electrical forces, coulombs law
electrical forces, coulombs law phys131
Popular in physics 131
verified elite notetaker
verified elite notetaker
verified elite notetaker
verified elite notetaker
verified elite notetaker
verified elite notetaker
Popular in Physics 2
This 9 page Class Notes was uploaded by James Lin on Monday February 15, 2016. The Class Notes belongs to phys131 at University of Maryland taught by Dr. Redish in Fall 2015. Since its upload, it has received 58 views. For similar materials see physics 131 in Physics 2 at University of Maryland.
Reviews for electrical forces, coulombs law
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 02/15/16
Effect of molecular shape on electric forces (solution) 4.2.4.P11 The electric attractions and repulsions of the ions on the surface of co mplex molecules play a critical role in many biochemical reactions. An i nteresting component of some of these processes is that chemical reactions or elec tron transfers result in a change of the shape of a molecule, exchanging chemical energy for electrical energy, and resulting in a different surface configurati on and therefore different interactions. In this problem we will explore the effect of a change of configuration of a molecule on the force between it and an ion in a neighboring molecule. Consider the electric force between four ions on the surface of two larg e molecules as shown in the figure at the right. The heavy black lines running out of the frame indicate bonds to other atoms in the mol ecule that we will not consider here. The purple ion in the molecule on the right has a charge +e, the red ion in the molecule on the left h as a charge -2e, and the two green molecules in the molecule on the left each have charges +e. A schematic diagram of these ions is shown o n the left below indicating relative distances. A.Qualitatively (without doing a calculation), would you expect the fo rce of the ions (ABC) on ion D to be attractive or repulsive? Explain your reasoning. I would expect if to be attractive. The three forces that the charges in the molecule are exerted on D is shown in the figure at the right. The attractive force from A is twice as big as each of the forces from B and C because the charge is bigger. But the forces from B and C are als o smaller than the force from A since they are farther away. Even more, the forces from B and C are not in the same direction, so they should partially cancel each other. As a result, A should definitely pr oduce a stronger (attractive) force than both B and C produce (repulsive) taken together. B. Now do the calculation to find the net force between (ABC) and D. E xpress your result as a multiple of the combinationF =k e /d . 0 c The force between any two charges is given by Coulomb’s law, with the direction along (or opposite to) the line between the two cha rgesS.ince A has charge -2e and D has charge e, the force between A and D is The magnitudes of the other two forces, shown in blue, they only have fa ctors of 1 in the charge (both have magnitude e), but their distance i s larger. By the Pythagorean theorem they are a distance of square root of 2 further away. Since Coulomb's law tells us that the magnitude of the force fall s as the square of the distance, be get http://umdberg.pbworks.com/...06/Effect%20of%20molecular%20shape%20on%20electric%20forces%20%28solutio n%29?mode=embedde[d 3/23/2016 8:39:03 AM] But these are vectors and the don't point in the same direction. So we c an't just add them. One way is to complete the parallelogram as shown in the figure at the right. Since t he angles are all right angles, and the sides are the same, this is a square. The length of the diagonal is squ are root of 2 times as big as the sides, so the sum of the two vectors points horizontally to the right and has a magnitude o √f2 F0/2. The result is therefore the difference -- an attraction pointing left with magnitude Fnet = 2F 0 √2F /0. = (2- √2/2) F = 0.29 F 0. (You can also break the two vectors up into vertical and horizontal com ponents. The vertical ones cancel, the horizontal ones add.) C. As a result of an electron being added to another part of the molecul e on the left, the configuration of the three ions (ABC) changes so that it looks like the figure on the right, drawing ion D in a bit closer as shown. Without doing the calculation, will the force that (ABC) exerts on D increase, decrease , or remain the same as it was in part B? Explain your reasoning. The force will definitely be more attractive. Now D is closer to A so the attraction is stronger, and the forces from B and C now cancel exactly since it is directly in between the two charges. Joe Redish 10/25/15 http://umdberg.pbworks.com/...06/Effect%20of%20molecular%20shape%20on%20electric%20forces%20%28solutio n%29?mode=embedde[d 