Homework #1 Solutions version 6 problems Exercises 1. Charges A and lie 5 cm apart, on a line. Charge A is q A = + µc, and charge is q = mc. (e careful of the metrix prefixes!) (a) efore doing any calculation, determine the direction of the force on both charges. Which will feel the larger force, A or? (b) Calculate the force (magnitude and direction) on both charges, using Coulomb s Law. +µc A 5cm mc (a) The charges have opposite charges, so they attract one another; the force on A is to the right and the force on is to the left. The magnitudes of the forces are equal, due to Newton s Third Law. (b) We already determined the directions above. The magnitude of the force on either charge is given by Coulomb s Law: F on A = F on = k q Aq r = (9 10 9 N m ) ( 10 6 C)( 10 C) (0.05 m) =. 10 4 N Thus the force on A is. 10 4 N to the right, and the force on is. 10 4 N to the left.. Three charges are fixed on a line as shown. Find the force (magnitude and direction) on charge. +1µC µc 1µC A C 0.01m 0.01m is the target; A and C are the sources. We find the force due to each source, and then add them together. The force on due to A points to the left (attraction), and has magnitude F A = (9 10 9 N ) m (1 10 6 C)( 10 6 C) (0.01 m) = 70 N The force on due to C also points to the left (repulsion), and has magnitude F C = (9 10 9 N ) m (1 10 6 C)( 10 6 C) (0.01 m) = 70 N And so the net force on is 70 N to the left plus 70 N to the left, or 540 N to the left.
. Coulomb s Law is F = k q 1q ˆr where ˆr is the unit vector pointing from the source charge to the target charge (that is, from the forcer to the forcee). The unit vector is defined as the vector from charge to charge, divided by its length: r ˆr = r r so we can alternatively write Coulomb s Law as F = k q 1q r r r = k q 1q r r This is more useful calculationally, but remember that the force decreases with the square of the distance, not the cube of the distance. Suppose a +1 µc charge (charge A) sits at coordinate (1,), and a + µc charge (charge ) sits at coordinate (4,5). (The coordinates are in meters.) (a) Find the vector r from A to. (b) Find the distance r (or r) from A to. (c) Find the unit vector ˆr from A to. (d) Find the force on due to A, in unit-vector notation. (e) Find the force on A due to, in unit-vector notation. 5 4 A 0 1 4 5 (a) The vector r from A to is meters to the right (call that ˆx) and two meters up (call that ŷ), so r = ˆx + ŷ. (b) We use the Pythagorean theorem here: r = + = 9 + 4 = 1. (c) The unit vector of r is ˆr = r ˆx + ŷ = = ˆx + ŷ r 1 1 1 (d) (Note: Unit-vector notation means something like ˆx 4ŷ, or î 4ĵ if you prefer that notation.) In this case, A is the source and is the target, so r is the vector from source to target, as it needs to be in Coulomb s Law. Let s do this in two slightly different ways. First, use F = k q Aq r ˆr. F = k q Aq r ˆr = (9 10 9 N m )(1 10 6 C)( 10 6 C) ( 1) = (.08 10 N)(0.8ˆx + 0.555ŷ) = (1.7 mn)ˆx + (1. mn)ŷ ( ˆx + ) ŷ 1 1 Now let s try the other form of Coulomb s Law F = k q Aq r r
= (9 10 9 N ) m (1 10 6 C)( 10 6 C) ( (ˆx + ŷ) 1) = (5.76 10 4 N)(ˆx + ŷ) = (1.7 mn)ˆx + (1. mn)ŷ Notice how the second way is a little less complicated-looking; you only have to deal with the 1 once, in the denominator. This is what I meant when I said that this second form was slightly easier to use in real calculations, particularly in cases when r does not point along a coordinate axis. (e) Newton s third law says that if feels a force F due to A, then A feels a force F due to. Thus the force on A is (1.7 mn)ˆx (1. mn)ŷ. (Don t feel bad if you went through the whole exercise again with the charges reversed; it was good practice. ut don t forget that Newton s Third Law might save you some time in the future, like on an exam.) 4. In the figure, the four particles form a square of edge length a = 5.00 cm and have charges q 1 = +10.0 nc, q = 0.0 nc, q = +0.0 nc, and q 4 = 10.0 nc. In unit-vector notation, find the force on charge q 1. q 1 q a y x q 4 q The charge q 1 is the target; the other three are the sources. We need to consider each source in turn. The trickiest part is coming up with r (the vector between source and target) for each. q : The vector from q to q 1 is r = aˆx, so r = a and ˆr = ˆx. (I only calculate ˆr here because it s obvious.) Thus F q = k q sq t r ˆr = k q q 1 (9 a ( ˆx) = 10 9 N ) m ( 0 10 9 C)(10 10 9 nc) (0.05) ( ˆx) = 7. 10 4 N(ˆx) q : To get from q to q 1, we go a distance a up, and a distance a to the left, so r = aŷ aˆx, and r = a + a = a. Thus F q = k q sq t r r r = k q q 1 (a (aŷ aˆx) ) a = kq q 1 (a )( (ŷ ˆx) ) (It s always good to simplify as much as possible before substituting numbers in.) With numbers, F q = (9 10 9 N ) ( m 0 10 9 C ) (10 10 9 C) (0.05) (ŷ ˆx) =.5 10 4 N(ŷ ˆx) = (.5 10 4 N)ŷ (.5 10 4 N)ˆx q 4 : The vector from q 4 to q 1 is r = aŷ, so r = a and ˆr = ŷ. Thus F q = k q sq t r ˆr = k q 4q 1 (9 a (ŷ) = 10 9 N ) m ( 10 10 9 C)(10 10 9 nc) (0.05) (ŷ) =.6 10 4 N( ŷ)
