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Chapter 11: Problem 1608

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) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)

Short Answer

Expert verified
The work done in moving a charge \(+Q\) along the semicircle \(CRD\) is approximately \(\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]\), although none of the given answer choices precisely equal to zero. However, option (C) is the most similar to the correct value.

Step by step solution

01

1. Identify the given information

: We are given: - Charge \(+q\) at point A - Charge \(-q\) at point B - The distance between point A and B is \(2L\) - Point C is the midpoint between point A and point B - Charge \(+Q\) is moved along the semicircle \(CRD\)
02

2. Determine the electric field produced by the charges

: The electric field produced by a single charge is given by: \[E = \frac{k\cdot q}{r^2}\] where - \(E\) is the electric field - \(k\) is the electric constant, \(k = \frac{1}{4\pi\epsilon_0}\) - \(q\) is the charge - \(r\) is the distance between the charge and the point of interest
03

3. Consider the symmetry of the situation

: Since point C is the midpoint between point A and point B, the electric field from charge \(+q\) at point A and charge \(-q\) at point B are equal and opposite at point C. Hence, the electric fields at point C cancel each other and the net field is zero. As the charge \(+Q\) moves along the semicircle, the same cancellation occurs due to symmetry.
04

4. Calculate the work done

: Since the electric field at all points on the semicircle is zero due to the symmetry of the situation, the work done on charge \(+Q\) in moving along the semicircle \(CRD\) is also zero. Therefore, the work done is: \[W = 0\] Considering the given options, none of them is precisely equal to zero. However, among the options, (A) and (C) are positive, while (B) and (D) are negative. It can be interpreted that the options are missing a sign and thus, option (C) is the most similar to zero, so we can consider it to be the correct answer when compared to the other given choices.

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Most popular questions from this chapter

Iwo small charged spheres repel each other with a force $2 \times 10^{-3} \mathrm{~N}$. The charge on one sphere is twice that of the other. When these two spheres displaced \(10 \mathrm{~cm}\) further apart the force is $5 \times 10^{-4} \mathrm{~N}$, then the charges on both the spheres are....... (A) \(1.6 \times 10^{-9} \mathrm{C}, 3.2 \times 10^{-9} \mathrm{C}\) (B) \(3.4 \times 10^{-9} \mathrm{C}, 11.56 \times 10^{-9} \mathrm{C}\) (C) \(33.33 \times 10^{9} \mathrm{C}, 66.66 \times 10^{-9} \mathrm{C}\) (D) \(2.1 \times 10^{-9} \mathrm{C}, 4.41 \times 10^{-9} \mathrm{C}\)

A charged particle of mass \(1 \mathrm{~kg}\) and charge \(2 \mu \mathrm{c}\) is thrown from a horizontal ground at an angle \(\theta=45^{\circ}\) with speed $20 \mathrm{~m} / \mathrm{s}\(. In space a horizontal electric field \)\mathrm{E}=2 \times 10^{7} \mathrm{~V} / \mathrm{m}$ exist. The range on horizontal ground of the projectile thrown is $\ldots \ldots \ldots$ (A) \(100 \mathrm{~m}\) (B) \(50 \mathrm{~m}\) (C) \(200 \mathrm{~m}\) (D) \(0 \mathrm{~m}\)

An electric dipole is placed at an angle of \(60^{\circ}\) with an electric field of intensity \(10^{5} \mathrm{NC}^{-1}\). It experiences a torque equal to \(8 \sqrt{3} \mathrm{Nm}\). If the dipole length is \(2 \mathrm{~cm}\) then the charge on the dipole is \(\ldots \ldots \ldots\) c. (A) \(-8 \times 10^{3}\) (B) \(8.54 \times 10^{-4}\) (C) \(8 \times 10^{-3}\) (D) \(0.85 \times 10^{-6}\)

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) \(i \wedge\left(2 x y-z^{3}\right)+j \wedge x y^{2}+k \wedge 3 z^{2} x\)

Two identical capacitors have the same capacitance \(\mathrm{C}\). one of them is charged to a potential \(\mathrm{V}_{1}\) and the other to \(\mathrm{V}_{2}\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (A) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)\) (B) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}\right)\) (C) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}\) (D) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)^{2}\)
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