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Chapter 11: Problem 1640

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and work done by the battery will be (A) \((1 / 2)\) (B) \((2 / 1)\) (C) 1 (D) \((1 / 4)\)

Short Answer

Expert verified
The ratio of the energy stored in the capacitor and the work done by the battery is \(\frac{1}{2}\).

Step by step solution

01

Write the formula for energy stored in a capacitor

The formula for the energy stored in a capacitor is given by: \[U_{cap} = \frac{1}{2}CV^2\] where \(U_{cap}\) is the energy stored in the capacitor, \(C\) is the capacitance, and \(V\) is the potential difference across the capacitor.
02

Write the formula for work done by the battery

The formula for the work done by the battery to charge the capacitor is given by: \[W_{battery} = QV\] where \(W_{battery}\) is the work done by the battery, \(Q\) is the charge on the capacitor, and \(V\) is the potential difference (EMF of the battery in this case) across the capacitor.
03

Express charge in terms of capacitance and potential difference

Recall that the charge on a capacitor can be expressed as: \[Q = CV\] Substitute this expression for charge into the work done by the battery formula: \[W_{battery} = (CV)V\] \[W_{battery} = CV^2\]
04

Calculate the ratio of energy stored and work done

Now we can find the ratio of the energy stored in the capacitor to the work done by the battery: \[\frac{U_{cap}}{W_{battery}} = \frac{\frac{1}{2}CV^2}{CV^2}\]
05

Simplify the ratio

Simplify the expression for the ratio: \[\frac{U_{cap}}{W_{battery}} = \frac{\frac{1}{2}CV^2}{CV^2} = \frac{1}{2}\] According to our calculations, the ratio of the energy stored in the capacitor and the work done by the battery is \(\frac{1}{2}\). Therefore, the correct answer is: (A) \(\frac{1}{2}\)

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Most popular questions from this chapter

If 3 charges are placed at the vertices of equilateral triangle of charge ' \(q\) ' each. What is the net potential energy, if the side of equilateral triangle is \(\ell \mathrm{cm}\). (A) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(3 q^{2} / \ell\right)$ (B) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(2 q^{2} / \ell\right)$ (C) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(q^{2} / \ell\right)\) (D) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(4 q^{2} / \ell\right)$

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) \(45.8^{\circ}\)

An inclined plane making an angle of \(30^{\circ}\) with the horizontal is placed in an uniform electric field \(E=100 \mathrm{Vm}^{-1}\). A particle of mass \(1 \mathrm{~kg}\) and charge \(0.01 \mathrm{c}\) is allowed to slide down from rest from a height of \(1 \mathrm{~m} .\) If the coefficient of friction is \(0.2\) the time taken by the particle to reach the bottom is $\ldots \ldots . .$ sec (A) \(2.337\) (B) \(4.337\) (C) 5 (D) \(1.337\)

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

A parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}\(. The separation between its plates is \)\mathrm{d}$. The space between the plates is now filled with two dielectrics. One of the dielectric constant \(\mathrm{K}_{1}=3\) and thickness \(\mathrm{d} / 3\) while the other one has dielectric constant \(\mathrm{K}_{2}=6\) and thickness \(2 \mathrm{~d} / 3\). Capacitance of the capacitor is now (A) \(1.8 \mathrm{pF}\) (B) \(20.25 \mathrm{pF}\) (C) \(40.5 \mathrm{pF}\) (D) \(45 \mathrm{pF}\)
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