Physics JEE(Advanced) Capacitance PDF

PHYSICS CAPACITANCE JEE(ADVANCED) TOPIC : CAPACITNCE DPP -14 1. An electron enters the re...

PHYSICS CAPACITANCE JEE(ADVANCED) TOPIC : CAPACITNCE DPP -14 1. An electron enters the region between the plates of a parallel plate capacitor at a point equidistant from either plate. The capacitor plates are 2 x 102 m apart and 101 m long. A potential difference of 300 volt is kept across the plates. Assuming that the initial velocity of the electron is parallel to the capacitor plates, calculate the largest value of the velocity of the electron so that they do not fly out of the capacitor at the other end. (take mass of electron = 9 × 10–31 kg) (Comment By Resonance : 2 x 102 m apart and 101 m long, these data not for practical capacitor). 2. Two capacitors A and B with capacitors 3µF and 2µF are charged to a potential difference of 100 V and 180 V respectively. The plates of the capacitors are connected as shown in fig. with one wire from each capacitor free. The upper plate of A is positive and that of B is negative. An uncharged 2µF capacitor C with lead wires falls on the free ends to complete the circuit. Calculate. (i) The final charge on the three capacitors and (ii) The amount of electrostatic energy stored in the system before and after the completion of the circuit. 3. A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 3d. The slab is equidistant from the plates. The capacitor is giv en some charge. As x goes from 0 to 3d. (A) The magnitude of the electric field remains the same (B) The direction of the electric field remains the same (C) The electric potential increases continuously (D) The electric potential increases at first, then decreases and again increases. 4. For the circuit shown, which of the following statements is true? (A) With S1 closed, V1 = 15 V, V2 = 20 V (B) With S3 closed, V1 = V2 = 25 V Download ATP STAR Page No. 1 APP And Practice For Free  (C) With S1 and S2 closed, V1 = V2 = 0 (D) With S1 and S2 closed, V1 = 30 V, V2 = 20 V Download ATP STAR Page No. 2 APP And Practice For Free 5. Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is (A) zero (B) q/2 (C) q (D) 2 q 6. Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is: (A) 1 1 C (V 2 - V 2) (B) C (V 2 + V 2) 1 2 1 2 4 4 1 1 C (V - V2)2 (D) C (V (C) + V2)2 4 4 1 1 7. Dotted line represents the charging of a capacitor with resistance X. If resistance is made 2X then which will be the graph of charging. (A) P (B) Q (C) R (D) S 8. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant d K = 2. The level of liquid is initially. Suppose the liquid level decreases at a constant speed V, the 3 time constant as a function of time t is Figure : C d R d Download ATP STAR Page No. 3 APP And Practice For Free  6 0 R (15d  9V t) 0 R (A) (B) 2d2 – 3d V t – 9V 2t2 5d 3V t 6 0 R (15 d – 9V t) 0 R (C) (D) 5d– 3V t 2d2  3d V t – 9V 2t2 Download ATP STAR Page No. 4 APP And Practice For Free 9. A capacitor of capacitance C0 is charged to a potential V0 and then isolated. A capacitor C is then charged from C0, discharged and charged again ; the process is repeated n times. Due to this, potential of the larger capacitor is decreased to V, then value of C is : (A) C0 [V0/V]1/n (B) C0[(V0/V)1/n – 1] (C) C0 [(V0/V) – 1] (D) C0 [(V/V0)n + 1] C 10. Each edge of the cube contains a capacitance C. The equivalent B C C capacitance between the points A and B will be C 6C 5C C C C (A) 5 (B) C 6 C C C 7C 12C (D) A C (C) 7 12 11. A parallel plate capacitor of capacitance C is as shown. A thin metal plate A is placed between the