162 Suppose the particle is at rest v0 0 when it is first released from the

162 suppose the particle is at rest v0 0 when it is

This preview shows page 8 - 11 out of 110 pages.

𝑔 ? = ?? ? = − ?? 𝑦 ? ? ̂ (1.6.2) Suppose the particle is at rest ( v 0 0) when it is first released from the positive plate. The final speed v of the particle as it strikes the negative plate is ? ? = √2|? ? |? = √ 2?? 𝑦 ? ? (1.6.3) where y is the distance between the two plates. The kinetic energy of the particle when it strikes the plate is ? = 1 2 ?? ? 2 = ?? ? ? (1.6.4) 1.7.Electric Dipole An electric dipole consists of two equal but opposite charges, +? an −? are separated by a distance 2a, as shown in Figure 1.7.1. Figure 1.7.1 Electric dipole The dipole moment vector ? which points from −? to +? (in the +y direction) is given by ? = 2???̂ (1.7.1) The magnitude of the electric dipole is ? = 2?? , where q > 0, For an overall charge-neutral system having N charges, the electric dipole vector ? is defined as ? = ∑ ? ? ? ? ?=? ?=1 (1.7.2) Where ? ? is the position vector of the charge ? ? . Examples of dipoles includes HCL, CO, H2O and other polar molecules. In principle, any molecule in which the centers of the positive and negative charges do not coincide may be approximated as a dipole. In Chapter 4 we shall also show that by applying an external field, an electric dipole moment may also be induced in an unpolarized molecule.
Image of page 8
9 1.8. Dipole in Electric Field What happens when we place an electric dipole in a uniform field ? ⃗⃗ = ??̂ , with the dipole moment vector ? making an angle with the x-axis? From figure 1.8.1, we see that the unit vector which points in the direction of ? ⃗⃗ is ???𝜽? + ?𝒊?𝜽? . Thus we have ? ⃗⃗ = ?𝒂?(???𝜽? + ?𝒊?𝜽? ) (1.8.1) Figure 1.8.1 Electric dipole placed in a uniform field. As seen from Figure 2.8.1 above, since each charge experience an equal but opposite force due to the field, the net force on the dipole is ? ??? = ? + + ? = 0 . Even though the net force vanishes, the field exerts a torque a toque on the dipole. The torque about the midpoint O of the dipole is 𝜏 = (? + × ? + ) + (? × ? ) = (????𝜃? + ????𝜃?) × ( ? + ?) + (−????𝜃? − ????𝜃?) × ( −? ?) = ????𝜃 ? + (− ? ) + ????𝜃 ? (− ? ) = 2?? ???𝜃 (− ? ) (1.8.2) where we have used ? ??? = ? + + ? . The direction of the torque is −? , or into the page. The effect of the torque 𝜏 is to rotate the dipole clockwise so that the dipole moment ? becomes aligned with the electric field ? . With F = qE, the magnitude of the torque can be written as 2 a ( qE ) sin (2 aq ) E sin pE sin and the general expression for toque becomes 𝜏 = ? × ? (1.8.3) Thus, we see that the cross product of the dipole moment with the electric field is equal to the torque.
Image of page 9
10 1.9. Charge Density The electric field due to a small number of charged particles can readily be computed using the superposition principle. But what happens if we have a very large number of charges distributed in some region in space? Let’s consi der the system shown in Figure 1.9.1: Figure 1.9.1 Electric field due to a small charge element q i . 1.9.1. Volume Charge Density Suppose we wish to find the electric field at some point P . Let’s consider a small volume element ∆? ? which contains an amount of charge ∆? ?
Image of page 10
Image of page 11

You've reached the end of your free preview.

Want to read all 110 pages?

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes