Problem Set 2: Solutions

Transcription

1 UNIVERSITY OF ALABAMA Department of Physics an Astronomy PH 102 / LeClair Summer II 2010 Problem Set 2: Solutions 1. The en of a charge rubber ro will attract small pellets of Styrofoam that, having mae contact with the ro, will move violently away from it. Describe why that happens. Solution: The electric fiel from the charge rubber ro will orient the charges in the overall electrically neutral Styrofoam pellets such that the opposite charges attracte closer to the ro, an the like charges farther away. This will cause the Styrofoam pellets to jump towar the rubber ro. Upon contact, charge from the rubber ro will neutralize opposing charges on the Styrofoam pellets, leaving the pellets with a net charge of the same sign as the ro. This will immeiately cause a repulsive interaction, which causes the pellets to move violently away. Say the rubber ro is charge negatively. This will cause the positive charges in the pellets to orient themselves closer to the ro, an the negative charges in the pellets to orient themselves farther away. This causes the pellets to be attracte to the ro. Once in contact, negative charge from the rubber ro will (through contact) move on to the pellets, neutralizing some of the positive charge. After neutralization, this leaves a net surplus of negative charge on the pellets, which will then be quickly repelle from the negatively charge ro. 2. A charge of q an a charge of 5q sit a istance away. Where coul a thir charge of magnitue 2q sit between them an experience no net force? Solution: Without calculating anything, we can first figure where the thir charge shoul be relative to the first two. Let us put the 5q charge at the origin, with the q charge a istance away along the +x axis (i.e., the 5q on the left, the q on the right). To the left of the 5q charge, the electric forces on the 2q charge from both the 5q an q charges woul be repulsive an towar x (left). There is no way that we can have two forces in the same irection cancel, so the 2q charge cannot be to the left of the origin. To the right of the q charge (so, x > ), the force from both charges on the 2q charge woul be repulsive to the right, an again there is no way we can have a net zero force. However, in between the two charges (so 0 < x < ), the repulsive forces are in opposite irections from the 5q an q charges, an cancellation is possible. Further, since the electric force epens on the magnitue of the charges involve, we know we will have to place the 2q charge closer to the q charge than the 5q. Let the 2q charge sit at a istance x from the origin, such that 0 < x <. From the arguments above, we know that we must fin x>/2 so that the 2q charge is closest to the q charge. The istance from the 2q

2 to the 5q charge is now just x, while the istance from the 2q to the q charge is x. The total force on the 2q charge is then just the sum of the iniviual forces from the 5q an q charges, taking into account that they point in opposite irections the force from the 2q charge shoul be negative, since it points towar x. The total force balance must be zero, hence F tot = k e (5q) (2q) x 2 k e (2q) (q) ( x) 2 = 0 (1) Rearranging an canceling, we solve for x, the istance between the 2q an 5q charges. k e (5q) (2q) x 2 = k e (2q) (q) ( x) 2 5 x 2 = 1 ( x) 2 5 x = 1 x 5 ( x) = x ( 1 + ) 5 x = x = (2) (3) (4) (5) (6) (7) Thus, the 2q must be a istance / ( ) from the 5q charge to feel no net force. You can verify that the forces are the same by plugging our result for x back into the expression for F tot above. 3. Three charges q sit at the vertices of an equilateral triangle whose sies are length. What is the net force on each charge? Roughly sketch the electric fiel lines aroun the set of charges. Solution: Here is the situation: F13 F12 F13x F12x 1 F13y F12y F13 30 o 30 o F12 1 /2 /2 2 3

3 Since the structure is symmetric, the force on each of the three charges will be the same each of the three charges is in the same situation as the others. From the geometry, we can see that any given charge will have two repulsive forces on it from the other two charges, an these forces will be irecte at a 60 angle with respect to each other. Consiering charge 1 above, the forces from charges 2 an 3 will be equal in magnitue. Further, from the symmetry of the problem, their x components (horizontal) will cancel, an the net force will be only in the vertical irection. Charges 2 an 3 are each a istance from charge 1, so we can reaily calculate the magnitue of either force if all three charges have magnitue q: F 12 = F 13 = k eq 2 2 (8) The net force along the y irection is simply aing the y components of the two forces: F y,net = F 12,y + F 13,y = F 12 cos 30 + F 13 cos 30 = 2F 12 cos 30 = 2 k eq 2 An, that is that = 3ke q 2 2 (9) Below is a quick calculation of the fiel, using tiny arrows to represent the electric force a positive charge woul feel at various points in space on a square gri (i.e., an arrow for the electric fiel at every point in space). These are not the fiel lines exactly that woul be connecting the arrows with smooth curves flowing out from the charges, an encoing the fiel strength in the ensity of lines.

