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Chapter 11: Problem 1653

Two identical capacitors have the same capacitance \(\mathrm{C}\). one of them is charged to a potential \(\mathrm{V}_{1}\) and the other to \(\mathrm{V}_{2}\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (A) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)\) (B) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}\right)\) (C) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}\) (D) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)^{2}\)

Short Answer

Expert verified
The short answer to the given problem is: The decrease in energy of the combined system is \( \frac{1}{4}C(V_1 - V_2)^2 \). The correct option is (C).

Step by step solution

01

Find initial energies of the capacitors

To find the initial energy of each capacitor, we can use the energy formula for capacitors: \[E = \frac{1}{2} CV^2\] The initial energy of the first capacitor is: \[E_1 = \frac{1}{2} CV_1^2\] The initial energy of the second capacitor is: \[E_2 = \frac{1}{2} CV_2^2\] The total initial energy of the system is: \[E_{initial} = E_1 + E_2 = \frac{1}{2}C\left(V_1^2 + V_2^2\right)\]
02

Find the final energy of the combined system

When the capacitors are connected, they will have the same voltage V across them. Since the capacitors are identical and have the same capacitance C, the total capacitance of the connected capacitors remains C. The charge conservation principle states that the sum of initial charges equals the sum of final charges. Therefore: \[Q_1 + Q_2 = Q_f\] Using the formula for the charge of a capacitor (Q = CV), we get: \[CV_1 + CV_2 = 2CV_f\] \[V_f = \frac{V_1 + V_2}{2}\] The final energy of the combined system is: \[E_{final} = \frac{1}{2}(2C)\left( \frac{V_1 + V_2}{2}\right)^2 = \frac{1}{4}C\left(V_1 + V_2\right)^2\]
03

Calculate the decrease in energy and find the correct answer

The decrease in energy of the combined system is: \[∆E = E_{initial} - E_{final} = \frac{1}{2}C\left( V_1^2 + V_2^2\right) - \frac{1}{4}C\left(V_1 + V_2\right)^2\] \[= \frac{1}{4}C\left[2(V_1^2 + V_2^2) - (V_1 + V_2)^2\right]\] \[= \frac{1}{4}C\left[2(V_1^2 + V_2^2) - (V_1^2 + 2V_1V_2 + V_2^2)\right]\] \[= \frac{1}{4}C\left(V_1^2 - 2V_1V_2 + V_2^2\right)\] \[= \frac{1}{4}C\left(V_1 - V_2\right)^2\] The decrease in energy of the combined system is given by option C, so the correct answer is C: \((1 / 4) C (V_1 - V_2)^2\).

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Most popular questions from this chapter

Two point masses \(\mathrm{m}\) each carrying charge \(-\mathrm{q}\) and \(+\mathrm{q}\) are attached to the ends of a massless rigid non-conducting rod of length \(\ell\). The arrangement is placed in a uniform electric field \(\mathrm{E}\) such that the rod makes a small angle \(5^{\circ}\) with the field direction. The minimum time needed by the rod to align itself along the field is \(\ldots \ldots .\) (A) \(t=\pi \sqrt{[}(2 M \ell) /(3 q E)]\) (B) $\mathrm{t}=(\pi / 2) \sqrt{[}(\mathrm{M} \ell) /(2 \mathrm{q} \mathrm{E})]$ (D) \(t=2 \pi \sqrt{[}(M \ell / E)]\)

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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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