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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. |
An electrical technician requires a capacitance of `2 muF` in a circuit across a potential difference of 1 kV. A large number of `1 mu F` capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors. |
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Answer» Total required capacitance , C=`2muF` potential difference, V=1kv =1000 v Capacitance of each capacitor , `C_(1)=1 muF` each capacitor can withstand a potential difference `V_(1)=400 V` Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. the potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. hence,number of capacitors in each row is given as `1000/400=2.5` Hence, there are three capacitors in each row. Capacitance of each row `=1/(1+1+1)=1/3 muF` Let there are n rows each hacing three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as `1/3+1/3+1/3+....n ` terms `=n/3` However, capacitance of the circuit is given as 2`muF` `:. n/3=2` `n=6` Hence, 6 rows of three capacitors are present in the circuit. a minimum of 6 x 3 i.e., 18 capacitors are required for the given arragements. |
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| 2. |
The capacitance of a parallel plate capacitor with air as medium is `3 muF.` with the introduction of a dielectric medium between the plates, the capacitance becomes `15mu F.` The permittivity of the medium isA. `5C^(2)N^(-1)M^(-2)`B. `15C^(2)N^(-1)m^(-2)`C. `0.44xx10^(-10)C^(2)N^(-1)m^(2)`D. `8.854xx10^(-11)C^(2)N^(-1)m^(-2)` |
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Answer» Correct Answer - C Capacitance of parallel pate capacitor with air is `C = (epsi_(0)A)/(d)` Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium is ` C = (K epis_(0)A)/(d)` where k is the dielectric constant of a medium ` or (C)/(C) = k or K = (15)/(3) = 5 or k = (epsi)/(epsi_(0))` or ` epsi = Kepsi_(0) = 5 x 8.854 xx 10^(-12) = 0.4xx 10^(-10) C^(2)N^(-1) m^(2)` |
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| 3. |
Assertion. Dielectric polarization means formation of positive and negative charges inside the dielectric. Reason. Free electrons are formed in this process.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertin and reason are ture but reason is not the correct explanation of assertion .C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C When an electric field is applied to the dielectri, each moleucles of dielectric gets polarised, centres of gravity of positive and negative charges get displaced form each other. Electric dipoles are produced inside. |
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| 4. |
A molecule of a substance has a permanent electric dipole moment of magnitude `10^(-29) `C m. A mole of this substance is polarized at low temperature by appling a strong elecrostatic field of magnitude `10^(6) V m^(-1)`. The direction of the field is suddenly changed by an angle of `60^(@)`. Estimate the heat released by the substance in aligning its dipole along the new direction of the field. For simplicity, assume `100%` polarisation of sample. |
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Answer» Here, dipole moment of each molecules` = 10^(-29) C m` As 1 mole of the substance contains `6 xx 10^(23)` molecules, total dipole moment of all the molecules, `p = 6 xx 10^(23) xx 10^(-29) C m` `= 6 xx 10^(-6) C m` Initial potential energy,`U_(i) = -pE cos theta = -6 xx 10^(-6) xx 10^(6) cos 0^(@) = -6J` Final potential energy (when `theta = 60^(@)`), `U_(f) = –6 xx 10^(-6) × 10^(6) cos 60^(@) = -3 J` Change in potential energy `= -3 J - (-6J) = 3 J` So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles. |
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| 5. |
Electric field inside a conductor can be zero only, if potential inside the onductor isA. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertin and reason are ture but reason is not the correct explanation of assertion .C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 6. |
Three concentric spherical shells have radii `a, b` and `c(a lt b lt c)` and have surface charge densities `sigma, -sigam` and `sigma` respectively. If `V_(A), V_(B)` and `V_(C)` denote the potentials of the three shells, then for `c = q + b`, we haveA. `V_(C)=V_(B)=V_(A)`B. `V_(C)=V_(A)ne V_(B)`C. `V_(C)= V_(B)ne V_(A)`D. `V_(C) ne V_(B) ne V_(A)` |
