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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The two metallic plates of radius `r` are placed at a distance `d` apart and its capacity is `C`. If a plate of radius `r//2` and thickness `d`of dielectric constant `6` is placed between the plates of the condenser, then its capacity will beA. 7C/2B. 3C/7C. 7C/3D. 9C/4 |
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Answer» Correct Answer - D |
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| 2. |
At which of the two points, `1` or `2`, of a charged capacitor with nonparallel plates is the surface charge density greater? |
| Answer» At point 1, distance between the plates is less, so attraction between the charge here can be more due to which concentration of charge at point 1 will be more. | |
| 3. |
Shows the variation of voltage `V` across the plates of two capacitors `A and B` versus incease in charge Q stored in them. Which of the capacitors has higher capacitance? Give reason for your answer. . |
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Answer» From the given graphs, find the voltages `V_(A)` and `V_(B)` on capacitors `A` and `B` corresponding to charge `Q` on each of the capacitors. Clearly, `V_(A)=Q/C_(A)` and `V_(B)=Q/C_(B)` or `(V_(B))/(V_(A)) = (Q//C_(B))/(Q//C_(A)) = (C_(A))/(C_(B))` Since, `V_(B) gt V_(A), C_(A) gt C_(B)`, i`.e., the capacitor `A` has the higher capacitance. |
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| 4. |
The charged plates of a capacitor attract each other. So work by some external force is required to pull the plates farther apart. What happens to the energy added by this work. |
| Answer» This energy is stored as electrostatic potential energy the plates. | |
| 5. |
A parallel plate air capacitor is connected to a battery. If the plates of the capacitor are pulled farther apart, then state whether the following statements are true or false. a. Strength of the electric field inside the capacitor remains unchanged, if the battery is disconnected before pulling the plates. b. During the process, work is done by the external force applied to pull the plates irrespective of whether the battery is disconnected or not. c. Strain energy in the capacitor decreases if the battery remains connected. . |
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Answer» a. True b. True c. True. |
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| 6. |
An isolated coductor, initially free from charge, is charge by repeated conacts with a plate, which afrer each contact has a charge Q due to some mechanism.If q is the charge on the conductor after the first operation, prove that the maximum charge that can be give to the conductor in this way is `Qq//Q-q`. |
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Answer» At each contact, the potential both becomes same. At the first contact, charges on the conductor and the plate are `Q-q` and `q`, respectively. So `V_(1)=(Q-q)/C_(1)=q/C_(2)` or `C_(1)/C_(2)=(Q-q)/q` (i) In the second contact, the charge transferred to the plate is `q_(2)`. Then `V_(2)=(Q-q_(2))/C_(1)=(q+q_(2))/C_(2)` or `(Q-q)/q=(Q-q_(2))/(q+q_(2))` or `(q+q_(2))(Q-q)=q(Q-q_(2))` or `qQ-q^(2)+q_(2)Q-q_(2)q=qQ-qq_(2)` or `q_(2)Q=q^(2)` `:. q_(2)=q^(2)/Q` The total charge transferred in a large number of contacts is `q_(max)=q+q^(2)/Q+q^(3)/Q_(2)+q^(4)/Q_(3)+...+oo` `=q/(1-q/Q)=(qQ)/(Q-q)`. |
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| 7. |
a. How many excess electrons must be added to one plate and removed from the other to give a `5.000 nF` parallel plate capacitor `25.0 J` of stored energy ? b. How could you modify the geometry of this capacitor so that can store `50.0 J` of energy without changing the charge on its plates? |