plates of the given capacitor in such a way that its edges touch the two plates as shown. The capacity now be comes. (A) 0 (B) 3C (C) 4C (D)  12. Rows of capacitors containing 1,2,4,8,.......... capacitors, each of capacitance 2F, are connected in parallel as shown in figure. The potential difference across AB = 10 volt, then : A Row1 B Row2 Row3 (A) Total capacitance across AB is 4F (B) Charge of each capacitor will be same (C) Charge on the capacitor in the first row is more than on any other capacitor (D) Energy of all the capacitors is 50 J 13. On a parallel plate capacitor following operations can be performed. P – connect the capacitor to a battery of emf V Q – disconnect the battery R – reconnect the battery with polarity reversed S – insert a dielectric slab in the capacitor (A) In PQR (perform P, then Q, then R), the stored electric energy remains unchanged and no thermal energy is developed (B) The charge appearing on the capacitor is greater after the action PSQ then after the action PQS Download ATP STAR Page No. 5 APP And Practice For Free  (C) The electric energy stored in the capacitor is greater after the action SPQ then after the action PQS (D) The electric field in the capacitor after the action PS is the same as that after SP Download ATP STAR Page No. 6 APP And Practice For Free 14. In an isolated parallel plate capacitor of capacitance C the four surfaces have charges Q1,Q2,Q3 and Q4 as shown in the figure. The potential difference between the plates is : Q1 Q3 Q2 Q4 1 (A) Q1  Q2 (B) Q2 (C) Q3 (D) Q )  Q )] [(Q (Q C C C 1 2 3 4 C 15. In the circuit shown in figure, each capacitor has a capacitance C. The emf of the cell is E. If the switch S is closed. C S C C E + – (A) some positive charge will flow out of the positive terminal of the cell (B) some positive charge will enter the positive terminal of the cell (C) the amount of charge flowing through the cell will be CE 4 (D) the amount of charge flowing through the cell will be   CE   3 16. A parallel plate capacitor has a dielectric slab in it. The slab just fills the space inside the capacitor. The capacitor is charged by a battery and the battery is disconnected. Now the slab is started to pull out uniformly at t = 0. If at time t, capacitance of the capacitor is C, potential difference across plate is V, and energy stored in it is U, then which of the following graphs are correct ? (A) (B) (C) (D) 17. We have a combination as shown in following figure. Choose the correct options : Download ATP STAR Page No. 7 APP And Practice For Free  (A) Total charge in this series combination is 600 C (B) The potential difference between the plates of C1 is 30 V (C) The potential difference between the plates of C2 is 20 V (D) The potential difference between the plates of C3 is 40 V Download ATP STAR Page No. 8 APP And Practice For Free 18. A parallel plate capacitor of plate area A & plate separation d is charged to a potential difference V & then the battery disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of the charge on each plate, the magnitude of the electric field between the plates (after the slab is inserted) & the magnitude of the work done on the system, in the process of inserting the slab, then :  0 (A) Q AV=  AV 2 1 0 KAV V 1  (B) Q = (C) E = (D) W = 0 d d Kd 2d  K 19. The figure shows a diagonal symmetric arrangement of capacitors and a battery. If the potential of C is zero, then (All the capacitors are initially uncharged). (A) VA = + 20 V (B) 4(VA – VB) + 2(VD – VB) = 2VB (C) 2(VA – VD) + 2(VB – VD) = 4VD (D) VA = VB + VD 20. Two capacitors