The foce between electic chages Coulomb s Law Two chaged objects, of chage q and Q, sepaated by a distance, exet a foce on one anothe. The magnitude of this foce is given by: kqq Coulomb s Law: F whee k is a constant: 9 9 k 8.99 0 N m / C 9 0 N m / C The diection of the foce is towad the second object if the objects have opposite signs, and away fom the second object if the signs ae the same. What does Coulomb s Law emind us of? Gavity Compaing gavity and the inteaction between chages In geneal, the foce of gavity is much weake than electostatic inteactions. Gavity is always attactive, while the foce between chages can be attactive o epulsive. The equations have simila foms, giving ise to simila behavio. GmM kqq F F G = 6.67e- N m /kg k = 8.99e9 N m /C The most abundant chaged paticles ae potons and electons, with poton heavie. Fo potons, q ~ e-9 C, m ~ e-7 kg. So F G /Fe ~ e(--x7)/e(9-x9) = e-36. This atio is even smalle Foces between two chages Two equal chages Q ae placed a cetain distance apat. They exet equal-and-opposite foces F on one anothe. Now one of the chages is doubled in magnitude to Q. What happens to the magnitude of the foce each chage expeiences?. oth chages expeience foces of magnitude F.. The Q chage expeiences a foce of F; the Q chage expeiences a foce F. 3. The Q chage expeiences a foce of F; the Q chage expeiences a foce F. 4. None of the above. fo electons. 3 4 Foces between two chages Let s examine this question fom two pespectives.. Newton s Thid Law can one object expeience a lage foce than the othe? No the objects expeience equal-and-opposite foces. kq ( )( Q). Coulomb s Law: F. Foces between two chages Figues and 3 show what happens to the foces when one of the chages is doubled in magnitude to Q. Figue 4 shows what happens when both chages ae doubled in magnitude. (Coulomb_mag_sim) 3 4 If we double one chage, the foce will be: kq ( )( Q) F 5 6
Supeposition If an object expeiences multiple foces, we can use: The pinciple of supeposition - the net foce acting on an object is the vecto sum of the individual foces acting on that object. one-dimensional situation with masses all, with a mass 4m, is placed on the x-axis at x = 0. all, which has a mass m, is placed on the x-axis at x = +4a. Whee would you place ball C, which also has a mass m, so that ball feels no net foce because of the othe balls? Is this even possible? 7 8 one-dimensional situation with masses (cont d) Initial: simila one-dimensional situation, but with chages (I) all, with chage +4q, is placed at x = 0. all, which has a chage -q, is placed at x = +4a. Whee would you place all C, which has a chage of magnitude q, and could be + o -, so that ball feels no net foce? y dawing the foce acting on due to (ed aow below), one may see that we should put ball C to the left of ball, at the same distance ball keeps fom ball. ns. If ball C has chage q, it should be placed at x = -4a. ns. If ball C has chage +q, it should be placed at x = +4a. 9 0 simila one-dimensional situation, but with chages (II) all, with chage +4q, is placed at x = 0. all, which has a chage -q, is placed at x = +4a. Whee would you place all C, which has a chage of magnitude q, and could be + o -, so that ball is the one that feels no net foce? ns. If ball C has chage +q, it should be placed at x = 6a. ns. If ball C has chage q, it should be placed at x = a. Case : Suppose ball C has chage +q (and so must be on the ight side of ): Let be the distance of ball C fom ball, and choose the diection pointing at ight to be positive. Total foce acting on, F =(- k)(4q)(q)/(4a) +(+k)(q)(q)/ The fist tem is the foce acting on ball due to ball. We assign this tem negative because the figue shows that this foce is pointing to the left. y assigning the sign of the foce, we have taken cae of whethe the chages ae like o unlike so we should not include the sign of the chages in the equation anymoe. Othewise we double-count the effect! C