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Answer» a. `U=Q^(2)//2C` `Q=sqrt(2UC)=sqrt(2(25.0 J)(5.000xx10^(-9)F))=5.00xx10^(-4) C` The number of electrons `N` that must be removed from one plate and added to the other is `N=Q//e=(5.00xx10^(-4)C)//(1.6xx10^(-19)C)` `=3.125xx10^(15)` electrons b. To double `U` while keeping `Q` constant, decrease `C` by a factor of `2`. `C=epsilon_(0)A//d`, halve the plate area or double the plate separation. |
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| 8. |
An uncharged parallel plate capacitor is connected to a battery. The electric field between the plates is 10V/m. Now a dielectric of dielectric constant 2 is inserted between the plates filling the entries space. The electric field between the plates now isA. 5 V/mB. 20 V/mC. 10 V/mD. none of these |
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Answer» Correct Answer - C |
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| 9. |
The plates of a parallel plate capacitor with no dielectirc are connected to a volatage source. Now a dielectric of dielectric constant `K` is inserted to fill the whose space between the plates with voltage source remainign connected to the capacitor-A. The energy stored in the capacitor will become will decrease k timesB. The electric field of inside the capacitor will decrease k timesC. The force of attration between the plates will become `K^(2)` timesD. The charge on the capacitor will become will k times |
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Answer» Correct Answer - A::C::D |
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| 10. |
The capacitor and the energy stored on a paralle plate concenser with air between its plates are respectively `C_(0)` and `W_(0)`.If the air is replaced by glass (dieletric constant =5) between the plates, the capacity of the plates and the energy stored in it will respectively beA. `5C_(0).5W_(0)`B. `5C_(0),(W_(0))/(5)`C. `(C_(0))/(5),5W_(0)`D. `(C_(0))/(5),(W_(0))/(5)` |
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Answer» Correct Answer - B |
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| 11. |
A parallel plate capacitor of capacity `C_(0)` is charged to a potential `V_(0), E_(1)` is the energy stored in the capacitor when the battery is disconnected and the plate separation is doubled, and `E_(2)` is the energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is dounled. find the ratio `E_(1)//E_(2)`. |
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Answer» Correct Answer - 4 (4) Initial energy : `E=1/2 C_(0)V_(0)^(2)` (i) `E_(1)=2E`, Because charge remains the same and the capacitance becomes half. (ii) `E_(2)=E/2`, because potential trmains the same and the capacitance becomes half. So` `E_(1)/E_(2)=4`. |
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| 12. |
A capacitor of capacitance `C_(0)` is charged to a patential `V_(0)` and then isolated. A small capacitor C is then charged from `C_(0)`, discharged and charge again, the process being is decreased to V. Find the value of C.A. `C_(0)(V_(0)/V)^(1/n)`B. `C_(0)[(V_(0)/V)^(1//n)-1]`C. `C_(0)[(V/V_(0))-1]^(n)`D. `C_(0)[(V/V_(0))^(n)+1]` |
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Answer» Correct Answer - B Potential of larger capacitor after the first chargin g is `V_(1)=(C_(0)V_(0))/(C+C_(0))` After second charging, potential is `V_(2)(C_(0)V_(1))/((C+C_(0)))=(C_(0)/(C+C_(0)))^(2)V_(0)` Aftre `n^(th)` Charging , potential is `V_(n)=(C_(0)/(C+C_(0)))^(n)V_(0)` But `V_(n)=V` So `C=C_(0)[(V_(0)/V)^(1//n)-1]`. |
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| 13. |
A capacitor of capacitance `C_(0)` is charged to a patential `V_(0)` and then isolated. A small capacitor C is then charged from `C_(0)`, discharged and charge again, the process being is decreased to V. Find the value of C. |