4 4. An ion milling machine uses a beam of gallium ions (m = 70 u) to carve microstructures from a target. A region of uniform electric fiel between parallel sheets of charge is use for precise control of the beam irection. Single ionize gallium atoms with initially horizontal velocity of m/s enter a 2.0 cm-long region of uniform electric fiel which points vertically upwar, as shown below. The ions are reirecte by the fiel, an exit the region at the angle θ shown. If the fiel is set to a value of E=90 N/C, what is the exit angle θ? v 0 +e E v θ Solution: A singly-ionize gallium atom has a charge of q = +e, an the mass of m = 70 u means 70 atomic mass units, where one atomic mass unit is 1 u= kg. What we really have here is a particle uner the influence of a constant force, just as if we were to throw a ball horizontally an watch its trajectory uner the influence of gravity (the only ifference is that since we have negative charges, things can fall up"). To start with, we will place the origin at the ion s initial position, let the positive x axi run to the right, an let the positive y axis run straight up. Thus, the particle starts with a velocity purely in the x irection: v 0 = ˆx. While the particle is in the electric-fiel-containing region, it will experience a force pointing along the +y irection, with a constant magnitue of qe. Since the force acts only in the y irection, there will be a net acceleration only in the y irection, an the velocity in the x irection will remain constant. Once outsie the region, the particle will experience no net force, an it will therefore continue along in a straight line. It will have acquire a y component to its velocity ue to the electric force, but the x component will still be. Thus, the particle exits the region with velocity v= ˆx + v y ŷ. The angle at which the particle exits the plates, measure with respect to the x axis, must be tan θ = v y Thus, just like in any mechanics problem, fining the angle is reuce to a problem of fining the final velocity components, of which we alreay know one. So, how o we fin the final velocity in the y irection? Initially, there is no velocity in the y irection, an while the particle is traveling between the plates, there is a net force of qe in the y irection. Thus, the particle experiences an acceleration a y = F y m = qe y m

5 The electric fiel is purely in the y irection in this case, so E y =90 N/C. Now we know the acceleration in the y irection, so if we can fin out the time the particle takes to transit the plates, we are one, since the the transit time t an acceleration a y etermine v y : v y = a y t Since the x component of the velocity is not changing, we can fin the transit time by noting that the istance covere in the x irection must be the x component of the velocity times the transit time. The istance covere in the x irection is just the with of the plates, so: x = t = 2.0 cm = t = x Putting the previous equations together, we can express v y in terms of known quantities: x v y = a y t = a y = qe y x = qe y x m m Finally, we can now fin the angle θ as well: tan θ = v y = qe y x m = qe y x mv 2 x An that s that. Now we plug in the numbers we have, watching the units carefully: [ ] [ ( θ = tan 1 qey x mv 2 = tan 1 19 C ) ] (90 N/C) (0.02 m) x ( kg) ( m/s) 2 [ ] = tan N kg m/s 2 note 1 N=1 kg m/s 2 = tan A charge of 1.8 nc sits at the center of a cube. What is the electric flux out of one face? Over the whole surface? Woul your answer change if the charge is not at the center of the cube? Solution: It is easier to answer the secon question first. If the cube encloses the point charge, then the

6 flux through the entire cube must be foun in accorance with Gauss law: i [ ]) N m Φ E,tot = 4πk e q = 4π ( ( [C] ) [ ] N m 2 = 204 C C 2 (10) Back to the first question. If the charge sits symmetrically at the center of the cube, then the flux must be the same over every face of the cube. Thus, each sie takes 1/6 of the total flux, since there are six sies: Φ E,sie = 1 [ ] N m 2 6 Φ E,tot = 33.9 (11) C If the charge were not in the center of the cube, the flux through the whole cube woul be the same, provie that the charge were still fully enclose by the cube. If the charge were partially penetrate by one sie of cube (say, half in an half out) then this woul not be the case, an we woul nee to figure out how much of the charge was insie the cube an how much was outsie. In either case, we coul not say anything very clever about the flux through any given sie figuring that out relie crucially on the charge being symmetrically place with respect to the six sies of the cube, so we coul say they all ha equal flux. If the charge were off-center, closer to one sie than the others, this is no longer true, an our answer to the first part of the question no longer hols. i Don t forget that n means 10 9.