of 2 F & 3 F are charged to 150 volt & 120 volt respectively. The plates of a capacitor are connected as shown in the fig. A discharged capacitor of capacity 1.5 F falls to the free ends of the wire and connected through the free ends of the wire, Then : (A) Charge on the 1.5 F capacitor will become 180 C at steady state. (B) Charge on the 2 F capacitor will become 120 C at steady state. (C) Positive charge flows through point A from left to right. (D) Positive charge flows through point A from right to left. 21. In the adjoining diagram all the capacitors are initially uncharged, they are connected with a battery as a shown in figure. Then Download ATP STAR Page No. 9 APP And Practice For Free  (A) Q = Q + Q and V = V + V V2  V3 (B) Q = Q + Q and V = V + 1 2 3 1 2 1 2 3 1 2 (C) Q1 = Q2 + Q3 and V = V1 + V3 (D) Q2 = Q3 and V = V2 + V3 Download ATP STAR Page No. 10 APP And Practice For Free 22. A parallel plate air capacitor is connected to a battery. The quantities charge, electric field and energy associated with this capacitor are given by Q0, V0, E0 and U0 respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous one as ; (A) Q > Q0 (B) V > V0 (C) E > E0 (D) U > U0 23. Three conducting plates of area 500 cm2 area kept fixed as shown. Distance between adjacent plates is 8.85 mm. A charge of 1.0 nC is placed on the middle plate. (a) What will be the charge on the outer surface of the upper plate ? (b) Find the potential difference developed between the upper and the middle plates. 24. Consider the arrangement of parallel plates of the previous problem. If 1.0nC charge is given to the upper plate instead of the middle, what will be the potential difference between (a) the upper and the middle plates and (b) the middle and the lower plates? 25. A spherical capacitor is made of two conducting spherical shells of radii a and b = 3a. The space between the shells is filled with a dielectric of dielectric constant K = 3 upto a radius c = 2a as shown. If the capacitance of given arrangement is n times the capacitance of an isolated spherical conducting shell of radius a. Then find value of n. 26. A parallel plate capacitor is filled with a dielectric up to one half of the distance between the plates. The manner in which the potential between the plates varies with distance is illustrated in the figure. Which half (1 or 2) of the space between the plates is filled with the dielectric and what will be the distribution of the potential after the dielectric is taken out of the capacitor provided that; (a) The charges on the plates are conserved or (b) The potential difference across the capacitor is constant. 27. Consider the situation shown in figure. The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf V. All surface are frictionless. Calculate the extension in the spring in equilibrium (spring is nonconducting). 28. The particle P shown in the figure has a mass m and a charge –q. Each horizontal plate has a surface area Download ATP STAR Page No. 11 APP And Practice For Free  A. What potential difference V should be applied to the combination to hold the particle P in equilibrium? Download ATP STAR Page No. 12 APP And Practice For Free 29. A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure. The width of each stair is a and the height is d. Find the capacitance of the assembly. 30. Find the equivalent capacitance of the combinations shown in the figure between the indicated points. 