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Answer» Correct Answer - `C_(0)[(V_(0)/V)^(1/n)-1)]` `q_(0)=C_(0)V_(0)` After charging, `q_(1)=q_(0)(C_(0)/(C_(0)+C))` After discharging and again charging first time, `q_(0_(1))=(C_(0)V_(0))C_(0)/((C_(0)+C))=(C_(0)^(2)V_(0))/((C_(0)+C))` i.e. `V_(0_(1))=(C_(0)V_(0))/((C_(0)+C))` `q_(0_(2))=(q_(0_(1)))(C_(0))/(C_(0)+C)=(C_(0)^(3)V_(0))/((C_(0)+C)^(2))` `V_(0_(2))=(C_(0)/C_(0)+C)^(2)V_(0)` After nth chargeing `V_(0_(n))=V=(C_(0)/(C+C_(0)))^(n)V_(0)` or `(C_(0)/(C+C_(0)))^(n)=(V/V_(0))` `C/C_(0)+1=(V_(0)/V)^(1//n)` or `C=C_(0)(V_(0)/V)^(1//n)-C_(0)=C_(0)[(V_(0)/V)^(1//n)-1]`. |
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| 14. |
A capacitor of capacitance `C` is charged to a potential differece `V_(0)`. The terminals of the charged capacitor are then connected to those of an uncharged capacitor of capacitance `C//2`. a. Compute the original charge of the system. b. Find the final potential differce across each capacitor. c. Find the final energy of the system. d. Calculate the decrease in energy when the capacitors are connected. e. Where did the 'lost' energy go? |
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Answer» `Q=CV_(0)` They have equal potential difference, and thir combined charge must add up to the original charge Therefore, `V=Q_(1)/C_(1)=Q_(2)/C_(2)` and also `Q_(1)+Q_(2)=CV_(0)` `C_(1)=C` and `C_(2)=C/2`, So `Q_(1)/C=Q_(2)/((C//2))` or `Q_(2)=Q_(1)/Q_(2)` or `Q=3/2Q_(1) or Q_(1)=2/3Q` So `V=Q_(1)/C_(1)=2/3Q/C=2/3V_(0)` c. `U=1/2(Q_(1)^(2)/C_(1)+Q_(2)^(2)/C_(2))=1/2[(2/3Q)^(2)/C+(2(1/3Q)^(2))/C] ` `=1/2Q_(2)/C=1/2CV_(0)^(2)` d. The original `U` was `U=1/2CV_(0)^(2)` or `DeltaU=(-1)/6CV_(0)^(2)`. Thermal energy of capacitor, wires, etc., and electromagnetic radiation. |
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| 15. |
A `2 muF` capacitor is charged to `100 V`, and then its plates are connected by a conducting Wire. The heat produced is .A. `0.001 J`B. `0.01 J`C. ` 0.1 J`D. `1J` |
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Answer» Correct Answer - B The whole amout of energy stored in capacitor will convert into heat. Hence `=1/2CV^(2)=1/2xx2xx10^(-6)xx(1000)^(2)=0.0 J`. |
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| 16. |
In the circuit shown in, `C_(1)=6 muF, C_(2)=3 muF`, and battery `B=20 V`. The switch `S_(1)` is first closed. It is then opened, and `S _(2)` is closed. What is the final charge on `C_(2)` .A. `120 muC`B. `80 muC`C. `40 muC`D. `20 muC` |
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Answer» Correct Answer - C After closing `S_(1)`, charge on `C_(1)` is `q=6xx20=120 muC`. Now, `S_(1)` is opene. On closing `S_(2)`, Charge q will be distributed between `C_(1)` and `C_(2)` according to their capacitances. So charge on `C_(2)` is `q_(2) = (C_(2)q)/(C_(1) + C_(2)) = (3 xx 120)/(3+6) = 40 muC` |
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| 17. |
A capacitor `4 muF` charged to `50 V` is connected to another capacitor of `2 muF` charged to `100 V` with plates of like charges connected together. The total energy before and after connection in multiples of `(10^(-2) J)` isA. `1.5 "and" 1.33`B. `1.33"and" 1.5`C. `3.0 "and" 2.67`D. `2.67"and" 3.0` |
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Answer» Correct Answer - A |
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| 18. |
For the congiuration of mede of permittiveties `epslon_(0), epsilon,` and `epsilon_(0)` between parallel plates each of area `A`, as shown in the equivalent capacitance is . .A. `epsilon_(0)A/d`B. `epsilonepsilon_(0)A//d`C. ` (epsilonepsilon_(0)A)/(d(epsilon+epsilon_(0)))`D. `(epsilonepsilon_(0)A)/((2epsilon+epsilon_(0))d)`. |
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Answer» Correct Answer - D `C_(eq)=(epsilon_(0))/(d/K_(1)+d/K_(2)+d/K_(3)` Hence, `K_(1)=K_(3)=1, K_(2)epsilon_(0)`. |
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| 19. |
