06 Physics 50: Worksheet 0 Name Concets: Electric Field, lines of force, charge density, diole moment, electric diole () An equilateral triangle with each side of length 0.0 m has identical charges of +q.0 C. What is the net electrostatic vector force on charge? () A oint charge q -3.00 C is located at x0. A second charge q +6.00 C is located at x.00 m. Find a oint other than infinity where the electric field is zero. (3) The electric diole consists of a ositive and a negative charge searated by a distance of a. Suose in this case, your diole had +q at xa and -q at x-a. Find an exression for the electric field along the y-axis. You should then be able to show that the electric field behaves as E x kqa y 3 at distant oints along the y-axis. (4) Suose in this case, your diole had +q at xa and -q at x-a. Find an exression for the electric field along the x-axis at x>a. You should then be able to show that the electric field behaves as E x 4kqa x 3 at distant oints along the x -axis. Then write the result in terms of the diole moment. (5) Suose that you have a ring of radius ra and total charge Q located in the x-y lane. What is the electric field for oints along the symmetry axis of this ring? How does this field behave along the axis at distant oints along the symmetry axis?
06 Physics 50: Worksheet 0 Name We have reviously defined the electric force between two charges and we have talked about the law of charges. It turns out that it is less imortant to talk about the electric force than another quantity called the electric field. The electric field is a real hysical entity and carries away energy from an accelerating electric charge. The electric field is defined by: E F electric q Here, by definition, q is a ositive test charge. E oints in the direction that a ositive test charge would move under the influence of an electric force. These lines of force can be sketched with a few rules: () They oint away from ositive charges. () They oint towards negative charges. (3) They don t intersect. (4) They oint normal to the surface of a conductor. (5) The density of these lines is an indicator of the electric field strength (6) A ositive charge laced on one of the field lines accelerates in the line s direction. I ll later show you how to draw these lines of force. In more elegant terms, then, using the definition of force that we had in the last lecture, we can write the electric field as: in k i q i in r r i r i i k q i r i r i Let s look at each of the symbols: n # of discrete charges in the system q i is the i th charge in the system. k is coulomb s constant r is the vector from the origin ointed towards the oint in sace. This would also be the location of the ositive test charge so the notation that we have develoed is really the same here: the test charge is now charge. r i is the vector from the origin ointed towards the charge q i in sace. r i is the unit vector directed from the charge q i towards the oint in sace. Don t get hungu on the fact that a articular charge might not be located at the origin: aly the rules I ve shown you in worksheet and you will correctly calculate the electric field.
06 Physics 50: Worksheet 0 Name One thing that you want to know is just how do you calculate r i. Here is the way although you have already seen this in worksheet. Firstly, r i r r i. We can now, from this find the unit vector retty easily. Again, in words: r i is the vector ointing from charge i toward oint in sace. The unit vector ointing in this direction is given by: r i r i r r i As a side note: what are the dimensions (units) associated with a unit vector? So, let me show you an examle calculation. Suose r x x+y ŷ+z ẑ and r i x i x+y i ŷ+z i ẑ. Then, r i ( x x i ) x+(y i y ) ŷ+ (z z i )ẑ (x x i ) +(y y i ) +(z z i ) Here are some numerical examles: Suose r x+ŷ+3ẑ and r 3 x+ŷ+ẑ. Then: r i r r i (3 ) x+( ) ŷ+( 3) ẑ x+0ŷ ẑ The unit vector is then: r i x ẑ + 8 ( x ẑ) ( x ẑ) Another really easy examle: Suose r i x+0ŷ+0ẑ and r 0 x+0ŷ+0ẑ. r i (0 ) x x and r i x x Note that r r i is simly the distance between the charge and the oint. It is not hard from this to see that for secial cases where the oint of interest () is located at the origin. we have: r i 0 r i r i. One detail about notation: I ll write: r i r r i occasionally and you ll robably do the same. While this is more technical, in rincile you might just find it an easier aroach than having to resolve electric field comonents each time. This alies to charges that are discrete.
