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Chapter 13: Problem 42

Four equal charges \(Q\) are placed at the four corners of a square of side \(a\). The work done in removing a charge \(-Q\) from the centre of the square to infinity is (A) Zero (B) \(\frac{\sqrt{2} Q^{2}}{4 \pi \varepsilon_{0} a}\) (C) \(\frac{\sqrt{2} Q^{2}}{\pi \varepsilon_{0} a}\) (D) \(\frac{Q^{2}}{2 \pi \varepsilon_{0} a}\)

Short Answer

Expert verified
The work done in removing a charge -Q from the center of the square to infinity is (B) \(\frac{\sqrt{2} Q^{2}}{4 \pi \varepsilon_{0} a}\).

Step by step solution

01

1. Find the distance between the charges

First, we need to find the distance between the charges in the center of the square and the charges at the corners. Since the charges are at the corners of a square, the diagonal of the square will be the line connecting the two opposite charges at the corners. We can use the Pythagorean theorem to determine the length of the diagonal and then find the distance between the charges. Length of the diagonal: \(d = \sqrt{2a^2}\) The distance between the charges, \(r = \frac{d}{2} = \frac{\sqrt{2a^2}}{2} = \frac{a\sqrt{2}}{2}\).
02

2. Find the potential energy of the system with the negative charge at the center

To find the potential energy, we need to multiply the Coulomb potential between the center charge and each corner charge by the magnitude of the center charge. Since we have four corner charges, we add up these potentials and then multiply by -Q. Potential energy at the center: \(U_1 = -Q \sum_{i=1}^{4} \frac{k Q}{(r_i)}\) Since all the distances \(r_i\) are equal, we can simplify this to: \(U_1= -4Q \cdot \frac{kQ}{a\sqrt{2}/2}\)
03

3. Find the potential energy of the negative charge at infinity

When the negative charge is removed to infinity, the potential energy will be zero because the separation between the charges becomes infinite. Potential energy at infinity: \(U_2 = 0\)
04

4. Compute the work done in moving the negative charge to infinity

The work done (W) in removing the negative charge from the center to infinity is the difference between the two potential energies: \(W = U_2 - U_1\) Substituting values from steps 2 and 3, we get \(W = 0 - (-4Q \cdot \frac{kQ}{a\sqrt{2}/2})\) Now, we know that \(k = \frac{1}{4\pi\varepsilon_0}\), so we can substitute this into the expression for work done: \(W = 4Q \cdot \frac{\frac{1}{4\pi\varepsilon_0}Q}{a\sqrt{2}/2}\) Simplifying this expression, we get \(W = \frac{\sqrt{2} Q^2}{4 \pi \varepsilon_{0} a}\) Therefore, the correct answer is (B) \(\frac{\sqrt{2} Q^{2}}{4 \pi \varepsilon_{0} a}\).

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Most popular questions from this chapter

Two capacitors of capacitance \(3 \mu \mathrm{F}\) and \(6 \mu \mathrm{F}\) are charged to a potential of \(12 \mathrm{~V}\) each. They are now connected to each other, with the positive plate of one to the negative plate of the other. Then (A) the potential difference across \(3 \mu \mathrm{F}\) is zero. (B) the potential difference across \(3 \mu \mathrm{F}\) is \(4 \mathrm{~V}\). (C) the charge on \(3 \mu \mathrm{F}\) is zero. (D) the charge on \(3 \mu \mathrm{F}\) is \(10 \mu \mathrm{C}\).

A non-conducting ring of radius \(R\) has charge \(Q\) distributed uniformly over it. If it rotates with an angular velocity \(\omega\), the equivalent current will be (A) Zero (B) \(Q \omega\) (C) \(Q \frac{\omega}{2 \pi}\) (D) \(Q \frac{\omega}{2 \pi R}\)

A spring block system undergoes vertical oscillations above a large horizontal metal sheet with uniform positive charge. The time period of the oscillation is \(T\). If the block is given a charge \(Q\), its time period of oscillation (A) remains same. (B) increases. (C) decreases. (D) increases if \(Q\) is positive and decreases if \(Q\) is negative.

\(A B C\) is an equilateral triangle. Charges \(+q\) are placed at each corner. The electric field intensity at centroid \(O\) will be (A) \(\frac{1}{2 \pi \varepsilon_{0}} \times \frac{q}{r^{2}}\) (B) \(\frac{1}{2 \pi \varepsilon_{0}} \times \frac{3 q}{r^{2}}\) (C) \(\frac{1}{2 \pi \varepsilon_{0}} \times \frac{\sqrt{3} q}{r^{2}}\) (D) Zero

The magnitude of electric intensity at a distance \(x\) from a charge \(q\) is \(E\). An identical charge is placed at a distance \(2 x\) from it. Then the magnitude of the force it experiences is (A) \(q E\) (B) \(2 q E\) (C) \(\frac{q E}{2}\) (D) \(\frac{q E}{4}\)
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