The plates of a parallel plate capacitor are charged up to `100 v`. Now, after removing the battery, a `2 mm` thick plate is inserted between the plates Then, to maintain the same potential deffernce, the distance betweem the capacitor plates is increase by `1.6 mm`. The dielectric canstant of the plate is .A. `5`B. `1.25`C. `4`D. `2.5` |
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Answer» Correct Answer - A As battery is disconnected, so charge will remain the same. It is given that final potential is the same. So final capacitance should be equal to initial capacitance, i.e. `(epsilon_(0)A)/d=(epsilon_(0)A)/((1.6+d)-t(1-1//K))` or `K = 5`. |
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| 20. |
`n` identical capacitors are connected in parallel to a potential difference `V`. These capacitors are then reconnected in sereis, their charges being left undisturbed. The potential difference obtained is |
| Answer» In series, potentials of all capacitors will be added, so net potential will be `NV`. | |
| 21. |
Three identical capacitors, each of capacitance `C`, are connected in series with a battery of emf `V` and get fully charged. Now the battery is removed and the capacitors are cannected in parallel with positive terminals at one point and negative terminals at other point. Then, The connon potintial will be.A. `V`B. `3V`C. `V//3`D. zero |
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Answer» Correct Answer - C In series, potential is divided. |
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| 22. |
A parallel plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In the process,A. no work doneB. work is done by the battery and the stored energy increases.C. work is done by the external agent, and the stored energy decteases. ltD. work is done by the battery as well as the external agent, but the stored energy done not charge. |
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Answer» Correct Answer - B Extra charge will flow through battery, so work is done by battery. External agent will do negative work. |
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| 23. |
Two identical capacitors with identical dielectric slabs in between them are connected in series as shown in. Now, the slab of one capacitor is pulled out slowly with the help of an external force `F` at steady state as shown. Mark the correct statement(s). .A. During the pring the process, charge (positive) flows frows from b to a.B. Diring the process, the charge of capacitor B is from b equal to the charge on A at all instants.C. Work done by `F` is positive.D. During the process, the battery has been charged. |
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Answer» Correct Answer - B::C::D As the dielectric slab is pulled out, then equivalent capacity of the system decrease, and hence charge supplied by the battery decreases as potential of the system remains constant. It means charging of battery takes place and a positive charge flows from `a to b`, As the two capacitors are connected in series, so charge on bothe capacitors remains the same at all instants. from energy conservation law, `U_(i)+W_("ext")=U_(f)+"work done on battery"+DeltaH` As dielectric slab is attracted by the plates of capcitors, to pull it out, F has to perform some work, i.e. `W_("ext")(F)gt0`. |
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| 24. |
In, the charges on `C_(1),C_(2)`, and `C_(3)`, are `Q_(1), Q_(2)`, and `Q_(3)`, respectively. .A. `Q_(2)=8 muC`B. `Q_(3)=12 muC`C. `Q_(1)=20 muC`D. `Q_(2)=12 muC` |
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Answer» Correct Answer - A::B::C Effective capacitance is `C=(4xx5)/5=(20)/9 muF` Charge on `C_(1)` is `Q_(1)=CV=(20)/(9)xx20 muC` Charge on `C_(2)` is `Q_(2)((C_(2))/(C_(2)+C_(3)))Q_(1)=2/(3)xx20=8 muC` Charge on `C_(3)` is `Q_(3) ((C_(3))/(C_(2)+C_(3)))Q_(1)=3/3xx20=12 muC` Since, `Q_(2)/C_(2)=Q_(3)/C_(3)` and `Q_(1)-Q_(2)+Q_(3)=Q_(2)+C_(3)/C_(2)Q_(2)=Q_(2)((C_(2)+C_(3))/C_(2))` Therfore, `Q_(2)=(C_(2)/(C_(2)+C_(3)))Q_(1)` `Q_(3)=(C_(3)/(C_(2)+C_(3)))Q_(1)` Thus, option `(a), (b)`, and `(c)` are correct. |
