Problem 3 Dielectric A dielectric rectangular slab has length s,width w,thickness d,and dielectric constant K.The slab is inserted on the right hand side of a parallel-plate capacitor consisting of two conducting plates of width w,length L,and thickness d.The left hand side of the capacitor of length L-s is empty.The capacitor is charged up such that the left hand side has surface charge densities +o,on the facing surfaces of the top and bottom plates respectively and the right hand side has surface charge densities tog on the facing surfaces of the top and bottom plates respectively.The total charge on the entire top and bottom plates is +0 and-O respectively.The charging battery is then removed from the circuit.Neglect all edge effects. Ls一s→ a)Find an expression for the magnitude of the electric field E on the left hand side in terms of o,og,K,s,w,L,and d as needed b)Find an expression for the magnitude of the electric field Eg on the right hand side in terms of o.og,K,s,w,L,8,and d as needed. c)Find an expression that relates the surface charge densities o,and og in terms of K,s,w,L,8,and d as needed. d)What is the total charge +0 on the entire top plate?Express your answer in terms of oL,oR,K,s,w,L,Eo,and d as needed. e)What is the capacitance of this system?Express your answer in terms of K,s, w,L,and d as needed. f)Suppose the dielectric is removed.What is the change in the stored potential energy of the capacitor?Express your answer in terms of K,s,w,L,E, and d as needed
Problem 3 Dielectric A dielectric rectangular slab has length s, width w , thickness d , and dielectric constant ! . The slab is inserted on the right hand side of a parallel-plate capacitor consisting of two conducting plates of width w , length L , and thickness d . The left hand side of the capacitor of length L ! s is empty. The capacitor is charged up such that the left hand side has surface charge densities ±! L on the facing surfaces of the top and bottom plates respectively and the right hand side has surface charge densities ±! R on the facing surfaces of the top and bottom plates respectively. The total charge on the entire top and bottom plates is +Q and !Q respectively. The charging battery is then removed from the circuit. Neglect all edge effects. a) Find an expression for the magnitude of the electric field EL on the left hand side in terms of ! L , ! R , ! , s, w , L , ! 0 , and d as needed. b) Find an expression for the magnitude of the electric field ER on the right hand side in terms of ! L , ! R , ! , s, w , L , ! 0 , and d as needed. c) Find an expression that relates the surface charge densities ! L and ! R in terms of ! , s, w , L , ! 0 , and d as needed. d) What is the total charge +Q on the entire top plate? Express your answer in terms of ! L , ! R , ! , s, w , L , ! 0 , and d as needed. e) What is the capacitance of this system? Express your answer in terms of ! , s, w , L , ! 0 , and d as needed. f) Suppose the dielectric is removed. What is the change in the stored potential energy of the capacitor? Express your answer in terms of Q , ! , s, w , L , ! 0 , and d as needed
Solutions: a)Find an expression for the magnitude of the electric field E,on the left hand side in terms of o,og,K,s,w,L,and d as needed. Using Gauss's LawE b)Find an expression for the magnitude of the electric field Eg on the right hand side in terms of o,og,K,s,w,L,8,and d as needed. Using Gauss's Law for dielectrics E E c)Find an expression that relates the surface charge densities o,and og in terms of K,s,w,L,and d as needed. The potential differencen the le side isEOn the right hand side it is Ed Since these must be equal we must have E。 d)What is the total charge +0 on the entire top plate?Express your answer in terms of o,og,K,s,w,L,E,and d as needed. =0(L-s)w+oRSw e)What is the capacitance of this system?Express your answer in terms of K,s, w,L,o,and d as needed. The potential difference is Edd -so the capacitance is C=_ -2u-[u-4]-c od f)Suppose the dielectric is removed.What is the change in the stored potential energy of the capacitor?Express your answer in terms of K,s,w,L,8, and d as needed. Since the battery has been removed,the charge on the capacitor does not change when we do this.So the change in the energy stored is
