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Chapter 11: Problem 1580

Two equal charges \(q\) are placed at a distance of \(2 \mathrm{a}\) and a third charge \(-2 q\) is placed at the midpoint. The potential energy of the system is ....... (A) $\left[\left(9 \mathrm{q}^{2}\right) /\left(8 \pi \epsilon_{0} \mathrm{a}\right)\right]$ (B) \(\left[q^{2} /\left(8 \pi \epsilon_{0} a\right)\right]\) (C) \(-\left[\left(7 q^{2}\right) /\left(8 \pi \epsilon_{0} a\right)\right]\) (D) \(\left[\left(6 q^{2}\right) /\left(8 \pi \in_{0} a\right)\right]\)

Short Answer

Expert verified
The short answer to the question is: (C) \(-\left[\left(7 q^{2}\right) /\left(8 \pi \epsilon_{0} a\right)\right]\)

Step by step solution

01

Identify charges and their positions

Denote the equal charges as \(q_1 = q_2 = q\), and the third charge as \(q_3 = -2q\). Their positions can be represented as follows: - Charge \(q_1\) at position \((-a,0)\) - Charge \(q_2\) at position \((a,0)\) - Charge \(q_3\) at position \((0,0)\)
02

Calculate distances between charges

We will find the distances between each pair of charges: - Distance between \(q_1\) and \(q_3\): \(r_{13} = a\) - Distance between \(q_2\) and \(q_3\): \(r_{23} = a\) - Distance between \(q_1\) and \(q_2\): \(r_{12} = 2a\)
03

Calculate potential energy between each pair of charges

Using the formula for potential energy between two point charges, \(U = \frac{k*q_1*q_2}{r}\) (with \(k = \frac{1}{4 \pi \epsilon_0}\)), calculate the potential energy between each pair of charges: 1. Potential energy between \(q_1\) and \(q_3\): \( U_{13} = \frac{k*q_1*q_3}{r_{13}}= \frac{k*q*(-2q)}{a} =-\frac{2kq^2}{a}\) 2. Potential energy between \(q_2\) and \(q_3\): \( U_{23} = \frac{k*q_2*q_3}{r_{23}} = \frac{k*q*(-2q)}{a} =-\frac{2kq^2}{a}\) 3. Potential energy between \(q_1\) and \(q_2\): \(U_{12} = \frac{k*q_1*q_2}{r_{12}} = \frac{k*q*q}{2a} = \frac{kq^2}{2a}\)
04

Calculate total potential energy of the system

Now, we will sum up the potential energies of each pair of charges to find the total potential energy of the system: \(U_{total} = U_{13} + U_{23} + U_{12} = -\frac{2kq^2}{a} -\frac{2kq^2}{a} + \frac{kq^2}{2a}\) \(U_{total} = -\frac{8kq^2}{2a} + \frac{kq^2}{2a} = -\frac{7kq^2}{2a}\) Recall that \(k = \frac{1}{4 \pi \epsilon_0}\), substituting this value into the expression: \(U_{total} = -\frac{7\left[\frac{1}{4 \pi \epsilon_0}\right]q^2}{2a} = -\frac{7q^2}{8 \pi \epsilon_0 a}\)
05

Match the answer to the options given

Comparing our result to the given options, we find that the total potential energy of the system matches option (C): \(-\left[\left(7 q^{2}\right) /\left(8 \pi \epsilon_{0} a\right)\right]\)

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Most popular questions from this chapter

Point charges \(4 \mu \mathrm{c}\) and \(2 \mu \mathrm{c}\) are placed at the vertices \(\mathrm{P}\) and Q of a right angle triangle \(P Q R\) respectively. \(Q\) is the right angle, \(\mathrm{PR}=2 \times 10^{-2} \mathrm{~m}\) and \(\mathrm{QR}=10^{-2} \mathrm{~m}\). The magnitude and direction of the resultant electric field at \(\mathrm{R}\) is \(\ldots \ldots\) (A) \(4.28 \times 10^{9} \mathrm{NC}^{-1}, 45^{\circ}\) (B) \(2.38 \times 10^{8} \mathrm{NC}^{-1}, 40.9^{\circ}\) (C) \(1.73 \times 10^{4} \mathrm{NC}^{-1}, 34.7^{\circ}\) (D) \(4.9 \times 10^{10} \mathrm{NC}^{-1}, 34.7^{\circ}\)

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) \(10 \mu \mathrm{c}\)

Charges \(+q\) and \(-q\) are placed at point \(A\) and \(B\) respectively which are a distance \(2 \mathrm{~L}\) apart, \(\mathrm{C}\) is the midpoint between \(\mathrm{A}\) and \(\mathrm{B}\). The work done in moving a charge \(+Q\) along the semicircle \(C R D\) is \(\ldots \ldots\) (A) $\left[(\mathrm{qQ}) /\left(2 \pi \mathrm{\epsilon}_{0} \mathrm{~L}\right)\right]$ (B) \(\left[(-q Q) /\left(6 \pi \in_{0} L\right)\right]\) (C) $\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]$ (D) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)

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) \((1 / 4)\)

The eftective capacitances of two capacitors are \(3 \mu \mathrm{F}\) and $16 \mu \mathrm{F}$, when they are connected in series and parallel respectively. The capacitance of each capacitor is (A) \(2 \mu \mathrm{F}, 14 \mu \mathrm{F}\) (B) \(4 \mu \mathrm{F}, 12 \mu \mathrm{F}\) (C) \(6 \mu \mathrm{F}, 8 \mu \mathrm{F}\) (D) \(10 \mu \mathrm{F}, 6 \mu \mathrm{F}\)
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