06 Physics 50: Worksheet 0 Name For the calculus eole, we have a more general definition which treats charges as a continuum: all charges k dq i ( r r i ) r i This means, if you know now to form the vectors, and to write the element of charge, calculation of electric fields for continuous charge distributions is fairly straight-forward. Also notice that I ve retained the redundant i subscrit here to make sure you know that we re talking about charges and that s what is being integrated over. Where the unit vector symbol really has the meaning that it is the unit vector directed from the charge oint dq towards the oint in sace. I encourage you to retain this notation in all your work in order to assure yourself that you know what is haening. One final imortant oint. I ll introduce in roblems 3 and 4 the electric diole moment. For a collection of j charges, we define the diole moment as: n q j r j j It s imortant not to confuse this with the which I m using to designate the oint in sace. There is also one additional term which is going to be introduced later, namely the olarization of a material which is designated by P and is defined as the electric diole moment er unit volume. There are some details to the diole moment which are worth noting. So long as the overall charge distribution is neutral, the diole moment is coordinate indeendent. The diole moment is related to the second term of a multiole exansion of the electric field. As it is normally used, the diole moment refers to the aroximate behavior of an ideal diole at large distances from the diole. By ideal, I mean that the distance between the ositive and negative charge is insignificant comared to the distance to the oint of interest in sace. This, in turn, means the multiole exansion is valid. Peole use the diole moment because it is often the largest contribution to the electric field. However, there are other terms in the multiole exansion that may be significant.
06 Physics 50: Worksheet 0 Name () An equilateral triangle with each side of length 0.0 m has identical charges of +q.0 C. What is the net electrostatic vector force on charge? Solution: I want to show you ways to do this roblem. The first uses symmetry and is quicker. The second is the brute force method. You will find this roblem on a sreadsheet. The force on any single charge is shown in red below: The little urle dots indicate the same angle which is 80/360 degrees. This is the angle that I m using below. This only works because each of the charges is the same and each distance is the same. I am indicating forces with the red arrows. Thus, the off-axis force is going to be given by: FF x x+f y ŷf (cos( ) x sin( )ŷ) The total force on the single charge is then given by: F net F x+f x x+f y ŷf (+cos( )) x sin( )ŷ] We find the magnitude of this force from Coulomb s law: F k q q x 0 8.99x09( 0.0 ) 0.899N r We can now find the force on this charge: F.349 x 0.779 ŷn. The magnitude of this force is F.349 +0.799.558N. The angle which this force makes with resect to the x-axis is given by: tan(ϕ) F y 0.779 F x.349 ϕ 30o. The correct result here is -30 degrees.
06 Physics 50: Worksheet 0 Name Now let me show you the second (and, in my oinion the more owerful) way to do this. I am utting labels on the charges as shown. I will let charge be at the origin, so it has coordinates (0,0). Probably the hardest art is to find the coordinates of charge #3. However you want to add it u, the x coordinate of the charge is 0.05 m. The length of this vector ointing to charge is 0.m. Thus, the y coordinate is: r 3 x 3 +y 3 y 3 r 3 x 3 y 3 r 3 x 3 y 3. (0.05) 0.0 0.0050.0867 The various vectors are then: r 0 x+0ŷ ; r 3 0.05 x+0.08667 ŷ; r 0. x+0ŷ; r 0 x+0ŷ Now we re going to need to calculate this: n F k q q i i r i r i i That means if we re calculating the force on charge, we need the following: n F () i i () k q i q r i r i () k q q r r +k q 3 q r 3 r 3 We are going to need to calculate the various vectors involved here. I m going to try to show this in detail here. r r r 0 x+0ŷ ] 0. x+0ŷ ]0. x+0ŷ : r r r x x x r 3 r r 3 0 x+0ŷ ] 0.05 x+0.0866ŷ ]0.05 x 0.0866 ŷ r 3 r 3 r 3 0.05 x 0.08667 ŷ 0.05 x 0.08667 ŷ ( 0.05) +(0.08667) 0.5 x 0.8667 ŷ0.5 x 3. ŷ The electric force at charge ( ) due to the other two charges is then: 3 q F k i q i r i r i i F kq q.0 x+ q 3.0 ( 0.5 x 3 ŷ)] kq 0.0.5 x 0.866 ŷ ] F 8.99x0 9 xx0 50 x 86.6ŷ ]349 x0 3 x 779x 0 3 ŷ.35 x 0.78ŷ N F (.35) +( 0.78).56N This force makes the same angle with resect to the +x axis as before: tan(ϕ) F y 0.78 F x.35 ϕ 30o We could have found the same result by calculating the electric field at charge due to charges and 3. The electric field at this oint is given by:
06 Physics 50: Worksheet 0 Name n F k q q i i r i r i i E k q.0 x+ q 3.0 ( 0.5 x 3 ŷ)] kq 0.0.5 x 0.866 ŷ ] E 8.99x0 9 x x0 6 50 x 86.6ŷ] 8990 0.0.5 x 0.866 ŷ ].35x0 3 x 0.78 x9 6 ŷ ] N C or.35x0 6 x 0.78x0 6 ŷ N C Then to get the force, multily the electric field by the charge at oint. Notice that electric fields can get retty large. But, a Newton of force from electrostatic charges is also retty large.
