ECE341 Test 2 Your Name: Tue 11/20/2018
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1 ECE341 Test Your Name: Tue 11/0/018 Problem 1 (1 The center of a soli ielectric sphere with raius R is at the origin of the coorinate. The ielectric constant of the sphere is. The sphere is homogeneously charge to a total charge Q. Express the electric fiel E(r an the isplacement D(r at all points in space efine by position vectors r. Sketch plots E(r an D(r. Which of the two is/are iscontinuous? Explain the reason for the iscontinuity. Taking the potential to be 0 at infinite istance from the origin, express the potential V(r at all points in space. How much energy is store by charging this sphere to total charge Q? The sphere is in ry air, the breakown (ionization electric fiel strength of which is V/m, which is much lower than the breakown fiel of the ielectric material of the sphere. Assume r = 4 an R = 0 cm for the sphere. What is the maximum energy that can be store? Note: Numerical accuracy is not require. One significant figure in your final answer suffices, therefore you on t have to use a calculator. You nee to look up a physical constant. ( Answer all the above questions with the ielectric sphere replace by a perfect conuctor sphere. Solution: (1
2 ECE341 Test Your Name: Tue 11/0/018 Electric fiel E(r is iscontinuous ue to polarization charge at the ielectric sphere surface.
3 ECE341 Test Your Name: Tue 11/0/018 Sanity check an visualization for the V(r results (for your interest; not require consistent with the above results calculate for the outer surface of the sphere. where it is interesting to compare the potential rop from r = 0 to r = R with that from r = R to r =. Also, compare this picture to the point charge case. The total store energy is the energy ensity integrate over the entire space, i.e., Here, w(r is the energy ensity an V is volume (not potential as use above an below. In the above, we alreay obtaine the electric fiel E(r insie an out of the sphere:
4 ECE341 Test Your Name: Tue 11/0/018 Therefore, An alternative solution (for your interest; not require: Here, the factor ½ is to eliminate ouble counting, as that in our parallel capacitor analysis. V is potential. The charge ensity outsie of the sphere is 0, an that insie the sphere is Therefore,
5 ECE341 Test Your Name: Tue 11/0/018 The strongest fiel is at the outer surface of the sphere. Upon breakown, therefore, To obtain the total store energy, let s first calculate = 4 J Then, the store energy is
6 ECE341 Test Your Name: Tue 11/0/018 4 J = 4. J ( For r > R, E(r, D(r, an V(r are each the same as they are in (1. For r < R, E(r = 0, D(r = 0, an The plot of D(r looks the same as E(r. Both E(r an D(r are iscontinuous at r > R, ue to the surface charge; net charge can only be on the surface of a perfect conuctor. The energy store in the fiel is the same as the part store outsie the sphere: As in (1, the maximum fiel, subject to breakown, is at the outer surface, which etermines the maximum reachable Q. The maximum store energy is the same as the part store outsie the sphere in (1 upon breakown:
7 ECE341 Test Your Name: Tue 11/0/018 Problem (1 Consier two point charges, +Q at (/, 0, 0 an Q at ( /, 0, 0, as shown in the figure below. Express Ex(x, 0, 0 E( x,0,0 xˆ for points on the x axis for all x. Schematically plot Ex(x, 0, 0 versus x. Assume Q > 0. Solution: E x ( x,0,0 4 ( x 0 Q sgn( x Q 4 ( x 0 sgn( x
8 ECE341 Test Your Name: Tue 11/0/018 (Problem continue ( Now, we have a point charge, +Q at (/, 0, 0, an an infinitely large ieal conuctor sheet at the x = 0 plane, as shown in the figure below. Express Ex(x, 0, 0 E( x,0,0 xˆ for points on the x axis for all x. Schematically plot Ex(x, 0, 0 versus x. Assume Q > 0. Solution: +Q 0 / x For x > 0, Q E x ( x,0,0 4 0 ( x Q sgn( x 4 ( x 0 For x < 0, Ex(x, 0, 0 = 0 sgn( x Q 4 ( x 0 Q 4 ( x 0 sgn( x
9 ECE341 Test Your Name: Tue 11/0/018 Problem 3 See the figure below. In a uniform magnetic fiel B (shown by ots, coming out of the paper, a metal bar (consiere an ieal conuctor is pushe to slie at a constant velocity v towars the right on a pair of parallel metal rails (also consiere ieal conuctors a istance apart. Electrons, each with charge q (q being positive, are responsible for the conuction in the metal bar. (1 Fin the magnitue an the irection of the current flowing through the resistor R connecte between the two rails. ( Does the magnetic fiel o any work in orer to rive this current in this circuit? If not, where oes the power issipate by the resistor as Joule heating come from? Hint: Assume no friction between the moving bar an the rails. What must be one to keep the bar moving at a constant velocity v? What is oing the work that is issipate by the resistor as heat? Show that the rate at which this work is one exactly equals the power issipate by the resistor. Solution: (1 As shown in the figure, the magnetic fiel exerts a force Fm to each electron in the bar. This force provies an emf, which is vb. Therefore the current is I = vb/r. Its irection is shown in the figure. ( The magnetic fiel oes not o any work. The power issipate by the resistor comes from the power provie by the force that pushes the bar to move. When the bar is moving at v, a current I is flowing. The magnetic fiel will apply a magnetic force to the bar: F = BI., opposite in irection to v. In orer for the bar to move at a constant v, a force of the same magnitue but in the same irection as v must push the bar. The rate of work one by this force is Fv.
10 ECE341 Test Your Name: Tue 11/0/018 The power issipation by the resistor is Obviously, energy is conserve. The force pushing the bar oes positive work at the rate (vb /R. The magnetic force resisting the motion of the bar oes negative work at the same rate. The magnetic force proviing the emf oes positive work at the same rate, to be consume by the resistor at the same rate. Now you see that the net work one by the magnetic fiel is 0.
11 ECE341 Test Your Name: Tue 11/0/018 Problem 4 The figure below shows a useful way to measure the ac current on a wire without breaking the circuit or using a current meter. Definitions of imensions are all shown in the figure. From Ampere s law, you know the magnetic fiel aroun the wire is H = I/( r, where r is the istance from the wire, whose iameter is negligible. The wire length is much longer than the imensions of the iron core, an therefore can be consiere infinite. The iron core has a relative permeability r. (1 Fin an expression for the fiel B(r insie the iron core. ( Fin the magnetic flux in the core (through the rectangular cross section that is b a wie an c high. Note: Since the fiel is not uniform, you nee to take an integral from a to b. (3 The coil woun aroun the core in the figure has N turns. The current carrie by the wire (along z axis is I = I0cos t. Fin an expression for the emf. (4 How o you use this setup to measure the current carrie by the wire? coil (1 ( (3 The negative sign signifies the phase relation between the emf with its polarity efine in the figure an the current I with its positive irection efine as +z.
12 ECE341 Test Your Name: Tue 11/0/018 (4 Measure the open circuit voltage of the coil, an then calculate the current accoring to the above equation, i.e., If the measure V is the effective value, then the calculate I is also the effective value.
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