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3. Three point charges --q , +q, and +q are on the x-axis as shown. The negative charge is at position (x, y, z) = (-a, 0, 0). One of the positive charges is at the origin and the other positive charge is at position (+a, 0, 0). b+ +q A) Write down an exact expression for the voltage everywhere V = V(x, y, z). Give the answer in cartesian coordinates (x, y, z). Use your expression to compute the voltage on the x-axis at r = (5a, 0. 0). (Give your answer in terms of q, a, and known constants.) B) What is the dipole moment of this system (magnitude and direction). Does the dipole moment depend on the choice of origin? Explain, briefly. C) Write down an expression for the voltage using the first two non-zero terms of the multipole expansion. You cau use either spherical coordinates or cartesian coordinates (up to you) . Use your expression to compute the voltage on the x-axis at r = (5a, 0, 0) D) Are your answers for the voltage at r = (5a, 0, 0) from parts (a) and (c) identical? Should they be? Explain, briefly.

          3. Three point charges --q , +q, and +q are on the x-axis as shown. The negative charge is at position (x, y, z) = (-a, 0, 0). One of the positive charges is at the origin and the other positive charge is at position (+a, 0, 0).
b+
+q
A) Write down an exact expression for the voltage everywhere V = V(x, y, z). Give the answer in cartesian coordinates (x, y, z). Use your expression to compute the voltage on the x-axis at r = (5a, 0. 0). (Give your answer in terms of q, a, and known constants.)
B) What is the dipole moment of this system (magnitude and direction). Does the dipole moment depend on the choice of origin? Explain, briefly.
C) Write down an expression for the voltage using the first two non-zero terms of the multipole expansion. You cau use either spherical coordinates or cartesian coordinates (up to you) . Use your expression to compute the voltage on the x-axis at r = (5a, 0, 0)
D) Are your answers for the voltage at r = (5a, 0, 0) from parts (a) and (c) identical? Should they be? Explain, briefly.

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3 three point charges q q and q are on the x axis as shown the negative charge is at position x y z a 0 0 one of the positive charges is at the origin and the other positive charge is at pos 37814

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University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Shu-Ting Huang

02/12/2022

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a) V = V(x, y, z) = qx + qy + qz Show more…

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3. Three point charges --q , +q, and +q are on the x-axis as shown. The negative charge is at position (x, y, z) = (-a, 0, 0). One of the positive charges is at the origin and the other positive charge is at position (+a, 0, 0). b+ +q A) Write down an exact expression for the voltage everywhere V = V(x, y, z). Give the answer in cartesian coordinates (x, y, z). Use your expression to compute the voltage on the x-axis at r = (5a, 0. 0). (Give your answer in terms of q, a, and known constants.) B) What is the dipole moment of this system (magnitude and direction). Does the dipole moment depend on the choice of origin? Explain, briefly. C) Write down an expression for the voltage using the first two non-zero terms of the multipole expansion. You cau use either spherical coordinates or cartesian coordinates (up to you) . Use your expression to compute the voltage on the x-axis at r = (5a, 0, 0) D) Are your answers for the voltage at r = (5a, 0, 0) from parts (a) and (c) identical? Should they be? Explain, briefly.

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00:02 Before we started the computation, first we set up some notation.

00:09 So we let this point to be a, let this point to be b, let this point to be c, this point to be d.

00:25 Then we look at the electric field at p, contributed from a, b, c, d separately.

00:35 So if we look at the contribution from b and b.

00:39 And d we know that the contribution from b is this direction, y direction.

00:51 Contribution from d is again only in the y direction or the j direction.

01:12 So now we write down contribution from b and d.

01:24 They are in a wide direction.

01:28 Now we look at contribution from a and c which is less trivial.

01:39 So the contribution from a, it is pointing to a because it is negative charge.

01:46 So we draw this line, ap and the cp.

02:01 Then this is the contribution of electric field from a.

02:07 And the contribution from c is pointing away from c.

02:16 We know that pa and pc are the same distance.

02:22 So, and the charge at a and c are opposite, but the same magnitude.

02:31 So the electric field are the same magnitude.

02:38 So from the symmetry, we know that that ea plus e, c are in x direction.

02:52 So now we know that in x direction, we only need to compute the contribution from a and the contribution from c.

03:08 So the contribution from a in the x direction, it is k -q over r squared, so k and the q, q is negative q, r square.

03:26 So we look at this triangle, the right triangle.

03:34 R is the hypotenuse, so it is r2 is a squared plus a over 2 square...

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