06 Physics 50: Worksheet 0 Name () A oint charge q -3.00 C is located at x0. A second charge q +6.00 C is located at x.00 m. Find a oint other than infinity where the electric field is zero. The electric field is defined by: where reresents a oint in sace. n F k q q i i r i r i i We locate the initial charge at r 0 x and the second charge at r x. The vector ointing to is given (in two dimensions) by: r x x+y ŷ The electric field at any oint in -D sace is then given by: k q r r +k q r r We can easily form each of these vectors now. r r r x x+y ŷ 0 x 0 ŷx x+y ŷ: r x x+y ŷ x +y r r r x x+y ŷ x 0 ŷ(x ) x+y ŷ : r ( x ) x+y ŷ (x ) +y We now form the electric field: x k q x+y ŷ (x +y ) x +y +k q ((x ) +y ) (x ) x+y ŷ (x ) +y We need to solve this for E 0 In this articular roblem, we notice q -q. Thus: x x+y ŷ 0q (x +y ) q (x ) x +y ((x ) +y ) x+y ŷ (x ) +y 0 x x+y ŷ (x +y ) x +y (x ) ((x ) +y ) x+y ŷ x x+y ŷ (x +y ) x +y ((x ) +y ) x x: x +y y ] 3/ (x ) (x ) +y ] 3/ (x ) +y (x ) x+y ŷ (x ) +y y ŷ : x +y ] 3/ (x ) +y ] 3/ It is easy to see that the y equation is satisfied with y 0. This could have also come from symmetry (in ast versions of this roblem I merely assumed this).
06 Physics 50: Worksheet 0 Name x x ( x ) 3 (x ) ] Use this in the x-equation: 3/ x x ] x 3 (x ) x ±(x ) +: x x x ( ) 0 x ( ) : x x + x (+ ) 0 x ( + ) The two solutions are thus obtained. Let me confirm that E is zero for each of these oints. If it s not we ll need to discard one solution. On to the actual answer for the solutions: if: x 0.44, the electric field is not zero. + if: x -.44 - the electric field is zero. You can check these results using the sreadsheet on our website by choosing a test charge of + located at each of the two solutions for x. I have rovided the sreadsheet showing the zero solution for you. What this sketch above shows is that between the two charges, it is imossible to have zero electric field (solution is between the two charges). Solution however is in a region where it is ossible to have zero electric field. The only other region where you might consider is outside (at ositive x). However, you will never get zero there because the +6 charge is not only larger than the -3 charge, but it is also always closer to the oint in that region. Now, how else could this be solved? This is a quicker (and not nearly so mathematically clean here): You could say this: the electric field at a oint along the x-axis is: k q r +k q x x r Now let s lose the vector notation. This will increase the number of incorrect solutions but that s the rice you ll need to ay for this. Thus, we have: q x x x x q 0 or q + q 0 here, q -q so this gives: + 0 or 0 x x x x We don't need to mess with the first result. It won't give real x values. Solving the second equation then gives: x ( x ) x (x ) x x The rest of the solution roceeds as before.
