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Electricity and Magnetism: PHY-204 Fall Semester 2015 Individual ASSIGNMENT Assignment 6: Select Circuit Elements This homework must be solved individually. Solution 1. A variable air capacitor used in a radio tuning circuit is made of N semicircular plates each of radius R and positioned a distance d from its neighbors, to which it is electrically connected. As shown in Fig. 1, a second identical set of plates is enmeshed with the first set. Each plate in the second set is halfway between two plates of the first set. The second set can rotate as a unit. Determine the capacitance as a function of the angle of rotation θ, where θ = 0 corresponds to the maximum capacitance. Fig. (1) Answer Recall first that capacitance of a parallel plate capacitor is given by C= ε0 A d where A is the area of each plate and d is the distance between two plates. This problem looks scary at first. But it really isn’t! It is just the case of a large number of parallel plates such that the area of adjacent plates inside capacitor is adjustable. More clearly, as we increase the value of θ between 0 and π, the overlap between the adjacent plates decreases. In fact, when θ = 0 we have the case of a large number of parallel plate capacitors with the shape of plates being that of a semi circle. When θ = π there is no area of overlap of the adjacent plates and the capacitance is zero. But before we compute the actual value of capacitance as a function of angle θ we find out what is the number of parallel plate capacitors formed by the two sets of N plates each. Let us tale N = 3 for convenience and count the number of capacitors formed. Date: 14 December, 2015, 5 : 00 pm 1 Electricity and Magnetism: PHY-204 Fall Semester 2015 1 2 3 + - 4 5 6 Here dotted lines show the set of moveable plates, while solid lines show the set of fixed plates. Clearly the number of capacitors formed is 5. It is easy to check for other values of N and conclude that the total number of capacitors formed by two sets of N plates is 2N − 1. So the total capacitance Ctotal in terms of capacitance of one parallel plate capacitance C is Ctotal = (2N − 1)C We add these capacitances because the resulting capacitors form a parallel network. Now we compute C as a function of θ. The distance between the plates is d/2 and area is a function of θ. So C= 2ε0 A(θ) · d To find A(θ) observe the following diagram of a semi-circular plate. θ A R The shaded region in this figure shows A(θ). Using the fact that area of a semi-circle of radius R is πR2 /2, we obtain, ( ) πR2 π − θ A(θ) = 2 π We have multiplied by π−θ π since this the ratio of angle subtended by shaded region to the entire angle subtended by the semi-circle.. Plugging this value in Ctotal we obtain, ( ( )) 2ε0 πR2 π − θ Ctotal = (2N − 1) d 2 π 2 (2N − 1)ε0 (π − θ)R · = d Date: 14 December, 2015, 5 : 00 pm 2 Electricity and Magnetism: PHY-204 Fall Semester 2015 2. The circuit in Fig.(2) consists of two identical, parallel metal plates connected by identical metal springs to a 100-V battery. With the switch open, the plates are uncharged, are separated by a distance d = 8.00 mm, and have a capacitance C = 2.00 µF. When the switch is closed, the distance between the plates decreases by a factor of 0.500. (a) How much charge collects on each plate? (b) What is the spring constant for each spring? HINT : Derive the relation for force exerted on each plate using F = C= Aε0 , x dU dx and capacitance where x is the arbitrary plate separation. Fig. (2) Answer (a) d = 8.00 mm d new d′ = = 4.00 mm 2 C0 = 2.00 µF A C0 d C0 = ε0 ⇒ A= d ε0 ε A ε A 0 0 =2 = 2C0 new C ′ = d′ d Q = C ′ △V = 2C0 △V = 400 × 10−6 = 4.00 × 10−4 C. (b) When the switch was open, there was no potential energy associated with the electric field. In steady state, the voltage across capacitor is 100 V. So Ue = Date: 14 December, 2015, 5 : 00 pm 3 Electricity and Magnetism: PHY-204 1 ′ 2 CV 2 Fall Semester 2015 = C0 V 2 = 2 × 10−6 (100)2 = 2.00 × 10−2 J. Each spring stretches by an amount △x = 0.4 2 = 0.2 mm. Hence for each spring its potential energy (elastic) must decrease by 1.00 × 10−2 J. So 1 k(△x)2 2 2 × 10−2 = 5.00 × 105 N/m. k = (0.2 × 10−3 )2 1.00 × 10−2 = This calculation is by virtue of conservation of energy. 