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Chapter 11: Problem 1637

A parallel plate condenser with dielectric of constant \(\mathrm{K}\) between the plates has a capacity \(\mathrm{C}\) and is charged to potential \(\mathrm{v}\) volt. The dielectric slab is slowly removed from between the plates and reinserted. The net work done by the system in this process is (A) Zero (B) \((1 / 2)(\mathrm{K}-1) \mathrm{cv}^{2}\) (C) \((\mathrm{K}-1) \mathrm{cv}^{2}\) (D) \(\mathrm{cv}^{2}[(\mathrm{~K}-1) / \mathrm{k}]\)

Short Answer

Expert verified
The net work done by the system in this process is \(W_{net} = CV^2\left(\frac{(K-1)}{K}\right)\), which corresponds to option (D).

Step by step solution

01

Calculate the capacitance with the dielectric

The capacitance of a parallel plate capacitor with a dielectric between the plates can be calculated using the formula: \[C_{d}=KC_0\] Where: - \(C_d\) is the capacitance with the dielectric - \(K\) is the dielectric constant - \(C_0\) is the capacitance without the dielectric, i.e., with air Since we are given that the capacitance \(C\) with the dielectric is already present, so we have: \[C_d = C\]
02

Calculate the capacitance without the dielectric

To calculate the capacitance without the dielectric, we can simply use the formula we derived above: \[C_0=\frac{C}{K}\]
03

Calculate the work done while removing the dielectric

While the dielectric is being removed from between the plates, the voltage remains constant. Therefore, the work done during this process can be calculated as: \[W_{removal}=\frac{1}{2}C_dV^2-\frac{1}{2}C_0V^2=\frac{1}{2}CV^2-\frac{1}{2}\left(\frac{C}{K}\right)V^2\]
04

Calculate the work done while reinserting the dielectric

Similarly, while reinserting the dielectric between the plates, the voltage remains constant. The work done during this process can be calculated as: \[W_{reinsertion}=\frac{1}{2}C_dV^2-\frac{1}{2}C_0V^2=\frac{1}{2}CV^2-\frac{1}{2}\left(\frac{C}{K}\right)V^2\]
05

Calculate the net work done

The net work done by the system is the sum of the work done during the removal and reinsertion processes: \[W_{net} = W_{removal} + W_{reinsertion}\] Plugging the values from Steps 3 and 4, we get: \[W_{net} = \frac{1}{2}CV^2-\frac{1}{2}\left(\frac{C}{K}\right)V^2+\frac{1}{2}CV^2-\frac{1}{2}\left(\frac{C}{K}\right)V^2\] Simplifying the expression, we get: \[W_{net} = CV^2\left(\frac{(K-1)}{K}\right)\] Comparing this result with the given answer choices, we find that the correct option is: (D) \(cv^2\left[\frac{(K-1)}{K}\right]\)

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Most popular questions from this chapter

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)

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A parallel plate capacitor has plate of area \(\mathrm{A}\) and separation d. It is charged to a potential difference \(\mathrm{V}_{0}\). The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is (A) $\left[\left(\mathrm{A} \in_{0} \mathrm{~V}_{0}^{2}\right) / \mathrm{d}\right]$ (B) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}^{2}\right) /(2 \mathrm{~d})\right]$ (C) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(3 \mathrm{~d})\right]$ (D) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]$

A small sphere whose mass is \(0.1 \mathrm{gm}\) carries a charge of $3 \times 10^{-10} \mathrm{C}\( and is tie up to one end of a silk fiber \)5 \mathrm{~cm}$ long. The other end of the fiber is attached to a large vertical conducting plate which has a surface charge of \(25 \times 10^{-6} \mathrm{Cm}^{-2}\), on each side. When system is freely hanging the angle fiber makes with vertical is \(\ldots \ldots \ldots\) (A) \(41.8^{\circ}\) (B) \(45^{\circ}\) (C) \(40.2^{\circ}\) (D) \(45.8^{\circ}\)

A parallel plate capacitor has the space between its plates filled by two slabs of thickness \((\mathrm{d} / 2)\) each and dielectric constant \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) If \(\mathrm{d}\) is the plate separation of the capacitor, then capacity of the capacitor is .......... (A) $\left[\left(2 \mathrm{~d} \in_{0}\right) / \mathrm{A}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]$ (B) $\left[\left(2 \mathrm{~A} \in_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right) /\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)\right]$ (C) $\left[\left(2 \mathrm{Ad} \epsilon_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]$ d] \(\left(K_{1}+K_{2}\right)\) (D) \(\left[\left(2 \mathrm{~A} \in_{0}\right) /\right.\)
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