of time, \(q = q_{0} \sin(2\pi nt)\). To find the time derivative of the charge, we will differentiate this function with respect to time, \(t\). \(\frac{dq}{dt} = \frac{d}{dt}(q_{0} \sin(2\pi nt))\) Using the chain rule, we get: \(\frac{dq}{dt} = 2\pi n q_{0} \cos(2\pi nt)\)
02

Calculate the displacement current

Now that we have the time derivative of the charge, we can plug it into the formula for the displacement current: \(I_d = \frac{dQ}{dt} \frac{1}{\varepsilon_{0} A}\) Substitute the expression for \(\frac{dq}{dt}\) we found in Step 1: \(I_d = \frac{2\pi n q_{0} \cos(2\pi nt)}{\varepsilon_{0} A}\) This is the expression for the displacement current through the parallel plate capacitor. Comparing it to the given options, we see that the answer is (D).

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