31. A finite ladder circuit is constructed by connecting several sections of 6 F, 8 F capacitor combinations as shown in the figure. Circuit is terminated by a capacitor of capacitance C. Find the value of C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between? Download ATP STAR Page No. 13 APP And Practice For Free  ANSWER’S & SOLUTIONS 2 1. 8 30 × 10 m/s V Sol. E= d eV F = qE = d eV acc (a) = (in y-direction) md ℓ time taken to cross capacitance, t = in y direction u d 1  eV  ℓ 2 2     u eVℓ2 = 1.6 10 –19  300  10 –   108m / s  = 2 2 2  md  u  md2 30 2. (i) QA = 90 µC, QB = 150 µC, QC = 210 µC (ii) 18 mJ Sol. (i) Charge on capacitor A, before joining with an uncharged capacitor qA = CV = (100) (3) µC = 300 µC Similarly, charge on capacitor B qB = (180) (2) µC = 360 µC After connection made Let q charge flow till steady state as shown by KVL 360 – q 300 – q q 2  3 –2 0  q = 210 C  QA = 90 µC, Q B = 150 µC, Q C = 210 µC (ii) (a) Electrostatic energy stored before, completing the circuit 1 1  1 2 Ui = –6 (3×10 ) (100) + 2 –6 (2×10 ) (180) 2  U  CV  2 2  2  = 4.74 × 10–2 J or Ui = 47.4 mJ (b) Electrostatic energy stored after, completing the circuit 1 (90  106 )2 12 (210 10 6 1  (150  106 )2 1 q2  ) + Uf = + 2  6 2 (2 10 ) 6 2 (2 106 ) U   = 18 × 10–2 J (3 10 )  2 C  or Uf = 18 mJ 3. (B,C) Sol. (B) and (C) are correct option. → In region I and III, there will be electric field E0 directed from +to –. In region II due to Download ATP STAR Page No. 14 APP And Practice For Free  orientation of → → → dipoles, there is an electric field Ek present in opposite direction of E0. But since E0 is also present, → the net electric field is E0 – Ek in the direction of E0 as shown in the diagram ( E0 – Vk) Download ATP STAR Page No. 15 APP And Practice For Free  Also, please note that when one moves opposite to the direction of electric field, the potential always increases. The stronger the electric field, the more is the potential increase. Since in region II, the electric field is less as compared to I and III therefore the increase in potential will be less but there has to be increase in potential in all the regions from x = 0 to x = 3d. 4. (D) Sol. If circuit does not completed than no flow of charge takes place through wire. Therefore potential difference across capacitor plates remains same V1 = 30 V and V2 = 20 V 5. (A) Sol. Due to incomplete circuit charge will not flow through the switch S. 6. (C) Sol. U = decrease in potential energy = Ui – Uf 2 1 1 V 1  V2 1 = C(V 2 2 +V )– (2C)   = C(V – V )2 1 2 2 2  2  4 1 2 7. (B) –t E Sol. logi = + log   ; Compare with y = mx +c RC  R 1  |m| = E RC ;c = log    R as R increases, |m| and C both decreases so (B) is the correct option. 8. (A) Sol. Time constant =  = RC Let thickness of dielectric filled is x at time t.  A0  KA0     C1C2  d  x  x  A0 C C= = A C1  C2  x0 d x d  d  R KA0 x  – Vt d    C=  3  x  K (d  x) d RK A 0 d  =  Vt    Vt Download ATP STAR Page No. 16 APP And Practice For Free  K d  3  3  A=1 K=2 3  2R0 = 6R 0 d  3Vt  6d  2d  = 5d  3Vt 6Vt Download ATP STAR Page No. 17 APP And Practice For Free 9. (B) Sol. Charge on C0, Q1 = C0V0, Initial charge on C1, Q2 = 0 2 C0V0 C Q1  Q2 Common potential V1   Q1 = C0V1 = 0 C  C0 V 0 C1  C  C0 = C2 2 3 C V  C  C Similarly V 0 1  0 V Q =C V = 0 V = 2  C  C0  C  C0  0 2 0 2 (C  C )2 0   0 n  C   1/ n  for n times, V  V  0 V =V  C=  0 – 1 C =   n 0  V   0  C  C0  10. (A) Sol. Due to symmetric charge distribution as shown for loop ACDB q V – – 5q q q q  V –V = V –V =  C = 6C – = VB A 3C 6C 3C A B 6C A B Ceq eq 5 11. (D) Sol. Theoritical capacitance = , because d become zero 12. (A,C) Sol. C = 2F C = C+ CCC C +...... eq 2 4 8 16 Ceq = C  1   1   1– 1/ 2  2 1/ 2 = 4F     Charge on first row capacitor is q1 = 2 × 10C = 20C Charge on second row capacitor is q2 = 1 × 10C = 10C 1 Download ATP STAR Page No. 18 APP And Practice For Free  Charge on third row capacitor is q3 = × 10C = 5C 2 Therefore charge on the capacitor in the first row is more than on any other capacitor. Energy stored in all capacitor is = 1 1 × 4 × 10–6 × (10)2 = 0.2 mJ Ceq V2 = 2 C = 2F 2 C =C+ C C C C    +...... eq 2 4 8 16 Ceq = C  1   1   1– 1/ 2  2 1/ 2 = 4F     Download ATP STAR Page No. 19 APP And Practice For Free 13. (B,C,D) Sol. In PQS process charge on capacitor is Q = CV In PSQ process charge on capacitor is Q = KCV 1 CV2 Electric energy stored in PQS is 2 = 1 KCV2 2 Electric energy stored in PSQ is = U >U PSQ PQS V Electric field in PS is E = d V Electric field in SP is E = d EPS = ESP 14. (B,C) Sol. Charge on outer surfaces are equal so Q1 = Q3 + Q2 + Q4................................................................ (i) and Q1 + Q2 + Q3 = Q4.............................................................................................................................................. (ii) Q2 Q1 – Q3 – Q4 V= C or V = C Q3 V= Q1 – Q2 – C or V = Q4 C Adding (i) and (ii) Q1 = Q4 and Q2 = – Q3 15. (A,D) 2C Sol. equivalent capacitance before switch closed is Ceq = , 3 2CE Total charge flow through the cell is q = 3 equivalent capacitance after switch S closed is Ceq = 2C Total charge flow through the cell is q = 2CE Therefore some positive charge flow through the cell after closing the switch is = q – q = 2CE – 2CE  4CE f i 3 3 16. (ABCD) Download ATP STAR Page No. 20 APP And Practice For Free Sol. Capacitance of capacitor is = C0 k 0 = a.L d Download ATP STAR Page No. 21 APP And Practice For Free  0 ax k 0 a(L – x) C= d  d a  C= x  k(L – x) a  a  kL – (k – 1)x kL – (k – 1)vt = d = d d So, C decreases linearly with time k0 aL Charge on capacitor Q = C0V0 V0 = constant d = 1 Potential difference across plate is V Q C V  V =  0 0 C C C V= a V0  [kL – (k – 1)vt] d 1 1 Potential energy U = C0V0-V  UV QV = 2 2 17. (ABCD) Sol. 1 1 1 32 60 20   1   4 Ceq C eq =  F 20 30 15 60 9 3 20 Total charge in this series combination is = × 90 q = 600C 3 q 600 Potential difference between the plate of C1 is = 20 = 30V C1 q 600 = Potential difference between the plate of C2 is = 30 = 20V C2 600 = q 15 = 40V Potential difference between the plate of C3 is = C3 1 1 1 1 32 Ceq  20  4   C = 60  20 F eq 9 3 30 15 60 18. (ACD) 0 A Sol. C= , C' = K A 0 KAV d 0 Q = CV = d d V E = V1 V Q = CV = C1V1  V1 = K d  Kd Download ATP STAR Page No. 22 APP And Practice For Free W=U –U = AV2 1 K 0 A  V 2  AV 21– 1  1 2 1C V 2 = 1 0 – 0 CV –  K  = f i 2 2 1 1 2 d 2 2 d 2d  K Download ATP STAR Page No. 23 APP And Practice For Free 19. (ABCD) Sol. given VC = 0 in AEFC VA – 20 = VC  VA = 20 V by KCL, at point D 2 (VA – VD) + 2 (VB – VD) + 4 (VC – VD) = 0 2 (VA – VD) + 2 (VB – VD) = 4 VD....................................................................................... (i) by KCL, at point B 4 (VA – VB) + 2 (VD – VB) + 2 (VC – VB) = 0 4 (VA – VB) + 2 (VB – VD) = 2VB............................................................................................... (ii) adding eq (i) and (ii) 2 (VA – VD) + 2 (VB – VD) + 4 (VA – VB) + 2 (VB – VD) = 4VD + 2 VB  6VA = 6 VD + 6VB  VA = VD + VB 20. (ABC) Sol. (300 – q) q 360 – q V + –  =V..........(i) A 2 1.5 3 A by solve this equation we get  q= 180C Charge on 1.5F capacitor is = 150C Charge on 2F capacitor is = 300 – 180 = 120C Therefore charge flows through A from left to right. 