Pset 3 Solutions

Pset 3 Solutions - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2010 Problem Set 3 Solutions Problem 1: Concept Questions. Explain your reasoning. Suppose an electrostatic potential has a maximum at point P and a minimum at point M. (a) Are either (or both) of these points equilibrium points for a negative charge? If so are they stable? Solution: The electric field is the gradient of the potential, which is zero at both potential minima and maxima. So a negative charge is in equilibrium (feels no net force) at both P & M. However, only the maximum (P) is stable. If displaced slightly from P, a negative charge will roll back “up” hill, back to P. If displaced from M a negative charge will roll away from the potential minimum. (b) Are either (or both) of these points equilibrium points for a positive charge? If so are they stable? Solution: Similarly, both P & M are equilibria for positive charges, but only M is a stable equilibrium because positive charges seek low potential (this is probably the case that seems more logical since it is like balls on mountains). Problem 2: Charges on a Square Three identical charges +Q are placed on the corners of a square of side a, as shown in the figure. (a) What is the electric field at the fourth corner (the one missing a charge) due to the first three charges? Solution: We’ll just use superposition: # & ! )3 i Q# Q % aˆ aˆ + aˆ aˆ ( i j j ij E= + + 3( = 1+ 2 2 & ˆ + ˆ 3 ' 4!" 0 % a 3 a ( 4!" 0 $ 2a % $ ' () () (b) What is the electric potential at that corner? Solution: A common mistake in doing this kind of problem is to try to integrate the E field we just found to obtain the potential. Of course, we can’t do that we only found the E field at a single point, not as a function of position. Instead, just sum the point charge potentials from the 3 points: q 1 Q# 1 #Q Q Q& 1& V= + (= * r i = 4!" % a + %2+ ( 4!" 0 i ) j ij 2 a a ' 4!" 0 a $ 2' 0$ (c) How much work does it take to bring another charge, +Q, from infinity and place it at that corner? Solution: The work required to bring a charge +Q from infinity (where the potential is 0) to the corner 1' Q2 $ is: W = Q!V = &2+ ) 4"# 0 a % 2( (d) How much energy did it take to assemble the pictured configuration of three charges? Solution: The work done to assemble three charges as pictured is the same as the potential energy of the three charges already in such an arrangement. Now, there are two pairs of charges situated at a distance of a, and one pair of charges situated at a distance of 2 a , thus we have # 1 Q2 & # 1 Q2 & 1 Q2 # 1& W = 2% +% = 2+ ( ( % ( $ 4!" 0 a ' $ 4!" 0 2 a ' 4!" 0 a $ 2' PS03-1 Alternatively we could have started with empty space, brought in the first charge for free, the second charge in the potential of the first and so forth. We’ll get the same answer. Problem 3: Estimation: High Voltage Power Lines Estimate the largest voltage at which it’s reasonable to hold high voltage power lines. Then check out this video http://web.mit.edu/8.02t/www/materials/ProblemSets/PS03_Video.mpeg care of a Boulder City, Nevada power company. Air ionizes when electric fields are on the order of 3 ! 106 V " m -1 . Solution: In order to answer this question we have to think about what happens if we go to very high voltages. What breaks down? The problem with high voltages is that they lead to high fields. And high fields mean breakdown. The voltage and field for a line charge are: E r = ! 2"# 0 r ; V r = ! 2"# 0 ln R0 r $ V r = E r r ln R0 r () () ( )( ) () () ( ) The strongest field, and hence breakdown, appears at r = R ~ 1 cm, the radius of a power line (that makes the diameter just under 1 inch – it might be 3 or 4 times that big but probably not ten times). The voltage is defined relative to some ground, either another cable (probably R0 ~ 1 m away) or at the most the real ground (R0 ~ 10 m away). So, Vmax = Emax R ln R0 R = 3 ! 106 V " m -1 1 cm ln 10 m 1 cm # 2 ! 105 V ( )( )( )( ) As it turns out, a typical power-line voltage is about 250 kV, about as large as we estimate here. Some high voltage lines can even go up to 600 kV though (or double that for AC voltages). They must use larger diameter cables. By the way, you can tell that breakdown is a real concern. In humid weather (during rainstorms for example) you will sometimes hear crackling coming from the power lines. This is corona discharge, a high voltage, low current breakdown, similar to the crackling you hear from the Van de Graff generator in class. The movie is of an arc discharge, a very high current phenomenon that can be very dangerous. Problem 4: Charged Sphere Consider a uniformly charged sphere of radius R and charge Q . Find the electric potential difference between any