06 Physics 50: Worksheet 0 Name x (x ) x ±(x ) +: x x x ( ) 0 x ( ) : x x + x (+ ) 0 x (+ ) The following roblem is quite imortant. Be sure you understand it. (3) The electric diole consists of a ositive and a negative charge searated by a distance of a. Suose in this case, your diole had +q at xa and -q at x-a. Find an exression for the electric field along the y-axis. You should then be able to show that the electric field behaves as E x kqa y 3 at distant oints along the y-axis. We begin with the definition of the electric field: k i r i r r i Now we need to obtain the various vectors involved. r a x : r a x : r y ŷ in r r r a x+y ĥ: r r r a x+y ŷ r r r a x+y ŷ : r a r +y r a x+y ŷ a +y Now we need to use these in the definition of the electric field. k q r r +k q r r so here: E k q i q a x+y ŷ a +y a +y +k q a x+y ŷ a +y a +y We thus have: (letting q be the same magnitude as q but of oosite sign): kq a +y a x+y ŷ a +y a x+y ŷ ] a +y a +y E kq a +y a x ] a +y kqa (a +y ) 3/ x This is the actual answer. Now let s look at how this behaves for y>>a. The exansion is: (±x) n ±nx+ n(n ) x ± n(n )(n ) x +...;x <! 3! Also note: (±x) n nx+ n(n ) x n(n )(n ) x +...;x <! 3! What you want to do is to divide by what is big. Here, that would be y.
06 Physics 50: Worksheet 0 Name kqa y + ( a ] 3/ 3 y ) x kqa y 3 3 ( a y ) +...] x kqa x y 3 The electric field at large distances along the erendicular bisector of the diole is: kqa x y 3 Both of these results are extremely imortant for systems involving electric dioles! It is also indeed very interesting to see that the diole falls off as /y 3 at large distances. The term qa is called the magnitude of the electric diole moment. With the aroximate field above, we can define the diole moment for equal and oosite charges with charges and locations: (+q;+a,0,0) and ( q; a,0,0) as: q j r j qa x+( q)( a x)q (a x )qd j where d is the vector ointing from the negative charge towards the ositive charge. With this, the electric field along the erendicular bisector of the electric diole becomes: k y 3 This is extremely imortant because it defines the electric diole. Students are warned that many (about ½) of the undergraduate general chemistry text books get this wrong. Now, what if we have the same diole but wanted to find the field at an arbitrary location in the x-y lane. r a x: r a x: r x x+y ŷ r r r (x a) x+y ĥ: r r r (x +a) x+y ŷ r r r ( x a) x+y ŷ kq (x a) +y (x a) +y : r r ( r x +a ) x+y ŷ (x +a) +y (x a ) x+y ŷ (x a ) +y (x +a ) x+y ŷ (x a) +y ] (x +a) 3/] +y (x +a ) +y ] kq ( x a ) x+y ŷ ( x +a ) x+y ŷ (x a ) +y ] 3/
06 Physics 50: Worksheet 0 Name kq ( x a ) x+y ŷ ( x +a ) x+y ŷ (x a ) +y ] 3/ (x +a ) +y ] 3/ ] To get the behavior at large distances, Note that if your diole is not located at the origin, then do a coordinate translation (and maybe a rotation) to ut it there. However, with the more strict formulation, this is not essential so let's see now what the electric field is for two equal but oosite charges located at two random oints in the x-y lane. :(+q,x, x )::( q,x, y )::(x,y ) r x x+y ŷ: r x x+y ŷ: r x x+y ŷ r (x x ) x+(y y )ŷ: r (x x ) x+(y y )ŷ kq (x x ) x+(y y )ŷ (x x ) +(y y ) ] 3/ (x x ) x+(y y )ŷ (x x ) +(y y ) ] 3/ ] At this oint, ste by ste, you become more restrictive but the result above is a general result. So let both charges lie along the x-axis. With this restriction, we have: kq (x x ) x+(y )ŷ (x x ) +(y ) ] 3/ (x x ) x+(y )ŷ (x x ) +(y ) ] 3/ ] kq (x a) x+(y )ŷ (x +a) x+(y (x a) +(y ) ] 3/ )ŷ ] (x +a) +(y ) ] 3/ r kq a r r a+a ] r + a 3/ let r a kq r 3/ kq r 3/ r a ]3/ r a + r ( a r ) r a cos ar + ( a (±x) n nx+ n(n )! r ) r + r a+a ] 3/ ] ]3/ + r + a ; aa x r a + r ( a r ) r + a ]3/ ] + cos ar + ( a r ) ]3/ ] x n(n )(n ) x +...;x < 3!