The net force on the target is the sum of these three vectors: F = 7. 10 4 Nˆx.5 10 4 Nˆx +.5 10 4 Nŷ.6 10 4 Nŷ = 4.7 10 4 N(ˆx) 1.1 10 4 N(ŷ) Thus F = 4.7 10 4 N(ˆx) 1.1 10 4 N(ŷ) Problems 5. Identical isolated conducting spheres 1 and have equal charges Q, and are separated by a distance which is large compared to their diameters (Fig. a). The electrostatic force acting on sphere due to sphere 1 is F. Suppose now that a third sphere, identical to the other but initially neutral and attached to an insulating handle, is touched first (b) to sphere 1, then (c) to sphere, and then (d) finally removed. The electrostatic force on sphere is now F. Find the ratio F / F. F (a) (b) 1 1 F 1 (c) (d) F 1 F In picture (a), spheres 1 and have charge Q, and sphere has charge 0. In picture (b), charge is allowed to flow from 1 to. ecause the spheres are exactly the same, they should both end up with the same charge. Therefore, after step b, sphere 1 and sphere both have charge Q/. In picture (c), charge is allowed to flow between and. The total charge in the two spheres is Q + Q/ = Q/. That charge, again, will distribute evenly between the two spheres. Therefore, after step (c), sphere has charge Q/4, sphere has charge Q/4, and sphere 1 still has charge Q/. If the distance between the two forces is R, then the initial force between the two is F = kq /R. The final force between the two is F = k (Q/)(Q/4) R = 8 k Q R = 8 F Therefore F /F = /8. 6. Consider the two fixed charges on an axis as shown. There is at least one point on the axis where one could place an electron and have it not feel a net force due to the other two charges. Find that point. (Note that you will have to consider three separate cases, depending on whether the electron is to the left of both charges, between them, or to the right of both.) +C C 1 0 1 4 5 m Since the figure shows a number line, suppose we place the electron at position x, and the force on it is zero. Our goal is to find the value of x. There are three possibilities to consider: either x < 0 (so the electron is to the left of both charges), 0 < x < 4 (between the charges), or x > 4 (on the right of both). 4
The magnitude of the force due to the C charge is F L = k e r, where r is the distance from x to 0; but this distance is simply r = x 0 = x, so F L = k e x. The distance from the electron at x to the right charge at 4 is r = x 4. The magnitude of the force due to the C charge is thus F R = k e (x 4). If the electron is between the two charges, it will be attracted to the left by the positive charge, and repelled to the left by the negative charge. Thus the electron feels a net force to the left, and can t be zero. If the electron is to the left of both charges, then it is attracted to the right by the positive charge and repelled to the left by the negative charge. Thus the force due to the left charge is F L = F L (ˆx) and F R = F R ( ˆx), so the net force is F = F Lˆx F Rˆx = (F L F R )ˆx The force on the electron is thus zero if F L = F R. F L = F R k e x = k e (x 4) = x = (x 4) (x 8x + 16) = x x 16x + = x = x + 16x = 0 = x = 16 ± 56 + 18 = 8 ± 4 6 = {1.80, 17.8} The first value of x, 1.80, lies between the two charges, so that doesn t make sense. ut x = 17.8 m does lie to the left of the charges, and so is indeed a point where the forces balance. (Notice, by the way, that the charge of the electron cancelled out pretty early. If the electron feels no force there, then no charge of either sign feels no force either; in fact, that is a point where the electric field of the two charges is zero.) What if the electron is to the right of the two charges? Then the electron is attracted to the left by the positive charge and repelled to the right by the negative charge. Thus the net force on the right hand side is F = F L ( ˆx) + F R (ˆx) = (F R F L )ˆx This is only zero if F L = F R. However, we showed that F L = F R only when x = 1.80 or x = 17.8; in neither case is the electron to the right of the two charges (i.e. x > 4). Therefore there is only one place on the number line where the forces cancel: at x = 17.8 m. (There is a way one could have guessed which side this point was on, without trial and error. If one moves far away from these two charges, to the left or right, then the two charges seem to blur into one point with net charge = 1 C: this is a negative charge, so far away the electron would tend to be repelled away from the center. If we put the electron just to the left of the positive charge, it will be attracted towards the center. As the electron moves to the left, then, it must transition from being attracted to being repelled: at the transition point, the force is zero.) 5