| Answer» Correct Answer - B | |
| 7. |
The magnitude of electric field `vecE` in the annular region of a charged cylindrical capacitor.A. is the same throughoutB. is higher near the outer cylinder than near the inner cylinderC. varies as `(1)/(r^(2))` where r is the distance from the axisD. varies as `(1)/(r^(3))` where r is the distance from the axis. |
| Answer» Correct Answer - C | |
| 8. |
Which of the following statements is/are true about the principle of Van de Graaff generator ?A. The actiion of sharp points.B. The charge given to a hollow conductor is transferrec to outer surface and it distibuted uniformly over it.C. It is used for accelerating uncharged particle.D. Both (a) and (b) are true. |
| Answer» Correct Answer - D | |
| 9. |
Two spheres of radius `a` and `b` respectively are charged and joined by a wire. The ratio of electric field of the spheres isA. `a/b`B. `b/a`C. `(a^(2))/(b^(2))`D. `(b^(2))/(a^(2))` |
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Answer» Correct Answer - B Let `q_(1)` and `q_(2)` be the charges and `C_(1)` and `C_(2)` be the capacitance of two spheres. The charge flows from the sphere at higher potential to the other at lower potentials because equal. After sharing the charge on two sphere would be. `(q_(1))/(q_(2)) = (C_(1)V)/(C_(2)V)" "......(i)` Also `(C_(1))/(C_(2)) = (a)/(b) " ".......(ii)` From (i) `(q_(1))/(q_(2)) = (a)/(b)` Ratio of surface charge on the two spheres `(sigma_(1))/(sigma_(2)) = (q_(1))/(4pia^(2)) .(4pib^(2))/(q_(2)) = (q_(1))/(q_(2))(b^(2))/(a^(2)) = (b)/(a)" "("using (ii)")` `therefore` The ratio of electric fields at the surfaces of tow spheres `(E_(1))/(E_(2)) = (sigma_(1))/(sigma_(2)) = (b)/(a)` |
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| 10. |
A parallel plate condenser is charged by connected it to a battery. The battery is disconnected and a glass slab is introduced between the plates. ThenA. Charge and potential difference.B. Potential difference and energy stored.C. Energy stored and capacitance.D. Capacitance and charge. |
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Answer» Correct Answer - B When battery is disconnected , charged remians constant . On introudcing glass slab , capacity increases. Potential differece and energy stored decrease . |
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| 11. |
A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase ro remain constant.A. capacitance will increase.B. energy stored will decrease.C. electric field will increase.D. voltage will decrease. |
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Answer» Correct Answer - C When the capacitor is connected to dc source , it get charged . The battery is then disconncted , so no more charge can flow in . On removing dielectric capacitance decrease. Energy stored `(u = (q^(2))/(2C))` will increase. Potential `(V = (q)/(C))` will also increase. Electric field `(E = (V)/(D))` will increase. |
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| 12. |
In the question number 4, work done in bringing a charge of `4xx10^(-9)` C form infinity to that point isA. `2.4xx10^(-4)J`B. `1.8xx10^(-4)J`C. `3.2xx10^(-5)J`D. `4.1xx10^(-5)J` |
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Answer» Correct Answer - B Work done , `W=q(V_f-V_i)=4xx10^(-9)xx4.5xx10^4` `=1.8xx10^(-4)` J |
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| 13. |
The electric potential at a point in free space due to a charge `Q` coulomb is `Q xx 10^(11)` volts. The electric field at that point isA. `12piepsi_(0)Qxx10^(22)Vm^(-1)`B. `4 pi epsi_(0)Qxx10^(22)Vm^(-1)`C. `12pi epsi_(0)Qxx10^(20)V m^(-1)`D. `4 pi epsi _(0)Qxx10^(20)V m^(-1)` |
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Answer» Correct Answer - B Here, `V=Q/(4piepsilon_0r) =Qxx10^11` `therefore 4piepsilon_0r=10^(-11)` …(i) Now, `E=Q/(4piepsilon_0r^2)=(Qxx4piepsilon_0)/((4piepsilon_0r)^2)=(4piepsilon_0Q)/((10^(-11))^2` (Using (i)) `=4piepsilon_0Qxx10^22 V m^(-1)` |
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| 14. |
A parallel plate capacitor has a uniform electric field E in the space between the the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor isA. `(1)/(2)epsi_(0)E^(2)`B. `(E^(2)Ad)/(epsi_(0))`C. `1/2epsi_(0)E^(2)Ad`D. `eisi_(0)E^(2)Ad` |