06 Physics 50: Worksheet 0 Name kq r ( r 3/ a) 3 ( a cos r + ( a r ) k r 3/ To lowest order: ( )] q a+q r 6a cos 3 r ( a r ) k r ( +q r 6a 3/ k So r 3/ (3 r ) r ] )] ( r + a ) 3 ( a r cos + ( a r cos ) ] k r ) )]] r 3/ (3 r ) r ] Essentially this aroximation of the diole has reduced the diole down to the extent that the distance between charges is insignificant comared to the oint in sace. It may not be valid for distributions where the net charge is not zero and also in the case that you are close to the diole, relative to the distance between the charges. The imortant thing to kee in mind here is this: the diole moment is only one of the terms in the multiole exansion and usually it is the redominate term. For a continuous, and electrically neutral charge distribution, the diole moment is calculated as: all charges r i ρ(r i )d 3 r i, which is certainly an untidy way to write this but it contains the exlicit understanding that r i oints towards satial regions containing charges that are art of the charge distribution, but kee in mind that discrete charge distributions and continuous charge distributions are different entities entirely.
06 Physics 50: Worksheet 0 Name (4) Suose in this case, your diole had +q at xa and -q at x-a. Find an exression for the electric field along the x-axis at x>a. You should then be able to show that the electric field behaves as E 4kqa x k at distant oints along the x -axis. 3 3 x x In the revious roblem, we had: kq (x a) x+(y )ŷ (x a) +(y ) ] 3/ (x +a) x+(y )ŷ (x +a) +(y ) ] 3/ ] Here, allow y to be zero. then this simlifies to become: Consider the case where x a>0 : (x kq a) x (x +a) x (x a) ] 3/ (x +a) ] 3/ ] E kq (x a) ] kq x x (x +a) (x a) (x +a) +x a+a x +x a a ] x kq k 3 3 x x (x a ) (x +a) 4x a] x 4kq a x You could also use the result above: E k r 3/ (3 r ) r ].In this case: E k x 3 Remember, however, our exression for the diole, E k r 3/ (3 r ) r ] ]., is really only valid for r>>a whereas doing the exact calculation is always valid (since it is without aroximation). This means that you can not always start with the field for the diole to reresent any diole you run into! However at those times when you are in the correct region for aroximation, it is aroriate to use this result.