3/23/2016 8:39:03 AM] Forces between charges (solution) 4.2.4.P1 Two small objects each with a net charge of Q (where Q is a positive nu mber) exert a force of magnitude F on each other. We replace one of the objects with another whose net charge is 4Q. (i) The original magnitude of the force on the Q charge was F; what is the magnitude of the force on the Q now? (b) Coulomb's law tells us that the force between two charges is so the force between the two is proportional to the product of the charg es. (ii). What is the magnitude of the force on the 4Q charge? (a) 16F (b) 4F (c) F (d) F/4 (e) other (b) The magnitude of the forces that the charges exert on each other are eq ual by Newton's 3rd law. And we note that the form of Coulomb's law satisfies this. Changing which charge you are paying attention to only switches q and Q. Since qQ = Qq, the magnitudes are the same. (iii) Next, we move the Q and 4Q charges to be 3 times as far apart as they were. Now what is the magnitude of the force on the 4Q? (a) F/9 (b) F/3 (c) 4F/9 (d) 4F/3 (e) other (c) Before moving them farther apart, the force is 4F. The force is proporti onal to the inverse of the distance squared. So if we move them 3 times farther apart, the force is reduced by a factor of 1/3 = 1/9. (iv) In the original state (2 charges Q) if the symbol Q were taken to have a negative value,w would the forces change compared to the original state? (a) stay the same (b) both would reverse (c) left one would reverse (d) right one would reverse (e) none of the above. (a) The product of two negative numbers is a positive number so the product of qQ is the same for both q and Q positive or both q and Q negative. We also know that "like charges repel" - whether they are both positive or both negative. * From the CSEM. http://umdberg.pbworks.com/w/page/102264094/Forces%20between%20charges%2 0%28solution%29?mode=embedde[d 3/23/2016 8:33:37 AM] Pushing a block on a block - quantitative (solution) 4.2.2.P6 A heavy block, labeled “A”, is sitting on a table. On top of that block is a lighter block, labeled “B” as shown in t he figure at the right. Consider three cases. (A) The finger is pushing but not hard enough. Neither block moves. (B) The finger is pushing hard enough that the two blocks are speeding up. (C) The blocks have sped up, and are now moving at a constant speed. The finger still h as to push to keep them going at a constant velocity. In all cases where the blocks are moving, they are moving together. (Bl ock B is not sliding on block A.) Block A has a mass of 0.6 kg; block B has a mass of 0.1 kg. the coefficient of friction between the block and the table is 0.3 and the coefficient of friction between the two blocks is 0.2. (You may use g = 10 N/kg and you may treat kinetic and static friction as the same.) 1. For case (A), what is the maximum force the finger can exert withou t the blocks beginning to move? From a FBD for block A, we see that the only horizontal forces acting on A are the push of the finger and friction from the table. (Block B doe sn't have any way to know that block A is being pushed so their surfaces are not trying to slide over one another and there will be no A-B friction force .) max Since the block is not yet moving, the friction force is static. The max imum static friction force between block A and the tabAT is = μ AT AT where μAT is the coefficient of friction between the block and the table (0.3) dAT is the normal force squeezing the two surfaces together. Since the downward forces on the block A include a normal force from B e qual to the weight of block B and the force of gravity on block A (= th e weight of block A), the Normal force from the table on A (the only upward fo rce) must be equal to the sum of the weights of the two blocks: N = W + W = m g + m g AB A B A B The maximum force the finger can apply without the blocks moving is equa l to the maximum static friction the block feels from the table, so Fapp = f max= μ N = μ (W + W ) = (0.3)(0.6 kg + 0.1 kg)(10 N/kg) = 2.1 N AT AT AT AT A B 2. For case (B), while the blocks are speeding up, the finger is pushi ng with a force of 3.C.an you find the acceleration of block A? If so, find it and show your work. If not, explain why not. net Yes we can. We can treat both blocks as a unit, so the problem becomes e asier. We just have to write the a = F/m equation for the combined object. (You get the same result if you write the F=ma equation for each block . You then just have to do a couple of substitutions to get the same res ult.) The net force on the blocks is the force of the finger pushing it forwar d minus the force of the friction from the table trying to prevent the s urfaces from sliding over each other. Our equation is therefore net 2 a = F /m = (3.5 N - 2.1 N)/(0.7 kg) = 2 m/s . Since we are treating static friction and kinetic as the same, the slidi ng friction force is the same as the maximum static friction force. 