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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

N identical drops of mercury are charged simultaneously to 10 volt. when combined to form one large drop, the potential is found to be 40 volt, the value of \(\mathrm{N}\) is \(\ldots \ldots\) (A) 4 (B) 6 (C) 8 (D) 10

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) \((\mathrm{Q} / \sqrt{3})\)

4 Points charges each \(+q\) is placed on the circumference of a circle of diameter \(2 \mathrm{~d}\) in such a way that they form a square. The potential at the centre is \(\ldots \ldots .\) (A) 0 (B) \((4 \mathrm{kd} / \mathrm{q})\) (C) \((\mathrm{kd} / 4 \mathrm{q})\) (D) \((4 \mathrm{kq} / \mathrm{d})\)

Two charged spheres of radii \(R_{1}\) and \(R_{2}\) having equal surface charge density. The ratio of their potential is ..... (A) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)\) (B) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)^{2}\) (C) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}\) (D) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)\)

A particle having a charge of \(1.6 \times 10^{-19} \mathrm{C}\) enters between the plates of a parallel plate capacitor. The initial velocity of the particle is parallel to the plates. A potential difference of \(300 \mathrm{v}\) is applied to the capacitor plates. If the length of the capacitor plates is $10 \mathrm{~cm}\( and they are separated by \)2 \mathrm{~cm}$, Calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of the particle is \(12 \times 10^{-24} \mathrm{~kg}\). (A) \(10^{4}(\mathrm{~m} / \mathrm{s})\) (B) \(10^{2}(\mathrm{~m} / \mathrm{s})\) (C) \(10^{-1}(\mathrm{~m} / \mathrm{s})\) (D) \(10^{3}(\mathrm{~m} / \mathrm{s})\)
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