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Chapter 11: Problem 1590

Two electric charges \(12 \mu \mathrm{c}\) and \(-6 \mu \mathrm{c}\) are placed \(20 \mathrm{~cm}\) apart in air. There will be a point \(P\) on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of \(P\) from \(-6 \mu c\) charge is ..... (A) \(0.20 \mathrm{~m}\) (B) \(0.10 \mathrm{~m}\) (C) \(0.25 \mathrm{~m}\) (D) \(0.15 \mathrm{~m}\)

Short Answer

Expert verified
The distance of point \(P\) from the \(-6 \mu C\) charge is 0.20 m. The correct answer is (A).

Step by step solution

01

Identify Known and Unknown Variables

The known variables from the problem are the electric charges \(Q_1 = 12 \mu C\) and \(Q_2 = -6 \mu C\) and the distance between them \(d = 20 cm\). The point \(P\) lies outside the region between them, where the electric potential is zero. The unknown variable is the distance \(r\) from charge \(Q_2\) to point \(P\).
02

Write Down the Formula

The formula for the electric potential at a point due to a point charge is \(V = \frac{kQ}{r}\). In this case, we need to consider the electric potentials from both charges and set the total potential to zero. This gives us the equation \(\frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} = 0\).
03

Substitute the Known Variables into the Equation

Substituting the known variables gives \(\frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} = 0\). Substituting \(Q_1 = 12 \mu C\), \(Q_2 = -6 \mu C\), and \(r_2 = r\) (the unknown distance from \(Q_2\) to point \(P\)) into the equation gives: \[ \frac{k \cdot 12}{d + r} - \frac{k \cdot 6}{r} = 0 \] Note that \(d = 20 cm = 0.2 m\) is the distance between the two charges, so \(r_1 = d + r = 0.2m + r\). The \(k\) terms look like they will cancel out in the next step.
04

Simplify the Equation and Solve for the Unknown

Dividing the whole equation by \(k\) and then multiplying by \(r(d + r)\) to eliminate the denominators gives: \[ 12r - 6(d + r) = 0 \] Solving this equation for \(r\) gives: \[ r = \frac{6d}{6} = 0.2m \] Therefore, the distance of point \(P\) from the \(Q_2\) charge is 0.20 m. The correct answer is (A).

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

Two parallel plate air capacitors have their plate areas 100 and $500 \mathrm{~cm}^{2}$ respectively. If they have the same charge and potential and the distance between the plates of the first capacitor is \(0.5 \mathrm{~mm}\), what is the distance between the plates of the second capacitor ? (A) \(0.25 \mathrm{~cm}\) (B) \(0.50 \mathrm{~cm}\) (C) \(0.75 \mathrm{~cm}\) (D) \(1 \mathrm{~cm}\)

Four equal charges \(\mathrm{Q}\) are placed at the four corners of a square of each side is ' \(\mathrm{a}\) '. Work done in removing a charge \- Q from its centre to infinity is ....... (A) 0 (B) $\left[\left(\sqrt{2} \mathrm{Q}^{2}\right) /\left(\pi \epsilon_{0} \mathrm{a}\right)\right]$ (C) $\left[\left(\sqrt{2} Q^{2}\right) /\left(4 \pi \epsilon_{0} a\right)\right]$ (D) \(\left[\mathrm{Q}^{2} /\left(2 \pi \epsilon_{0} \mathrm{a}\right)\right]\)

Three particles, each having a charge of \(10 \mu \mathrm{c}\) are placed at the corners of an equilateral triangle of side \(10 \mathrm{~cm}\). The electrostatic potential energy of the system is (Given $\left.\left[1 /\left(4 \pi \epsilon_{0}\right)\right]=9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{c}^{2}\right)$ (A) \(100 \mathrm{~J}\) (B) \(27 \mathrm{~J}\) (C) Zero (D) Infinite

Two point charges \(100 \mu \mathrm{c}\) and \(5 \mu \mathrm{c}\) are placed at points \(\mathrm{A}\) and \(B\) respectively with \(A B=40 \mathrm{~cm}\). The work done by external force in displacing the charge \(5 \mu \mathrm{c}\) from \(\mathrm{B}\) to \(\mathrm{C}\) where \(\mathrm{BC}=30 \mathrm{~cm}\), angle \(\mathrm{ABC}=(\pi / 2)\) and \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\) \(=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{c}^{2}\). (A) \(9 \mathrm{~J}\) (B) \((9 / 25) \mathrm{J}\) (C) \((81 / 20) \mathrm{J}\) (D) \(-(9 / 4) \mathrm{J}\)
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