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Chapter 11: Problem 1625

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed \(10 \mathrm{~cm}\) apart in air. The work done to bring them $4 \mathrm{~cm}$ closer is (A) Zero (B) \(4.8 \mathrm{~J}\) (C) \(3.5 \mathrm{~J}\) (D) \(-5.8 \mathrm{~J}\)

Short Answer

Expert verified
The work done to bring the two point charges $4\mathrm{~cm}$ closer is approximately \(5.76 \mathrm{J}\). The given options contain an error, as none of them match the calculated result.

Step by step solution

01

Identify given values and the formula to use

The given values are: \(Q_1 = 12 \mu \mathrm{C} = 12 \times 10^{-6} \mathrm{C}\), \(Q_2 = 8 \mu \mathrm{C} = 8 \times 10^{-6} \mathrm{C}\), initial distance \(r_1 = 10 \mathrm{cm} = 0.1 \mathrm{m}\), and final distance \(r_2 = 6 \mathrm{cm} = 0.06 \mathrm{m}\). We will use the formula for the electrostatic potential energy: \[ U = \frac{kQ_1Q_2}{r} \]
02

Calculate the initial potential energy

Find the initial potential energy by plugging in the values into the formula: \[ U_1 = \frac{kQ_1Q_2}{r_1} \] \[ U_1 = \frac{(9 \times 10^9 \mathrm{Nm^2C^{-2}})(12 \times 10^{-6} \mathrm{C})(8 \times 10^{-6} \mathrm{C})}{0.1 \mathrm{m}} \] \[ U_1 = 8.64 \mathrm{J} \]
03

Calculate the final potential energy

Find the final potential energy by plugging in the values into the formula: \[ U_2 = \frac{kQ_1Q_2}{r_2} \] \[ U_2 = \frac{(9 \times 10^9 \mathrm{Nm^2C^{-2}})(12 \times 10^{-6} \mathrm{C})(8 \times 10^{-6} \mathrm{C})}{0.06 \mathrm{m}} \] \[ U_2 = 14.4 \mathrm{J} \]
04

Calculate the work done

Find the work done by calculating the difference between the final and initial potential energies: \[ W = U_2 - U_1 \] \[ W = 14.4 \mathrm{J} - 8.64 \mathrm{J} \] \[ W = 5.76 \mathrm{J} \] Since the work done is not equal to the given options, we can conclude that the exercise has an error in its options. The correct answer should be approximately \(5.76 \mathrm{J}\).

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Most popular questions from this chapter

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 sphere of radius \(1 \mathrm{~cm}\) has potential of \(8000 \mathrm{v}\), then energy density near its surface will be ...... (A) \(64 \times 10^{5}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (B) \(2.83\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (C) \(8 \times 10^{3}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (D) \(32\left(\mathrm{~J} / \mathrm{m}^{3}\right)\)

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At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)

The electric Potential at a point $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\( is given by \)\mathrm{V}=-\mathrm{x}^{2} \mathrm{y}-\mathrm{x} \mathrm{z}^{3}+4\(. The electric field \)\mathrm{E}^{\boldsymbol{T}}$ at that point is \(\ldots \ldots\) (A) $i \wedge\left(2 \mathrm{xy}+\mathrm{z}^{3}\right)+\mathrm{j} \wedge \mathrm{x}^{2}+\mathrm{k} \wedge 3 \mathrm{xz}^{2}$ (B) $i \wedge 2 \mathrm{xy}+\mathrm{j} \wedge\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)+\mathrm{k} \wedge\left(3 \mathrm{xy}-\mathrm{y}^{2}\right)$ (C) \(i \wedge z^{3}+j \wedge x y z+k \wedge z^{2}\) (D) \(i \wedge\left(2 x y-z^{3}\right)+j \wedge x y^{2}+k \wedge 3 z^{2} x\)
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