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Answer» Correct Answer - C Capacitacne of a parallel plate capacitor is `C = (epsi_(0)A)/(d) " "………(i)` Potential difference between the plates is V = Ed The energy stored in the capacitor is `U = (1)/(2) CV^(2) = (1)/(2) ((epsi_(0)A)/(d)) (Ed)^(2) " "("using (i) and (ii)")` ` = (1)/(2) epsi_(0) E^(2) Ad` |
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| 15. |
A metallic sphere of radius 18 cm has been given a charge of `5xx10^(-6)C.` The energy of the charged conductor isA. `0.2J`B. `0.6J`C. `1.2J`D. `2.4J` |
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Answer» Correct Answer - B Here ` r= 18 cm = 18 xx 10^(-2) m , q = 5 xx 10^(-6)C` As `C = 4 pi epsi_(0)r =(18 xx 10^(-2))/(9 xx 10^(9)) = 2 xx 10^(-11) F` Energy of charged conductor is `U = (q^(2))/(2C) = ((5 xx 10^(-6))C)/(2 xx 2 xx10^(-11)F) = 0.625 J` |
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| 16. |
The potential at a point due to charge of `5xx10^(-7)C` located 10 cm away isA. `3.5xx10^(5)V`B. `3.5xx10^(4)V`C. `4.5xx10^(4)V`D. `4.5xx10^(5)V` |
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Answer» Correct Answer - C Here, `q=5xx10(-7)` C , r=10 cm =0.1 m Potential, `V=1/(4piepsilon_0) q/r = (9xx10^9 xx 5 xx10^(-7))/0.1=4.5xx10^4` V |
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| 17. |
The work done in carrying a charge q once round a circle of radius r with a charge Q at the centre isA. `(qQ)/(4pi epsi_(0)a)`B. `(qQ)/(4pi epsi_(0)a^(2))`C. `(q)/(4pi epsi_(0)a)`D. zero |
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Answer» Correct Answer - D The electric potential at any point on circle at radius a due to charge Q at its centre is `V=1/(4piepsilon_0)Q/a` It is an equipotential surface. Hence, work done in carrying a charge q round the circle is zero. |
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| 18. |
As per this diagram a point charge `+q` is placed at the origin `O`. Work done in taking another point charge `-Q` from the point `A(0, a)` to another point `B(a, 0)` along the staight path `AB` is: A. `(qQ)/(4piepsi_(0))((a-b)/(ab))`B. `(qQ)/(4piepsi_(0))((b-a)/(ab))`C. `(qQ)/(4piepsi_(0))((b)/a^(2)-(1)/(b))`D. `(qQ)/(4piepsi_(0))((a)/(b^(2))-(1)/(b))` |
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Answer» Correct Answer - A Potential at point A is `V_A=1/(4piepsilon_0) q/a` Potential at point B is `V_B=1/(4piepsilon_0) q/b` Work done in taking a charge Q from A to B is `W=Q(V_B-V_A)=(Qq)/(4piepsilon_0)[1/b-1/a]=(Qq)/(4piepsilon_0)["a-b"/"ab"]` |
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| 19. |
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å: (a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation? |
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Answer» The distance between electron-proton of a hydrogen atom, `d = 0.53 Å` Charge on an electron, `q_(1) = -1.6 xx 10^(-19)C` Charge on proton, `q_(2) =+1.6 xx 10^(-19)C` (a) Potential at infinity is zero. Potential energy of the system, p-e = Potential energy ay infinity - Potential energy at `= 0-(q_(1)q_(2))/(4pi in_(0)d)` Where, `in_(0)` is the permittivity of three space `(1)/(4pi in_(0)) = 9 xx 10^(9) Nm^(2) C^(-2)` `:.` Potential energy `= 0 -(9 xx 10^(9) xx (1.6 xx 10^(-19))^(2))/(0.53 xx 10^(10)) = -43.7 xx 10^(-19)J` Since `1.6 xx 10^(-19)J = 1eV`. `:.` Potential energy `= (-43.7 xx 10^(-19))/(1.6 xx 10^(-19)) = -27.2 eV` Therefore, the potential energy of the system is `-27.2 eV`. (b) Kinetic energy is half ofthe magnitude of potential energy, Kinetic energy `= (1)/(2) xx (-27.2) = 13.6 eV` Total energy `= 13.6 - 27.2 =13.6 eV` Therefore, the minimum work required to free the electron is `13.6 eV`. (c) When zero of potential energy s taken, `d_(1) = 1.06 Å` `:.` Potential energy of the system = Potential energy at `d_(1)` - Potential energy at d `= (q_(1)q_(2))/(4pi in_(0)d_(1)) - 27.2 eV` `= (9 xx 10^(9) xx(1.6 xx 10^(-19))^(2))/(1.06 xx 10^(-10)) - 27.2 eV` `= 21.73 xx 10^(-19) J -27.2 eV` `= 13.58 eV - 27.2 eV` `= -13.6 eV` |
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| 20. |
An electric dipole is placed at the centre of a sphere. Mark the correct options:A. Electric field is zero at every point of the sphereB. Electic field is zero anywhere on the sphereC. The flux of electric field is not zero through the sphereD. All of these |