Fom the above, the total foce acting on, F =(- k)(4q)(q)/(4a) +(+k)(q)(-q)/ The second tem i.e., the foce acting on due to C is similaly obtained. Solve the equation fo and obtain: F =0 4/(4a) =/ = 4a = a. Since ball C must be on the ight side of, we should choose = +a. That gives the position of ball C to be x = 4a + a = +6a. Case : all C has chage -q (and so must be on the left side of ): C Total foce acting on, F =(- k)(4q)(q)/(4a) +(+k)(q)(q)/ y setting F = 0, we obtain the same equation as the one we wote befoe fo case, so it should have the same solution, i.e., = a. Since ball C in this case must be on the left side of, we should choose = -a. That gives the position of ball C to be x = 4a + (-a) = +a. 3 4 two-dimensional situation Case : Thee is an object with a chage of +Q at the cente of a squae. Can you place a chaged object at each cone of the squae so the net foce acting on the chage in the cente is diected towad the top ight cone of the squae? Each chage has a magnitude of Q, but you get to choose whethe it is + o. Case let me count the ways. Thee is an object with a chage of +Q at the cente of a squae. Can you place a chaged object at each cone of the squae so the net foce acting on the chage in the cente is diected towad the top ight cone of the squae? Each chage has a magnitude of Q, but you get to choose whethe it is + o. How many possible configuations can you come up with that will poduce the equied foce?. 5. eithe 0 o moe than 4 5 Fist, obseve that all must be negative and ball 4 must be positive. n unknown chage 7 With that, by having eithe both alls &3 negative (below left) o positive (below positive) will not affect the net foce as the foces they act on the ball at the cente would cancel. Two chages and ae placed on the cicumfeence of a cicle. Chage is positive while the sign of chage is unknown. Suppose the aow indicates the diection of the net foce due to the two chages on a positive test chage at the cente of the cicle. What is the sign of the unknown chage? How is its magnitude compaed to that of chage? (unknown) 7 8
n unknown chage The foce due to chage must be pointing ight so as to be away fom chage as shown by the ed aow. The foce due to chage must lie along the line joining chage and the test chage (i.e., the geen dash line). The net foce is a vecto sum of the foces fom chage and on the positive test chage. To have the net foce points in the diection given, the foce due to chage must point towads chage. That means the unknown chage is negative. Since the net foce is close to the foce vecto due to chage than it is to that due to chage, the contibution fom the fome must be bigge. (The thee foce vectos in the actual situation ae shown in the diagam above.) So, the magnitude of chage must be lage than the unknown chage. Note that we can ignoe the distance in ou consideation of the foces because the distance Diection of the net foce What is the diection of the net foce acting on the -q chage in the squae shown at ight? Solution We fist conside the foces due to the two +q chages. They should poduce a net foce along the diagonal towads the -3q chage. On the othe hand, the q chage poduces a foce along the diagonal away fom itself (see the blue aow). The magnitude of the vecto sum of the foces due to the two +q chage is: [(kq /L ) ] / = kq /L. The magnitude of the foce due to the -3q chage is 3kq /L (Note: The distance between the -3q chage and -q chage is (L)). Since this is bigge, the net foce on the -q chage points away fom the -3q chage. between the test chage and chages and ae the same. 9 0 two-dimensional situation Case : The net foce on the positive cente chage is diected staight down. What ae the signs of the equal-magnitude chages occupying each cone? How many possible configuations can you come up with that will poduce the desied foce? Case let me count the ways. Thee is an object with a chage of +Q at the cente of a squae. Can you place a chaged object at each cone of the squae so the net foce acting on the chage in the cente is diected staight down? Each chage has a magnitude of Q, but you get to choose whethe it is + o. How many possible configuations can you come up with that will poduce the equied foce?. 5. eithe 0 o moe than 4 Fist, obseve that both chages 3 & 4 need to be negative. two-dimensional situation With that, both chages & must be positive to maintain the diection of the net foce (see figue at left.) Case 3: Thee is no net foce on the positive chage in the cente. What ae the signs of the equal-magnitude chages occupying each cone? How many possible configuations can you come up with that will poduce no net foce? 3 4
Case 3 let me count the ways. Thee is an object with a chage of +Q at the cente of a squae. Can you place a chaged object at each cone of the squae so thee is no net foce acting on the chage in the cente? Each chage has a magnitude of Q, but you get to choose whethe it is + o. How many possible configuations can you come up with that will poduce no net foce?. 5. eithe 0 o moe than 4 Case 3 let me count the ways. So long as the two chages along each diagonal have the same sign, the foces acting on the chage at the cente would be canceled. So, thee ae x = 4 ways. signs diagonals 5 6