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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 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) \(t=2 \pi \sqrt{[}(M \ell / E)]\)

Two identical metal plates are given positive charges \(\mathrm{Q}_{1}\) and \(\mathrm{Q}_{2}\left(<\mathrm{Q}_{1}\right)\) respectively. If they are now brought close to gather to form a parallel plate capacitor with capacitance \(\mathrm{c}\), the potential difference between them is (A) \(\left[\left(Q_{1}+Q_{2}\right) /(2 c)\right]\) (B) \(\left[\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right) / \mathrm{c}\right]\) (C) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) /(2 \mathrm{c})\right]\) (D) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) / \mathrm{c}\right]\)

A simple pendulum consists of a small sphere of mass \(\mathrm{m}\) suspended by a thread of length \(\ell\). The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength \(\mathrm{E}\) directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is (A) $\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}-(\mathrm{q} \mathrm{E} / \mathrm{m})\\}]^{(1 / 2)}$ (B) \(\mathrm{T}=2 \pi(\ell / \mathrm{g})^{(1 / 2)}\) \(\left.\left.\left.\mathrm{m}_{\mathrm{}}\right\\}\right\\}\right]^{(1 / 2)}\) (D) \(\mathrm{T}=2 \pi[(\mathrm{m} \ell / \mathrm{qE})]^{(1 / 2)}\) (C) \(\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}+(\mathrm{qE} / \mathrm{t}\)

A parallel plate capacitor of capacitance \(5 \mu \mathrm{F}\) and plate separation \(6 \mathrm{~cm}\) is connected to a \(1 \mathrm{~V}\) battery and charged. A dielectric of dielectric constant 4 and thickness \(4 \mathrm{~cm}\) is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is (A) \(2 \mu \mathrm{c}\) (B) \(5 \mu \mathrm{c}\) (C) \(3 \mu \mathrm{c}\) (D) \(10 \mu \mathrm{c}\)

Three identical spheres each having a charge \(\mathrm{q}\) and radius \(R\), are kept in such a way that each touches the other two spheres. The magnitude of the electric force on any sphere due to other two is \(\ldots \ldots \ldots\) (A) $(\mathrm{R} / 2)\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{5} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (B) $\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{2} / 3)(\mathrm{q} / \mathrm{R})^{2}$ (C) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{3} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (D) $-\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{3} / 2)(\mathrm{q} / \mathrm{R})^{2}$
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