3. Find the time Δt needed for this force to bring the block from rest to a speed of 10 cm/ s. In this situation we have the acceleration and the change in speed, so t he equation for acceleration will suffice. Since the force is assumed co nant, will be constant and we get a = Δv/Δt or Δt = Δv/a. Since vi= 0 and v f 0.1 m/s, Δv = 0.1 m/s and 2 Δt = Δv/a = (0.1 m/s)/(2 m/s ) = 0.2 s. 3. For case (B), we are assuming that the blocks are moving together. This means that block B is speeding up. But the finger is not touching b lock B. What force is responsible for the acceleration of block B? Find its magnitude and direction. Is our assumption correct? Will block B indeed be able to spe ed up with block A without slipping? Explain. http://umdberg.pbworks.com/...7/Pushing%20a%20block%20on%20a%20block%20-%20quantitative%20%28solution% 29?mode=embedde[d 3/23/2016 8:36:59 AM] In order to accelerate together with block A, block B needs a net force pushing it to the left of Fnet= m a = (0.1 kg)(2 m/s ) = 0.2 N B The only horizontal force on block B is the friction force between block s A and B. The maximum value of this force is given by fABmax= μ AB AB = μAB WB= (0.2)(0.1 kg)(10 N/kg) = 0.2 N Just enough (given the approximation for g you were told to make). 4. Suppose the bottom block has a mass of 0.4 kg and the top block has a ma ss of 0.1 kg What force do you need to exert to keep the blocks moving a t a constant speed of 10 cm/s? In this case, assuming it's still moving in one direction, a constant sp eed implies a constant velocity, which implied an acceleration of 0. If the acceleration is 0, Newton's 2nd law tells us that F= 0, so the push of the finger needs to exactly balance the friction fo rce with the table (the only other external horizontal force acting on the combined blocks). F app= f = μ N = μ (W + W ) = (0.3)(0.4 kg + 0.1 kg)(10 N/kg) = 1.5 N AT AT AT AT A B Joe Redish 10/25/15 http://umdberg.pbworks.com/...7/Pushing%20a%20block%20on%20a%20block%20-%20quantitative%20%28solution% 29?mode=embedde[d 3/23/2016 8:36:59 AM] Repelling pith balls (solution) 4.2.4.P16 Two styrofoam spheres having a mass of about 1 g each are charged by add ing some excess electrons to each sphere. They are then suspended from a rod with two insulating strings. They are observe d to hang at equilibrium as shown in the figure on the right, which is drawn approximately to scale. Estimate the amount of charge that is found on each sphere. Let's estimate that the charges we have placed on each sphere is the same and etl Q. They might be different, but the spheres are identical and we presume we have rubbed them the same way, so they are likely to have similar cha rges. How can we figure out how big the charges are? Let's look first at the p icture. The balls are not hanging straight down. Why not? Because there is an electric force between them pushing them outward. P resumably, the stronger the electric push, the farther out the balls would hang, so the angle must tell us something about the str ength of the electric force. If we knew the electric force and if we knew the distance between the balls, we could calculate the electric charge from the electric force law, F = k Q /d . We can C estimate the angles and the distance from looking at the figure. Theref ore, our chain of reasoning is as follows. Figure out the angles and distances from looking at the picture. Figure out the magnitude of the electric force by using a free-body diag ram and the condition that the force on each ball must balance. Figure out the magnitude of the charge from using Coulomb's law. From the figure, we can estimate that the distance d is about 9 cm or (for estimation purposes) ~0.1m. This means that the sid es of the triangle with the angle marked θ are about 3 cm [= (9 cm - 3 cm)/2] and 10 cm. (I did this by eye. Better estimates might be done by measuring with a ruler, but since we are only estimating -- that is, we only want accuracy to on e significanftigure -- this should be OK.) This gives us that tanΘ = opposite/adjacent = (3 cm)/(10 cm) = 0.3 We can get the hypotenuse by the Pythagorean theorem to be . From this we can get the sine and cosine of the angle as well (sin = 3/10.5 = 0.29 and cos = 10/ 10.