point lying on a sphere of radius r and the point at the origin, i.e. V ( r ) ! V (0) . Choose the zero reference point for the potential at r = 0 , i.e. V (0) = 0 . How does your expression for V ( r ) change if you chose the zero reference point for the potential at r = ! , i.e. V ( ! ) = 0 . Solution: In order to solve this problem we must first calculate the electric field as a function of r for the regions 0 < r < R and r > R . Then we integrate the electric field to find the electric potential difference between any point lying on a sphere of radius r and the point at the origin. Because we are computing the integral along a path we must be careful to use the correct functional form for the electric field in each region that our path crosses. There are two distinct regions of space defined by the charged sphere: region I: r < R , and region II: r > R . So we shall apply Gauss’s Law in each region to find the electric field in that region. For region I: r < R , we choose a sphere of radius r as our ! ! Gaussian surface. Then, the electric flux through this closed surface is " EI ! dA = E I ! 4# r 2 . The "" sphere has a uniform charge density ! = Q / (4 / 3)" R3 . Because the charge distribution is uniform, the charge enclosed in our Gaussian surface is given by Qenc "(4 / 3)# r 3 Q r 3 . Now we apply = = !0 !0 ! 0 R3 PS03-2 !Q ! Q r3 which we can solve for the electric EI ! dA = enc . to arrive at: E I ! 4" r 2 = " "" #0 # 0 R3 ! Qr ˆ ˆ field inside the sphere EI = E I r = r , 0 < r < R. 4!" 0 R3 ! Q ˆ ˆ For region II: r > R : We can solve for the electric field EII = E II r = r, r > R. 4!" 0 r 2 We can now find the electric potential difference between any point lying on a sphere of radius r and the origin, i.e. V (r ) ! V (0) . We begin by considering values of r such that 0 < r < R . We shall r != r ! ! calculate the potential difference by calculating the line integral V (r ) " V (0) = " $ EI # dr! ; 0 < r < R Gauss’s Law: We use as integration variable r ! and integrate from r ! = 0 to r ! = r : r #= r r #= r Qr # Qr # Qr 2 ˆ % dr #r = $ & ˆ #=$ ; 0<r < R V (r ) $ V (0) = $ & dr r 4!" 0 R 3 4!" 0 R 3 8!" 0 R 3 r #=0 r #=0 For r > R : we are taking a path form the origin through regions I and regions II and so we need to use both functional forms for the electric field in the appropriate regions. The potential difference between any point lying on a sphere of radius r > R and the origin is given by the line integral expression ! r != R ! ! r =r ! ! ! " $ EII # dr! ; r > R . Using our results for the electric field we get that V (r ) " V (0) = " $ EI # dr r !=0 r != R r !=0 V (r ) $ V (0) = $ r #= R Qr # Q ˆ ˆ ˆ ˆ &0 4!" 0 R3 r % dr #r $ r#&R 4!" 0 r #2 r % dr #r ; r > R This becomes = r #= r #= r r #= r Qr # Q V (r ) $ V (0) = $ % dr # $ % dr # ; r > R Integrating yields 3 4!" 0 R 4!" 0 r #2 r #=0 r #= R r #= R Qr #2 V (r ) $ V (0) = $ 8!" 0 R 3 r #= R r #=0 Q + ; r > R Substituting in the endpoints yields 4!" 0 r # r #= R r #= r Q Q #1 1 $ + & % ' ; r > R A little algebra then yields 8!" 0 R 4!" 0 ( r R ) Q 3Q V (r ) # V (0) = ; r>R # 4!" 0 r 8!" 0 R We plot V (r ) vs. r in the figure. Note that the graph of the electric potential function is continuous at r = R. V (r ) % V (0) = V (r ) % V (0) = % PS03-3 When we set r = ! , the potential difference between the sphere at infinity and the origin is 3Q . If we had chosen the zero reference point for the electric potential at r = ! , V (#) $ V (0) = $ 8!" 0 R 3Q with V (!) = 0 . With that choice, we have that V (0) = . This amounts to just adding the 8!" 0 R 3Q constant to the above results for the potential function V (r ) giving 8!" 0 R # 3Q Qr 2 $ ; 0<r < R % 3 % 8!" 0 R 8!" 0 R . V (r ) = & % Q ; r>R % 4!" 0 r ' In the above expression we can easily check that V (!) = 0 . Equivalently we shift our previous graph up by 3Q / 8!" 0 R as shown in the graph below. Problem 5: Charged Washer A thin washer of outer radius b and inner radius a has a uniform negative surface charge density "! on the washer (note that ! > 0 ). a) If we set V (!) = 0 , what is the electric potential difference between a point at the center of the washer and infinity, V ( P ) ! V (") ? Solution: The potential difference V ( P) ! V (") between infinity and the point P at the center of the washer is given by k (#! )da" V ( P ) # V ($ ) = % ! ! . Choose as an r # r" source integration element a ring of radius r ! and width dr ! with charge dq" = (#! )da" where da" = 2! r "dr " . !! Because the field point P is at the origin r = 0 and the vector from the origin to the any point on the ! ˆ ring is r! = r !r , therefore in the above expression the distance from the integration element, the ring, to 1 1 the field point P is ! ! = . So