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| 25. |
Four identical capacitors are connected as shown in diagram. When a battery of `6 V` is connected between `A` and `B`, the charges stored is found to be `1.5 muC`. The value of `C_(1)` is A. `1.5muF`B. `15muF`C. `1.5muF`D. `0.1muF` |
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Answer» Correct Answer - c |
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| 26. |
Two identical parallel plate capacitors are connected in series to a battery of `100V`. A dielectric slab of dielectric constant `4.0` is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectivelyA. `50V, 50V`B. `80V, 20V`C. `20V, 80V`D. `75V, 25V` |
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Answer» Correct Answer - a |
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| 27. |
In the given network of capacitors as shown in, given that `C_(1) = C_(2)C_(3)=400pF`, and `C_(4)=C_(5)=C_(6)=200pF`, The effective capacitance of the circuit between `X` and `Y` is. A. `810 pF`B. `205 pF`C. `600 pF`D. `410 pF` |
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Answer» Correct Answer - D Start with `C_(3)` and `C_(4)` in parallel, then `C_(2)` in series, then `C_(5)` in parallel, then `C_(1)` in series, series, and finally `C_(6)` in parallel. |
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| 28. |
The equivalent capacitance of three capacitors of capacitance `C_(1):C_(2)` and `C_(3)` are connected in parallel is `12` units and product `C_(1).C_(2).C_(3) = 48`. When the capacitors `C_(1)` and `C_(2)` are connected in parallel, the equivalent capacitance is `6` units. then the capacitance areA. `2, 3, 7`B. `1.5, 2.5, 8`C. `1, 5, 6`D. `4, 2, 6` |
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Answer» Correct Answer - d |
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| 29. |
Two capacitors of equal capacitance `(C_(1)=C_(2))` are as shown in the figure. Initially, while the swithc is open (as shown) one of the capacitors is uncharged and the other carries charge `Q_(0)`. The energy stored in the charged capacitor is `U_(0).` Sometime after the switch is closed, the capacitors `C_(1)"and" C_(2)` carry charged `Q_(1)"and"Q_(2)` respectively, the energy stored in the capacitors are `U_(1)"and"U_(2)` respectively. Which of the following expression is correct? A. `Q_(0) =1/2(Q_(1)+Q_(2))`B. `U_(0)=U_(1)+U_(2)`C. `Q_(0)=1/2((U_(1))/(Q_(1))+U_(2)/Q_(2))D. `Q=Q_(2)` |
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Answer» Correct Answer - `(A,D)` |
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| 30. |
Several capacitors are connected as shown in. If the charge on the `5 muF` capacitor is `120 muC`, the potential between points A and D` is . |
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Answer» Correct Answer - `5.33 V` Charge across `5muF` capacitor is `10muC`, Therefore, potential difference across `5 muF`. Cpacitor is `q/C=(10xx10xx10^(-6))/(5xx10^(-6))-=2 V` As `3,4`, and `5muF` capacitors are in parallel, so potential difference across each of the them equals `2V`. Charge on `3 muF` capacitor is `CV=3xx10^(-6)xx2=6 muC` Charge on `4 muF` capacitor is `4xx10^(-6)xx2=8 muC` Total charge flowing in upper branch of circuit is `10 muC+6muC+ 8muC=24 muC` Potential difference across `C_(2)` is `24 muC/F=6 V`. Total potential differecnce across AB is `6+2=8 V`. Equivalent capacitance of lower branch of circuit is `(6xx3)/(6+3)=2 muF` So, charge flowing is `2xx10^(-6)xx8=16 muC` herefore, potential difference between `A` and `C`is `(16muC)/(3 muF)=(16)/3=5.33 V`. |
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| 31. |