06 Physics 50: Worksheet 0 Name (5) Suose that you have a ring of radius ra and total charge Q located in the x-y lane. What is the electric field for oints along the symmetry axis of this ring? How does this field behave along the axis at distant oints along the symmetry axis? Non-calculus version This icture showing this articular situation is to the left. The symmetry of the roblem allows me to say that the only comonents of the electric field which survive will lie along the z-axis (i.e. the offaxis comonents of the electric field cancel) at oints along the z- axis. In this case, then, we have Etotal j E j E j r j. j The angle is the same no matter where on the ring you look from the symmetry axis (at a fixed z ). Also, the distance from the ring to the oint z is the same for every oint along the ring. To determine the electric field, write the charge on the ring in terms of the charge density on the ring. If you consider that the ring has a total length given by: π a, the total charge Q on the ring is given by: Q( π a) λ where I am reresenting a linear charge density here by λ. The electric field from a very small section at the to of the ring is given by: E j+ kq j z ẑ aŷ where the subscrit + means I ve icked the oint from (a +z ) z +a the to of the ring. The electric field coming from a oint on the bottom of the ring (exactly oosite from the revious osition) is given by: E j- k q j z ẑ+a ŷ (a +z ) z +a If I add these two electric fields, I get the result: E j+ + E j- kq j a +z ] z 3/ ẑ Now you need to determine how many such charge airs there are on the ring. If you let the small charge q j be reresented by q j λ (a(δ ϕ)) where Δ ϕ reresents a small angle, then we can rewrite the electric field as: E j+ + E k λ a (Δϕ ) j- a +z ] z 3/ ẑ If we now let Δ ϕ reresent ½ of the total angle of the ring (which is π and remember, I m adding u charge airs here), the electric field from the entire ring becomes: k λ a(π) E j+ + E j- a +z ] z 3/ ẑ In terms of the total charge Q laced uon the ring, we thus have: E j+ + E j- kq z a +z ] ẑ 3/ Let s also look at how this behaves as x gets large. (a +z ) 3/ 3/ 3 z ( +a z z ) 3( + 3 a z +... ) z 3 At large distances, E k Q ẑ (the ring looks a lot like a oint charge). z
06 Physics 50: Worksheet 0 Name Calculus version For calculus eole, this is your first examle of how to integrate over a continuous charge distribution. I ll do it quite directly without looking at symmetry. The differential charge density is given by: dqλ adϕ with the charge density defined as reviously. You can verify that integrating over this charge density gives you the total charge if λ Q πa We find the electric field by: de all charges Now the magnitude of the electric field arising from dq is given by: d E k dq z +a ] The vector ointing towards a charge location is given by: r i x i x+y i ŷa cos(ϕ) x+a sin(ϕ)ŷ Note: How did I get this? Look at the icture and remember that when you convert from Cartesian coordinates to olar coordinates, the transformation is: xa cos ϕ & ya sin ϕ where I am using to reresent the olar angle in the x-y lane. It is very imortant also to notice that I ket the unit vectors in Cartesian coordinates. This is a rule you do not want to break when integrating over unit vectors since it is only the Cartesian unit vectors which are constant in sace! The vectors that we need are given by: r i r r i acos(ϕ) x asin(ϕ)ŷ+z ẑ r i acos(ϕ) x asin(ϕ)ŷ+z ẑ z +a I am now ready to calculate the integral: ϕ π k ϕ0 λ adϕ acos(ϕ) x asin(ϕ)ŷ+z ẑ ] z +a ] 3/ The x and y comonents then integrate to zero and what remains is then: k(π)a λ z z +a ] 3/ ẑ kq z z +a ] 3/ ẑ Do study the stes that I ve used to do this roblem. If you follow these stes in this way, the roblem of integration over continuous charge distributions will be straight-forward. The most difficult art of the roblem is actually setting u the differential charge distribution. In advanced Electricity and Magnetism, there is a mechanical method of doing this which would take us too far astray for now. Here is a nice alication of what you have learned that also ties some things together!