3. A parallel-plate capacitor consists of a fixed plate and a movable plate that is allowed to slide in the direction parallel to the plates. Let x be the distance of overlap, as shown in Fig. (3). The separation between the plates is fixed. (Movable) (Fixed ) x Fig. (3) (a) Assume that the plates are electrically isolated, so that their charges ±Q are constant. In terms of Q and the (variable) capacitance C, derive an expression for the leftward force on the movable plate. HINT : Consider how the energy of the system changes with x. (b) Now assume that the plates are connected to a battery, so that the potential difference ϕ is held constant. In terms of ϕ and the capacitance C, derive an expression for the force. (c) If the movable plate is held in place by an opposing force, then either of the above two setups could be the relevant one, because nothing is moving. So the forces in (a) and (b) should be equal. Verify that this is the case. Answer I’ll discuss this in the class. 4. A conducting sphere of radius R initially has a uniform surface-charge density σ)0 . ( t Beginning at t = 0, this charge is drained off over a period t0 such that σ = σ0 1− . t0 Date: 14 December, 2015, 5 : 00 pm 4 Electricity and Magnetism: PHY-204 Fall Semester 2015 R σ Ic (a) Find the magnitude of the displacement current Id just outside the surface of the sphere. (b) Calculate the current Ic carried off by the wire. Compare the displacement current Id to Ic . Answer Area Vector Id R σ Ic Dashed line is an open area whose edge is the red circle. This is similar to an inflated balloon with an open neck. We use Maxwell-Ampere’s Law in this question. We choose a very small circular path encircling the wire carrying the conduction current. This path is shown in red in the corresponding diagram. For part (a) we consider the spherical surface just outside the conducting sphere as the open surface corresponding to this path. Note that Ic through this spherical surface (shown as a dashed circle in the diagram above) is zero. (a) d Id = ε0 dt ∫ ⃗ dA ⃗ E. Open Area = ε0 Date: 14 December, 2015, 5 : 00 pm dE (4πR2 ). dt 5 Electricity and Magnetism: PHY-204 Fall Semester 2015 dE . Using Gauss’s Law for Electric Charges we find the electric dt field just outside the sphere in terms of the electric charge density: ∫ ⃗ dA ⃗ = Qenc E. ε0 We need to find Closed Area E(4πR2 ) = Q ε0 Q 4πε0 r2 σ = , ε0 = from which we obtain dE 1 dσ = dt ε0 dt ( ) 1 d t = (σ0 ) 1− ε0 dt t ( )0 1 1 (σ0 ) − = ε0 t0 σ0 = − . ε0 t0 (4πR2 )σ0 (4πR2 )σ0 and |Id | = . The negative sign of the displacet0 t0 ment current shows that it is directed towards (into) the charged sphere whose Hence Id = − area is pointing outwards. 2 Ιd = - 4π R σ0 t0 2 Ιc = + 4π R σ0 t0 (b) Ic = − dQ dt (as current comes from charge depletion) ) ( d t 2 = −(4πR )σ0 1− dt t0 2 (4πR )σ0 = + . t0 Date: 14 December, 2015, 5 : 00 pm 6 Electricity and Magnetism: PHY-204 Fall Semester 2015 Therefore, |Ic | = |Id | and Ic = −Id . Note that this was expected from the Maxwell-Ampere’s Law. Both currents equal the path integral of the magnetic field around the small circular red path, and therefore must be equal in magnitude. Their signs are perfectly consistent with the definition of the area vectors. Think about this. If in confusion, come and talk to me. 5. The induced emf in an inductor of inductance L varies with time according to εind (t) = −2At, where A is a positive constant. (a) If there is no current through the inductor at t = 0, calculate the current as a function of time for t > 0. (b) Is the current increasing or decreasing when t > 0? (c) Discuss how your answer to part b is consistent with the sign of εind and what you know about the behavior of inductors? Answer (a) εind (t) = −2At, A > 0 di εind (t) = −L dt di −2At 2At = = dt −L L ∫ t 2A i(t) = t′ dt′ + i(t = 0) L 0 A At2 = t2 + 0 = · L L (b) The current is increasing with time. (c) The current increases with time, so the induced emf acts in a direction to arrest this increase. So it grows in the ‘negative’ direction with respect to time. This is in line with the behavior of inductors. 6. The toroid in the figure below has 200 rectangular windings, and the toroid redii are Rin = 160 mm and Rout = 240 mm. The height of each winding is h = 20 mm, such that the rectangular cross section of each winding is (Rout − Rin ) × h = (80 mm) × (20 mm). What is the inductance of the toroid? Date: 14 December, 2015, 5 : 00 pm 7 Electricity and Magnetism: PHY-204 Fall Semester 2015 Rout Rin h Answer d |ε| = N dt ∫ ⃗ · dA ⃗ B Now the magnetic field inside the toroid depends upon the distance from the center r. A coil is shown in the cross section view. r Rin h Rext ⃗ is constant at a given r. From symmetry |B| (B)(2πr) = µ0 N i µ0 N i B = · 2πr Date: 14 December, 2015, 5 : 00 pm 8 Electricity and Magnetism: PHY-204 Fall Semester 2015 Hence over the area inside the loop shown, we have ∫ ∫ µ0 N i Rout 1 ⃗ ⃗ dr B · dA = 2π r=Rin r ( ) µ0 N ih Rout = ln 2π Rin ( ) 2 µ0 N h Rout di ln |ε| = 2π Rin dt ( ) 2 µ0 N h Rout ⇒ L = ln 2π Rin ( ) −7 4π × 10 × (200)2 × 0.02 80 = ln 2π 20 −4 = 2.22 × 10 H. Note that L has units of Henry (H). 1 H= 1 VsA−1 and it depends on N 2 . Date: 14 December, 2015, 5 : 00 pm 9