Two identical capacitors, have the same capacitance C. One of them is charged to potential `V_1` and the other `V_2`. The negative ends of the capacitors are connected together. When the poistive ends are also connected, the decrease in energy of the combined system isA. `1/4 C(V_(1)^(2) - V_(2)^(2))`B. `1/4 C(V_(1)^(2) + V_(2)^(2))`C. `1/4 C(V_(1) - V_(2))^(2)`D. `1/4 C(V_(1) + V_(2))^(2)` |
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Answer» Correct Answer - C |
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| 32. |
Two parallel plate capacitors differ only in the spacing between their (very thin) plates , AB has a spacing of `5` mm and a capacitance of `20 pF`, while CD has a spacing of `2 mm` . Plates `A` and `C` carry charges of `+1 nC`, while B and D each carry `-1 nC`, What are the potential diffences ` V_(AB)` and `V_(CD)` after the capacitor `CD` is slid centrally between and parallel ot the plates of AB without touching them ? Would it make any difference if CD was not centrally paced between A and B ? . |
| Answer» The net charges on the plates cannot change, but the charges on the plates on either of any of spaces must be equal and opposite. Consequently, the charges on `C` and `D` must be `-1nC` and `+nC`, respectively, on their outside surfaces and` +2nC` and `-2 nC`, respectively, on their surfaces. The capacitance of any pair plates is inversely proportional to their separation,with `5 mm` corresponding to `20 pF`. Thus if `AC` is `x mm` and `DB` is `(3-x) mm`, the capacitances of the three successive capacitors are `100//x, 50`, and `100//(3-x)pF`. The valtage `V_(CD)` is therefore, `40 V`, and `V_(AB)` is `10x+40+10(3-x)=70 V`, independent of the value of `x`. | |
| 33. |
A `10 muF` capacitor and a `20 muF` capacitor are connected in series across a `200 V` supply line. The chraged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. what is the potential difference across each capacitor ?A. `(400)/(9) V`B. `(800)/(9) V`C. `400 V`D. `200 V` |
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Answer» Correct Answer - B |
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| 34. |
Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant K. The potential differences across the capacitors now becomes...........A. `(2V)/(K+2)`B. `V/(K+2)`C. `(3V)/(K+3)`D. `(3V)/(K+2)` |
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Answer» Correct Answer - D `V_(C)=(Q_(1)+Q_(2))/(C_(1)+C_(2))=(CV+2CV)/(KC+2C)=(3V)/(K+2)`. |
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| 35. |
One plate of a capacitor is fixed, and the other is connected to a spring as shown in. Area of both the plates is `A` In strady state (equilibrium), separation between the plates is `0.8d` (spring was unstretched , and the distance between the plates was `d` when capacitor was uncharged). The force conntant of the spring is approximately. A. `(125)/(32) (epsilon_(0)AE^(2))/d^(3)`B. `(2epsilon_(0)AE^(2))/d^(3)`C. `(6epsilon_(0)E^(2))/(Ad^(2))`D. `(epsilon_(0)AE^(3))/(2d^(3))` |
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Answer» Correct Answer - A `F=Kx` `Q^(2)/(2Aepsilon_(0))=K(0.2d)` or or `((epsilon_(0)A)/(0.8d)E)^(3)/(2Aepsi_(0))=0.2 Kd` or `K=3.9 epsilon_(0)AE^(2)//d^(3)` `K=4epsilon_(0)AE ^(2)//d^(3)`. |
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| 36. |
A, B C and D are large conduting plates kept parallel each other,A and D are fixed. Plates B and C connected to each other by a rigied conduction rod, can slide over friction less rails as shown,Initially the distence between `T^(1)` ates C and D.If now the rod (along with plates B and C) is slightly moved toward right, the capacitance between the terminals 1 and 2 A. remains unchangedB. increaseC. decreaseD. nothing can be said |
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Answer» Correct Answer - A::C |
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| 37. |