the integral becomes r " r! r ! k ($! )da# k ($! )2" r #dr # = $ k! 2" (b $ a ) V ( P ) $ V (% ) = & ! ! =& r# r $ r# source r #= a r #=b b) An electron of mass m and charge q = !e is released with an initial speed v0 from the center of the hole (at the origin) in the upward direction (along the perpendicular axis to the washer) experiencing no forces except repulsion by the charges on the washer. What speed does it ultimately obtain when it is very far away from the washer (i.e. at infinity)? Solution: By conservation of energy (note that V (#) $ V ( P) = k! 2" (b $ a ) >0) 0 = #K + #U = #K + q (V ($) % V ( P)) = #K % ek! 2" (b % a ) . If we denote the initial speed of the electron by v0 and the speed of the electron when it is very far 2 2 away by v f then !K = (1/ 2)mv 2 " (1/ 2)mv0 . Hence (1/ 2)mv 2 # (1/ 2)mv0 = ek! 2" (b # a) > 0 . f f PS03-4 We can now solve for the final speed of the electron when it is very far away from the washer 2 v f = v0 + ek! 4" (b # a ) / m . Problem 6: Read Experiment 2: Faraday Ice Pail Consider two nested cylindrical conductors of height h and radii a & b respectively. A charge +Q is evenly distributed on the outer surface of the pail (the inner cylinder), -Q on the inner surface of the shield (the outer cylinder). (a) Calculate the electric field between the two cylinders (a < r < b). For this we use Gauss’s Law, with a Gaussian cylinder of radius r, !! Q 1Q Q height l " E ! dA = 2# rlE = inside = l % E (r )a < r < b = "" $0 $0 h 2# r$ 0 h (b) Calculate the potential difference between the two cylinders: The potential difference between the outer shell and the inner cylinder is a ' b* a Q Q Q dr $ = " ln r $ b = ln ) , !V = V ( a ) " V ( b) = " & 2#% 0 h 2#% 0 h ( a + 2# r $% 0 h b (c) Calculate the capacitance of this system, C = Q/ΔV C = 2"# o h |Q | |Q | = = | !V | $ b' |Q | $ b ' ln & ) ln & ) 2"# 0 h % a ( % a( (d) Numerically evaluate the capacitance for your experimental setup, given: h ≅ 15 cm, a ≅ 4.75 cm 2!" o h 1 15 cm and b ≅ 7.25 cm C = = + 20 pF 9 -1 # b & 2 ) 9 * 10 m F # 7.25 cm & ln % ( ln % $ a' $ 4.75 cm ( ' (e) Find the electric field energy density at any point between the conducting cylinders. How much energy resides in a cylindrical shell between the conductors of radius r (with a < r < b ), height h , thickness dr , and volume 2! rh dr ? Integrate your expression to find the total energy stored in the capacitor and compare your result with that obtained using U E = (1 / 2)C ( !V )2 . The total energy stored 1 1#Q& in the capacitor is uE = ! 0 E 2 = ! 0 % Then 2 2 $ 2" r! 0 h ( ' 1#Q& Q 2 dr . Integrating we find that dU = uE dV = ! 0 % 2" rh dr = 4"! 0 h r 2 $ 2" r! 0 h ( ' U= U= 2 2 ! ! b a dU = ! ! b a Q 2 dr Q2 = ln( b / a ) . From part (d), C = 2!" o h / ln( b / a ) , therefore 4"# 0 h r 4"# 0 h b a dU = b a Q 2 dr Q2 Q2 1 = ln( b / a ) = = C $V 2 which agrees with that obtained above. 4"# 0 h r 4"# 0 h 2C 2 PS03-5 For further study: Electrostatic Shielding Below are three captured images. The center is blanked out so that you can’t see what is going on inside the conductor. For each, you should be able to describe where the charge is (ROUGH angle and distance), tell whether these pictures are field lines (grass seeds) or equipotential streaks (“Electric Potential”) and indicate whether the figure shows the total field, or just the external or induced field. You may want to play with the visualization to try to duplicate these patterns. (a) (b) (c) (a) These are electric fields lines (grass seeds) of the entire field. We can tell because they come in perpendicular to the equipotential surface of the conductor, which is only true for the total field (not the individual parts). The charge is clearly below the conductor (θ = 270º) and just off the screen (R = 11.5). (b) Here the lines are neither perpendicular nor parallel to the conductor, so it can’t be for the entire field. They loop around, looking like a dipole, so they are associated with the induced charges, not the external charge. Are they field lines or equipotentials though? Without seeing the center this is non-trivial. If the charge were below, the field lines would look very much like this. But since the left and right “lobes” are not symmetric, it must be equipotentials created by a charge on the left (R = 6, θ = 180º). (c) This one is easier. The lines wrap around the conductor, so they are clearly equipotential lines associated with the entire field. The charge is on the right (R=11, θ=0º) PS03-6 ...
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This note was uploaded on 10/05/2010 for the course MIT 802FALL taught by Professor Stephens during the Fall '10 term at MIT.

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