= 3 0.97). We don't actually need to be this accurate for an estimation problem. It is enough to estimate - by eye - that the vertical component is somewhere between 2 and 4 times the horizontal one. Now we need to create a free body diagram and balance the forces. This is shown in the figure at the right. We don't know either the tension nor the electric force, but by balancing the up force (the vertical component of the tension) against the down (the weight) and the left force (the horizontal component of the tension) against the right (the electric force) we get two equation for our two unknowns and can solve for everything. It works like this. Since we know the mass of each sphere is 1 g = 0.001 kg, the weight is http://umdberg.pbworks.com/w/page/101597848/Repelling%20pith%20balls%20% 28solution%29?mode=embedde[d 3/23/2016 8:38:01 AM] W = mg = 0.001 kg x 10 N/kg = 0.01 N. The tangent of the angle is about 0.3 so our electric force must be abo ut 0.003 N. Assuming both charges are equal tQ o, and approximating d as 10 cm, we get -7 Taking the square root, we get that the charge is about 0.5 x 10 C, or about 50 nanoCoulombs. Joe Redish 10/8/15 http://umdberg.pbworks.com/w/page/101597848/Repelling%20pith%20balls%20% 28solution%29?mode=embedde[d 3/23/2016 8:38:01 AM] With a grain of salt (Solution) 4.2.4.P2 Our knowledge of electric forces between charges gives us the opportunit y to begin to understand the forces that hold matter together. As a simple example , consider a crystal of salt - Sodium Chloride (NaCl). Each molecule of salt separ ates into a + - positive sodium ion (Na ) and a negative chloride ion (Cl ). In a crystal of salt, these ions are arrayed alternately in a three-dimensional cubical array as sh own in the figure at the right. We ask whether or not the electrostatic forces bet ween the ions might be able to hold it together. To simplify our analysis for the purpose of this problem, consider a single plane of this 3D structure. Let's look at an edge of one of these planes, shown in the figure at the left. We will analyze whether in the indicated arrangement, the net electrostatic force on a charge on the e dge of the crystal is into the crystal (holding the crystal together) or out of the crystal (tending to break the crystal up). (a) First consider the force between an ion on the edge and one of its neigh bors on the edge, for example, the sodium ion labeled A and the chloride ion labeled B. If the charge o n the ions each have a magnitudee = 1.6 x 10 -19C and the lattice spacing is d (the distance between neighboring ions along the indicated F . Express your lines), write an expression for the magnitude of the force between nei ghboring ions,0 answer in terms of the symbols e, d, and whatever constants you require The magnitude of the force between the ions is given by Coulomb's law: (b) Now consider the attraction on the sodium ion at A from its three nearest chloride neighbors, labeled B, C, and D. Find the magnitude and direction of the net force on A due to these three charges. Express your answer in terms of F for ease of interpretation. 0 Now we have to pay attention to direction. But the magnitudes for each of the forces F B→A = F C→A = FD→A = F0since each ion B, C, D is the same distance (d) from A. The arrangements of forces is shown below. We can see that the force from C on A and B on A are equal and opposite so they will cancel. This leaves the net force on A from these ions as equa l to F pulling inward as shown. 0 (c) Then, consider the repulsion on the sodium ion at A from itstwo nearest sodium neighbors, labeled E and F. Find the magnitude and direction of the net force on A due to these two charges. Again, express your answer in terms of F for ease of interpretation. 0 The forces from E and F are as shown in the figure at the left below. The distance between A and the ions E and F is the diagonal of a square of side, http://umdberg.pbworks.com/w/page/84540157/With%20a%20grain%20of%20salt% 20%28Solution%29?mode=embedded[3/23/2016 8:42:17 AM] so it is the square root of two times d. This means that the reciprocal of the distance squared will yield a factor of 2 in the d enominator. So the result will be half as big as F . But they are not in the same direction. We have to add then as vectors as shown in the figure at the right below . 0 Since the net of these forces is the diagonal of a square of sideF /2, it has the length square root of 2 times as big asF /2, or F /Sqrt(2) = 0.707 F 0 0 0 0 pushing outward as shown. So the force pulling in is bigger than the force pushing out. (d) We only considered nearest neighbors. Why do you think we did this? Do you think this is a reasonable approximation? Give a brief justificat ion for your answer. We did it because the force falls off as the square of the distance so that the farther away ones are not as important individually -- but ther e are a lot of them, so maybe this is not such a good approximation -- just a first suggestion. We really should do a much better (and much harder) calculation (and one in 3D!). Joe Redish 4/8/09 http://umdberg.pbworks.com/w/page/84540157/With%20a%20grain%20of%20salt% 20%28Solution%29?mode=embedded[3/23/2016 8:42:17 AM]
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'