06 Physics 50: Worksheet 0 Name Suose you have a crystal which has two ositive charges located as shown and an electron is located along the symmetry axis between the two charges at a distance z from the center which is very small comared to a. Let s see what haens. This roblem is unlike the diole roblem in that each of the charges is the same. However, looking at the non-calculus aroach to the ring roblem (roblem 5), it is immediately aarent what the electric field is along the symmetry axis. The electric field is given by: E j+ + j - k(q j ) a +z ] z z +a ẑ kq z a +z ] ẑ 3/ Now we re going to look at this exression in the limit that z a. We again use the binomial exansion but we need to rewrite 3 the denominator slightly. a +z ] 3/ a ( + z ]3/ a ) 3( a + 3 ( z a ) +...) The leading term is then a 3 which gives us the aroximate electric field at the center as: keq z ẑ a 3 Now let s find the electrostatic force on the electron which is traed in such a situation. This is easily seen to be given by: F(q electron ) E e E keq z a 3 ẑ This force is linear in the dislacement variable and restoring. If you comare this force to the Hooke s law force ( F Κx x) then you would exect to see the electron oscillate with simle harmonic oscillation and thus would have a frequency given by: ω Κ keq f keq m e a 3 m 3 π a 3 m e You often hear that molecules act like srings connected to masses but this really shows the effect. The electron will oscillate (and thus, it will store energy). The roblem is that this is a classical calculation. It is, however, very easy at this oint, with a little bit of quantum mechanics to obtain an energy sectrum for the electron traed between two ositively charged ions like this! The energy sectrum will be given by: E n ħ ω ( n+ ) ;n,,3,...;ω Κ m e Here, you also see Planck s constant which is given by: ħ h π.0546x 0 34 J s
06 Physics 50: Worksheet 0 Name Here is another nice alication related to the electric diole. Suose that we aly a uniform electric field along the y-axis of the diole in roblem 4. Assume an external electric field is given by: EE ŷ τ + r + X F + The angle between and E is. The angle between E and is. These two angles are related by: ϕ+ 360 0. The angle between the ositive x-axis and P is. The coordinates of the charges are: r + acos( ) x+a sin( )ŷ: r - acos( +80 0 ) x+a sin( +80 0 )ŷ The torque on the ositive charge is given by: x ŷ ẑ acos( ) asin( ) 0 0 E q 0 x(0) ŷ(0)+ẑ ( E q acos( ))E q acos( )ẑ Since cos(β) cos(β+80 0 ), and the fact that the negative charge has a negative sign, the torque from the negative charge is the same as for the ositive charge. τ - τ + Γ Ea q cos( )ẑ E cos( )ẑ Now, there is also another connection: +90 0 90 0 360 0 ϕ 360 0 +90 0 450 0 ϕ 450 0 ϕ We thus have the net torque on the diole given as: Γ E cos(450 0 ϕ)ẑ E cos(450 0 )cos(ϕ)+sin(450 0 )sin(ϕ) ]ẑ Γ E sin(ϕ)ẑ X E where the angle is measured starting with the ositive axis and rotating around in the ositive manner (counterclockwise). On the other hand, if you want to relate this to the angle which starts along the Positive E direction and rotates counterclockwise towards, then you have sin(ϕ)sin(360 0 )sin(360 0 )cos( )sin( )cos(360 0 ) sin( ) Thus the torque is given by: Γ E sin( )ẑ X E Now here is why I worry so much about the sign of this torque: if the sign is wrong, simle harmonic oscillation won t result from the analysis below. In articular, you want to fix yourself onto the electric field vector and watch the diole oscillate about your reference frame, rather than fixing yourself on the diole and watching the electric field oscillate. According to Newton s laws, we have that a torque roduces an angular acceleration: ΓIα So the equation of motion is given by: E E sin( )Iα α sin( )0 I Calculus students write this as:
06 Physics 50: Worksheet 0 Name d E + dt I sin( )0 Now if you consider only small angles, then: sin( ) This means that simle harmonic oscillation will result with a frequency of oscillation given by: ω E E qae I ma ma qe π f f qe ma π ma As before, the energy sectrum of the oscillating diole would be quantized and thus: E n ħ ω ( n+ ) ;n,,3,...;ω qe ma Here, you also see Planck s constant which is given by: ħ h π.0546x0 34 J s This is yet one more examle of where concets from the first semester are very imortant in the second semester of hysics for a more comlete icture. Incidentally, you ll also need to know something about the electric olarization. The electric olarization of a material P is defined as the diole moment er unit volume of the material. This can be difficult to calculate but it is a vector quantity. You can also calculate the work required to orient a diole from some angle (as I have defined it above) to the x-axis to some angle (where 0). 0 W 0 Γd Ecos( )d E sin( )] 0 Esin( ) Since 90 0 +90 0 sin( )sin( +90 0 )sin cos(90 0 )+sin(90 0 )cos cos( ) we can rewrite this result in terms of the dot roduct. Thus, in terms of the angle between E and, we have: U E Which would corresond to the energy of a diole in an external electric field. This is imortant classically for a lot of dioles in an external electric field. You can do an average over angles using Boltzman statistics to obtain an average angle (this leads to an equation known as the Langevin equation). Note that the negative sign insures that when the diole moment is anti-aligned with the electric field, the energy is at a maximum.