21. (ABC) Sol. In shown fig. C2 and C3 are parallel capacitor therefore V2 = V3. Charge Q1 flow through battery and gone to C1 and divided into C2 and C3 V2  V3 Q1 = Q2 + Q , total potential V = + V2 = V1 + V3 = V1 + 3 2 V1 22. (AD) Sol. Download ATP STAR Page No. 24 APP And Practice For Free  Potential difference = V0 Potential difference = V0 Capacitance = C Capacitance = KC [K is the dielectric constant of Slab K > 1] Download ATP STAR Page No. 25 APP And Practice For Free  Q0 = CV0 New charge = KC V0 1 1 02 KC V 02 Potential Energy = CV New potential energy = 2 2 Correct options are (A), (D). 23. (a) 0.50 nC (b) 10 V Sol. Charge after distribution q 1 A 0 500 10 – 4  8.85  10–12  charge on outer plate 2  nC ;C = = 50 pF 2 d 8.85  10–3 = charge on outer surface of upper plate is 1 nC = 0.5nC = 2 q 0.5 10 –9  = 10V V= 50 10 –12 C 24. (a) 10 V (b) 10 V Sol. a charge on outer most plate = q 1 2  2 nC b V = q  1/ = 10 V 2nC ab C 50pF c V = q  1/ = 10V 2nC cd C 50pF 25. n=3 Sol. c q dr 1c  1b b q dr q 1 q  d V  E.dr V=  4 kr 2   4 2 = r Download ATP STAR Page No. 26 APP And Practice For Free  4  4   –   –  k r r a 0  a 0 c 0 Download ATP STAR Page No. 27 APP And Practice For Free  0  c q 11 1 1 1 =    – – – q bc – ab – kab  kac  4     0 k  a c  b  = 4 0  kabc   c   q ba(b – c)  b(c – a) q V=  ; C = 4 0  kab   eq V c Download ATP STAR Page No. 28 APP And Practice For Free  40 kabc 40 (3)(a)(3a)(2a) = Ceq = ka(b – c)  b(c – 3a(3a – 2a)  3a(2a – a) a) 0 Ceq = 3 (4 a ) = 3 (capacitance of capacitor of isolated conducting shell of radius a)  n=3 (a) f A 1 2 26. 1st part has di-electric; (b) B Q (a) by E ; Q remain constant so electric field in region 1 varies same as region 2, so now A  = electric field graph of obtain by extend E of region 2 V (b) now V is constant so by E ; electric field graph directly obtain by joining A and B. D = 0 bv 2(K  1) 27. 2dKS 1 1 o o K(ℓ – x)b  xb 2    V2 Sol. Potential energy = (C1 + C 2) V = d 2  d  2 U V 2 dC F=– =– 2 dx x –1 d C F= 2 dx V2 dC where, dx o b {1 – K} = d 1 F= V 2 o b(K – 2 1) d at equilibrium, F = Ks x F 0 bv 2(K  1) x= = ks 2dKS 2mg 0 A Download ATP STAR Page No. 29 APP And Practice For Free 28. V= qC C Sol. Ceq. = 2 CV Q= 2 Q CV surface charge density =  = = 2A A Download ATP STAR Page No. 30 APP And Practice For Free     CV Electric field between the plates of + = 2 o 2 o = o 2A o capacitor = C Vq Force = qE = 2A o for equilibrium this electrostatic force must be equal to CVq mg  = mg 2A o 2mg A o V= Cq 8 o A 29. 15 d Sol. This is combination of 5 capacitors connected parallel. (A / 5) A 8 8o A C = C1 (A / 5) (A / 5) + C2 + C3 + C4 + C5 = o =o  3 = 15 d  + o2  + o2 d 2d 3d 5d 20 F (b) 27 30. (a) 7 7 F Sol. Charge on capacitor is assume as according to reverse symmetry. (a) q1 + q2 = Ceq V..............................................................(i) from loop AB EFA q1 q2 – q1 =0 2q q  –2   Download ATP STAR Page No. 31 APP And Practice For Free  –3q = 0...............(ii) 2 1 2 4 4 for loop ABCA q1 q2 – – + V = 0  4V = 2q + q.....(iii) 2 4 1 2 Download ATP STAR Page No. 32 APP And Practice For Free  from (i), (ii) and (iii) we get  2q 3q1   1   3  2  20 q1 + = Ceq  C = F  4  q eq 7  1 2  (b) Equivalent circuit diagram  from part (a) 20 27 Resultant capacitance = 1 F 7 7 31. 12 F Sol. If capacitance of C is equal to 12F. Then equivalent capacitance of the ladder between points A and B is becomes in dependent of the number of section in between points. 6C  6C Download ATP STAR Page No. 33 APP And Practice For Free +8=C  C2 – 8C – 48 = 0  C = 12 µF Download ATP STAR Page No. 34 APP And Practice For Free