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Chapter 11: Problem 1586

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are \(10 \mathrm{~cm}\) apart each other. The work done in bringing them $4 \mathrm{~cm}$ closer is .... (A) \(5.8 \mathrm{~J}\) (B) \(13 \mathrm{eV}\) (C) \(5.8 \mathrm{eV}\) (D) \(13 \mathrm{~J}\)

Short Answer

Expert verified
The work done in bringing the two point charges 4 cm closer is 5.8 J or 3.62x10¹⁹ eV.

Step by step solution

01

List the given information

- Charge 1: \(q_1 = 12 \mu C\) - Charge 2: \(q_2 = 8 \mu C\) - Initial distance: \(r_1 = 10 cm = 0.1 m\) - Decrease in distance: \(dr = 4 cm = 0.04 m\) - Final distance: \(r_2 = r_1 - dr = 0.1 - 0.04 = 0.06 m\)
02

Calculate the initial and final electric potential energy

Using the formula for electric potential energy between two point charges, we have: \(U = \frac{k \cdot q_1 \cdot q_2}{r}\) where \(k = 8.9875 \times 10^9 \; \frac{Nm^2}{C^2}\) is the electrostatic constant. We need to calculate the initial and final potential energies and then find the difference to determine the work done. Initial potential energy, \(U_1\): \(U_1 = \frac{k \cdot q_1 \cdot q_2}{r_1}\) Final potential energy, \(U_2\): \(U_2 = \frac{k \cdot q_1 \cdot q_2}{r_2}\)
03

Calculate the work done

To calculate the work done, we find the difference between the initial and final potential energies: Work done (\(W\)) = \(U_2 - U_1\)
04

Convert the work done to Joules and electron volts

We can now calculate the work done and also convert it to electron volts, using the conversion factor: \(1eV = 1.602 \times 10^{-19} J\) \(W = U_2 - U_1\) in Joules \(W_{eV} = \frac{W}{1.602 \times 10^{-19}}\) in electron volts Now substitute the given values and calculate the work done to choose the correct option from the given choices.

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Most popular questions from this chapter

Two point positive charges \(q\) each are placed at \((-a, 0)\) and \((a, 0)\). A third positive charge \(q_{0}\) is placed at \((0, y)\). For which value of \(\mathrm{y}\) the force at \(q_{0}\) is maximum \(\ldots \ldots \ldots\) (A) a (B) \(2 \mathrm{a}\) (C) \((\mathrm{a} / \sqrt{2})\) (D) \((\mathrm{a} / \sqrt{3})\)

A long string with a charge of \(\lambda\) per unit length passes through an imaginary cube of edge \(\ell\). The maximum possible flux of the electric field through the cube will be ....... (A) \(\sqrt{3}\left(\lambda \ell / \in_{0}\right)\) (B) \(\left(\lambda \ell / \in_{0}\right)\) (C) \(\sqrt{2}\left(\lambda \ell / \in_{0}\right)\) (D) \(\left[\left(6 \lambda \ell^{2}\right) / \epsilon_{0}\right]\)

For the system shown in figure, if the resultant force on q is zero, then \(q=\ldots \ldots \ldots\) (A) \(-2 \sqrt{2} \mathrm{Q}\) (B) \(2 \sqrt{2} \mathrm{Q}\) (C) \(2 \sqrt{3} \mathrm{Q}\) (D) \(-3 \sqrt{2} Q\)

The electric flux for gaussian surface \(\mathrm{A}\) that enclose the $\ldots \ldots\( charged particles in free space is (given \)\left.q_{1}=-14 n c, q_{2}=78.85 \mathrm{nc}, q_{3}=-56 n c\right)$ (A) \(10^{4} \mathrm{Nm}^{2} / \mathrm{C}\) (B) \(10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (C) \(6.2 \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (D) \(6.3 \times 10^{4} \mathrm{Nm}^{2} / \mathrm{C}\)

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) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]$
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