Two metallic bodies separated by a distance 20 cm are given equal and opposite charged on magintude 0.88 `muC`.The component of electric field along the line AB,between the plates, varies as `E_(x)=3x^(2)+0.4N//C,` where x (in meter) is the distance from one body towards the other body as shown A. The capacitance of the system is 10 `muF`B. The capacitance of the system is 20 `muF`C. The potential difference between A and C is 0.088 voltD. The potential difference between A and C natnot be determined from the given |
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Answer» Correct Answer - A::C |
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| 38. |
Consider the circuit shown where `C_(1) = 6 mu F, C_(2) = 3 mu F` and `V=20 V.` Capacitor `C_(1)` is first charged by closing the switch `S_(1).` Switch `S_(1)` is then opened, and the charged capacitor is connected to the uncharged capacitor `C_(2)` by closing `S_(2).` A. Total charge that has flown through the battery is `120 mu C.`B. Final charge on `C_(1)` after opening switch `S_(1)` and closing switch `S_(2)"is" 80 mu C.`C. Final charge on `C_(2)` after opening switch `S_(1)` and closing switch `S_(2)"is" 40 mu C`D. Total heat produced after closing switch `S_(2)"is" 1.8 mJ.` |
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Answer» Correct Answer - `(A,-B,C)` |
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| 39. |
Two capacitors `A` and `B` are connected in series with a battery as shown in the figure. When the switch `S` is closed and the two capacitors get charged fully, then A. The potential difference across the plates of A is 4 V and across the plates of B is 6VB. The potential difference across the plates of A is 6V and across the plates of B is 4 VC. The retio charge in A and B is `2:3`D. The ratioof charge inA and B is `3:2` |
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Answer» Correct Answer - B |
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| 40. |
The resultant capacitance of given circuit is A. 3CB. 2CC. CD. `C/3` |
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Answer» Correct Answer - A |
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| 41. |
In the figure shown. Find the e.m.f. `varepsilon` for which charge on `2mu F` capacitor is `4 mu C.` |
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Answer» Correct Answer - `varepsilon = 6V "or" varepsilon = 34 V` |
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| 42. |
In three capacitors `C_(1),C_(2)`, and `C_(3)`, are joined to a battery, With symbols having their usual meaning, the correct conditions will be .A. `Q_(1)=Q_(2)=Q_(3)"and"V_(1)=V_(2)+V_(3)=V`B. `Q_(1)=Q_(2)+Q_(3)"and"V=V_(1)+V_(2)=V_(3)`C. `Q_(1)=Q_(2)+Q_(3)"and"V=V_(1)+V_(2)`D. `Q_(2)=Q_(3)"and"V_(2)=V_(3)` |
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Answer» Correct Answer - C |
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| 43. |
Ten capacitors are joined in parallel and charged with a battery up to a potential `V`. Thery are then disconnected from the combination will be .A. `1V`B. `10V`C. `5V`D. `2V` |
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Answer» Correct Answer - B In series, all the potentials will be added. |
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| 44. |
In three capacitors `C_(1),C_(2)`, and `C_(3)`, are joined to a battery, With symbols having their usual meaning, the correct conditions will be .A. `Q_(1)=Q_(2)=Q_(3)` and `V_(1)=V_(2)=V_(3)=V`B. `Q_(1)=Q_(2)+Q_(3)` and `V=V_(1)+V_(2)+V_(3)`C. `Q_(1)=Q_(2)+Q_(3)` and `V=V_(1)+V_(2)`D. `Q_(2)=Q_(3)` and `V_(2)=V_(3)` |
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Answer» Correct Answer - c. At junction `A, Q_(1)` Will be divided into `Q_(2)` and `Q_(3)`. Hence, `Q_(1)=Q_(2)+Q_(3). C_(2)` and `C_(3)` are in parallel, So potentian on them will be same, i.e. `V_(2)=V_(3)`. V will be divided into `V_(2)` and `V_(2)`or `V_(3)`. Hence, `V=V_(1)+V_(2)` or `V_(2)+V_(3)`. |