Solutions: a) Find an expression for the magnitude of the electric field EL on the left hand side in terms of ! L , ! R , ! , s, w , L , ! 0 , and d as needed. Using Gauss’s Law EL = ! L " o b) Find an expression for the magnitude of the electric field ER on the right hand side in terms of ! L , ! R , ! , s, w , L , ! 0 , and d as needed. Using Gauss’s Law for dielectrics R R o E σ κε = c) Find an expression that relates the surface charge densities ! L and ! R in terms of ! , s, w , L , ! 0 , and d as needed. The potential difference on the left side is L L o d E d σ ε = . On the right hand side it is R R o d E d σ κε = . Since these must be equal we must have R L σ σ κ = d) What is the total charge +Q on the entire top plate? Express your answer in terms of ! L , ! R , ! , s, w , L , ! 0 , and d as needed. Q L s w sw =σ σ L R ( − ) + e) What is the capacitance of this system? Express your answer in terms of ! , s, w , L , ! 0 , and d as needed. The potential difference is L L o d E d σ ε = , so the capacitance is ( ) ( ) ( ) L R o o R L L o Q L s w sw w w C Ls s Ls s C V dd d σ σ ε ε σ κ σ σ ε − + ⎡ ⎤ == = ⎢ ⎥ − + = ⎡ ⎤ − + = ⎣ ⎦ ∇ ⎣ ⎦ f) Suppose the dielectric is removed. What is the change in the stored potential energy of the capacitor? Express your answer in terms of Q , ! , s, w , L , ! 0 , and d as needed. Since the battery has been removed, the charge on the capacitor does not change when we do this. So the change in the energy stored is
enag是-是-- d e定g亿-+sg4H 1 28w L (L-s)+ks change in energy= oa( od (K-1)s 2ew[[(L-s)+]
( ) ( ) 2 22 2 2 1 1 11 change in energy 22 2 1 1 2 2 o o o o o Q QQ C C CC Q d d Qd ε εκ Lw w L L s s ε κ wLs s ⎛ ⎞ = − = ⎜ ⎟ − ⎝ ⎠ ⎛ ⎞ ⎡ ⎤ = ⎜ ⎟ − = ⎢ ⎥ − ⎡ ⎤ − + − + ⎝ ⎠ ⎣ ⎦ ⎣ ⎦ ( ) ( ) ( ) ( ) 2 2 1 change in energy 2 2 o o Qd Qd Ls sL s w w LLs s LLs s κ κ ε ε κ κ ⎡ ⎤⎡ ⎤ − + − − = = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎡ ⎤⎡ ⎤ ⎡⎤ ⎡⎤ − + − + ⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣⎦ ⎣⎦
Problem 4:Capacitors and Dielectrics (a)Consider a plane-parallel capacitor completely filled with a dielectric material of dielectric constant k.What is the capacitance of this system? The capacitance is C=K d -=KCo (b)A parallel-plate capacitor is constructed by filling the space between two square plates with blocks of three dielectric materials,as in the figure below.You may assume that t>>d.Find an expression for the capacitance of the device in terms of the plate area A and d,K,K,,and K3. K2 d/2 K 业 K3 —0/2 The capacitor can be regarded as being consisted of three capacitors, C-K6 4/2.C,=K 4/2 d d/2 and C=Ke 4/2 with C and C3 conected in series,and the combination connected in parallel with C1.Thus,the equivalent capacitance is c-c-c+62 +
Problem 4: Capacitors and Dielectrics (a) Consider a plane-parallel capacitor completely filled with a dielectric material of dielectric constant κ . What is the capacitance of this system? The capacitance is C = !" 0A d =!C0 (b) A parallel-plate capacitor is constructed by filling the space between two square plates with blocks of three dielectric materials, as in the figure below. You may assume that ℓ >> d. Find an expression for the capacitance of the device in terms of the plate area A and d, ! 1 , ! 2 , and ! 3 . The capacitor can be regarded as being consisted of three capacitors, C1 = ! 1 " 0 A 2 d ,C2 = ! 2 " 0 A 2 d / 2 and C3 = ! 3 " 0 A 2 d / 2 , with C2 and C3 connected in series, and the combination connected in parallel with C1. Thus, the equivalent capacitance is C = C1 + 1 C2 + 1 C3 ! " # $ % & '1 = C1 + C2 C3 C2 + C3 = ( 1 ) 0 A 2 d + ) 0A d ( 2 ( 3 ( 2 +( 3 ! " # $ % & = ) 0A d ( 1 2 + ( 2 ( 3 ( 2 +( 3 ! " # $ % &
Problem 5:Conductors and Potential A charge-O is at the center of a neutral metal(conducting)sphere of radius R that is in turn centered in a larger metal (conducting)sphere of radius R,which carries a net charge of +O.The potentials of the inner and outer spheres with respect to infinity are respectively: +6 01 - Inner Outer A.kO( 1 B.-k B Q 0 C.0, R D.k.O( 1 0 ,R Explain your reasoning. Reason:The induced charge distributions are shown in the figure below.The overall charge is zero,so there is not charge on the outer surface of the outer sphere
Problem 5: Conductors and Potential A charge !Q is at the center of a neutral metal (conducting) sphere of radius R1 that is in turn centered in a larger metal (conducting) sphere of radius R2 , which carries a net charge of +Q. The potentials of the inner and outer spheres with respect to infinity are respectively: Inner Outer A. ke Q( 1 R2 ! 1 R1 ) ke Q R2 B. !ke Q R1 0 C. 0, ke Q( 1 R1 ! 1 R2 ) D. ke Q( 1 R2 ! 1 R1 ) 0 Explain your reasoning. Reason: The induced charge distributions are shown in the figure below. The overall charge is zero, so there is not charge on the outer surface of the outer sphere