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Chapter 11: Problem 1587

The displacement of a charge \(Q\) in the electric field $E^{-}=e_{1} i \wedge+e_{2} j \wedge+e_{3} k \wedge\( is \)r^{-}=a i \wedge+b j \wedge$ The work done is \(\ldots \ldots\) (A) \(Q\left(e_{1}+e_{2}\right) \sqrt{\left(a^{2}+b^{2}\right)}\) (B) \(Q\left[\sqrt{ \left.\left(e_{1}^{2}+e_{2}^{2}\right)\right](a+b)}\right.\) (C) \(Q\left(a e_{1}+b e_{2}\right)\) (D) \(\left.Q \sqrt{[}\left(a e_{1}\right)^{2}+\left(b e_{2}\right)^{2}\right]\)

Short Answer

Expert verified
The short answer is: (C) \(Q(a e_{1}+b e_{2})\)

Step by step solution

01

Identify the force on the charge in the electric field

Since the problem gives us the electric field \(E^{-} = e_{1} i\wedge + e_{2} j\wedge+e_{3} k\wedge\), we can find the force on the charge by multiplying the electric field by the charge, Q. Therefore, the force can be written as: \(F^{-} = Q(e_{1} i\wedge + e_{2} j\wedge+ e_{3} k\wedge)\)
02

Identify the given displacement

The problem gives us the displacement of the charge as \(r^{-} = a i\wedge + b j\wedge\).
03

Compute the dot product of the force and displacement

The work done on the charge is the dot product of the force and displacement: \(W = F^{-} \cdot r^{-}\) Substitute the force and displacement expressions: \(W = Q(e_{1} i\wedge + e_{2} j\wedge+ e_{3} k\wedge) \cdot (a i\wedge + b j\wedge)\) To find the dot product, we can multiply component by component and sum them: \(W = Q (a e_{1} (i\wedge \cdot i\wedge) + b e_{2} (j\wedge \cdot j\wedge))\) Since \(i\wedge \cdot i\wedge =1\) , \(j\wedge \cdot j\wedge =1\) and \(i \cdot j = j \cdot k = k \cdot i = 0\), we get: \(W = Q(a e_{1} + b e_{2})\)
04

Identify the correct answer

Since we found that the work done is given by \(W=Q(a e_{1}+b e_{2})\), the correct answer is choice (C).

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Most popular questions from this chapter

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) \(0.85 \times 10^{-6}\)

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