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| 45. |
An uncharge parallel plate capacitor having a dielectric of dielectric constant `K` is connected to a similar air coare parallel plate capacitor charged to a potential `V_(0)`. The two share the charge, and the common potential becomes `V`. The dielectric constant `K` is`A. `V_(0)/V-1`B. `(V_(0))/V+1`C. `V/V_(0)-1`D. `V/V_(0)+1` |
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Answer» Correct Answer - A `C_(1) =C,V=_(1)V_(0),C_(2)=KC,V_(2)=0`, and`V_(common)=V` `V_("common")=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2)` `:. V=(CV_(0)+KCxxO)/(C+KC)` or `K+V_(0)/V-1`. |
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| 46. |
Three capacitors are arranged as shown in. Find the equivalent capactiy across the points A and B. . |
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Answer» Since there is no capacitor in the path `APB`, the points `A`, `P`, and `B` are electrically same, i.e., the input and output points are directly connected (short circuited). Thus, the entire charge will prefer to flow along the path `APB`. It means that the capacitors connected in the circuit will not receive any charge for storing. Thus, the equivalent capacitance of this circuit is zero. |
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| 47. |
A capacitor is connected across a battery . a. Why done each plate receive a charge of exactly the same magnitude? b. Is this true even if the plates are of different size? |
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Answer» a. The charge from one plate gets transferred to another plate through battery. The battery pumps the charge from one plate to another. b. Yes size of plates does not matter. |
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| 48. |
In the previous problem, when both the keys are closed, charge that will flow through key `K_(1)` will be :A. `(t)/((d+t))Q`B. `(d)/((d+t))Q`C. `((d+t))/(t)Q`D. `((d+t))/(d)Q` |
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Answer» Correct Answer - a |
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| 49. |
Three plates A, B and C are placed close to each other with `+Q` charge given to the middle plate. The inner surfaces to A and C can be connected to earth through plate D and keys `K_(1)` and `K_(2)` . The plates D is a dielectric slab with dielectric constant `K_(1)` then the charge that will flow through plate D and keys `K_(1)` and `K_(2)` . The plate D is a dielectric slab with dielectric constant K, then the charge that will flow though plate D when `K_(2)` is closed and `K_(2)` is open is A. `-Q`B. `-Q//2`C. `Q(1-(1)/(K))`D. `(Q)/(2)(1-(1)/(K))` |
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Answer» Correct Answer - c |
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| 50. |
Consider the situation shown in figure (31-E23 ) . The switch S is open for a long time and then closed . (a) Find the charge flown through the battery when the switch S is closed (b)Find the done by the battery . |
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Answer» a. Since a switch is open for long time. The capacitor is in series. Hence equivalent capacity is `C_(eq)=C//2`. Initial charge on each capacitor is `q_(0)=Cepsilon//2`. When switch is closed, find charge on capacitor `C_(1)`, is `q_(1)=Cepsilon` and find charge in `C_(2)`, is `q_(2)=0`. Hence, when the switch is closed, charge flow through battery is `Deltaq=(Cepsilon-(Cepsilon)/2)=(Cepsilon)/2`. b. Work done by battery is `Deltaq.epsilon=(Cepsilon)/2xxepsilon=(Cepsilon^(2))/2` c. Change in energy stored is `1/2Cepsilon^(2)-1/2xx(C-2)xxepsilon^(2)=1/4Cepsilon^(2)` d. Heat developwd is `W_("battery")-DeltaU=(Cepsilon^(2))/2-(Cepsilon^(2))/4=(Cepsilon^(2))/4`. |
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