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Answer» Correct Answer - B When electric dipole is held in the sphere, electric field is not zero anywhere on the sphere.However, net electric flux through the sphere is zero. |
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| 21. |
Which of the following is not ture ?A. For a point charge, electrostatic potential varies as 1/r.B. For a dipole, the potential depends on the magnitude of potition vector and pdipole moment vector.C. The electric dipole potential varties as 1/r at large distance.D. For a point charge, the electrostatic field varies as `1//r^(2).` |
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Answer» Correct Answer - C The electric dipole potential varies as `1/r` at large distance is not true . |
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| 22. |
Figure shows the field lines of apoint negative charge. In going from B to A, the kinetic energy of a small negatice charge will A. increaseB. decreaseC. remain constantD. data insufficient |
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Answer» Correct Answer - B Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases going from B to A . |
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| 23. |
Work done by an electrostatic field in moving a given charge from one point to another …… upon the chosen.A. zeroB. positiveC. negativeD. data insufficient |
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Answer» Correct Answer - C In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative . |
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| 24. |
The distance between `H^(+) and Cl^(-)` ions in HCl molecules is `1.38Å.` The potential due to this dipole at a sistance of `10Å` on the axis of dipole isA. `2.1` VB. `1.8` VC. `0.2` VD. `1.2` V |
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Answer» Correct Answer - C Here, `2a=1.38xx10^(-10)m, r=10xx10^(-10)` m charge , q=`1.6xx10^(-19)`C As potential , `V=P/(4piepsilon_0r^2)=(q(2a))/(4piepsilon_0r^2) = (9xx10^9xx1.6xx10^(-19)xx1.38xx10^(-10))/((10xx10^(-10))^2)` |
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| 25. |
Which among the following statements is true about the work done in bringing a unit positive charge from point P to Q in an electrostatic field ? A. Minimum work is done in case of path II.B. Maximum work is done in case of path I.C. Work done is same in all the three paths.D. Work done is zero in case of path II. |
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Answer» Correct Answer - C Work done on a unit charge +q by the electrostatic field due to any given charge configuration is independent of the path, and depends in only on its initial and final positions. |
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| 26. |
A hollow conducting sphere is placed in an electric field produced by a point charge placed at `P` as shown in figure. Let `V_(A), V_(B), V_(C )` be the potentials at points `A, B and C` respectively. Then A. `V_(C)gt V_(B)`B. `V_(A)gt V_(B)`C. `V_(B)gt V_(C)`D. `V_(A)=V_(C)` |
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Answer» Correct Answer - D In the figure given , the hollow conducting sphere becomes an equipotential surface. Therefore `V_A=V_C` |
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| 27. |
The work done to move a charge along an equipotential from `A` to `B`A. must be defined as `-underset(P)overset(Q)intvecE.vec(dl)`B. is zeroC. can have a non-zero valueD. both (a) and (b)are correct |
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Answer» Correct Answer - D Work done to move a unit charge along an equipotential surface from P to Q, `W=-oversetQundersetP intvecE.dvecl` On equipotential surface `vecE bot dvecl` `W=-undersetPoversetQ int E(dl)cos90^@=0` |
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| 28. |
The angle between the equipotential surface and the electric field (or line of force) at any point on the equipotential surface isA. `90^(@)` alwaysB. `0^(@)` alwaysC. `0^(@)` to `90^(@)`D. `0^(@)` to `180^(@)` |
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Answer» Correct Answer - A Electric field is always perpendicular to the equipotential surface at any point. |
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| 29. |
What do you understand by potential gradiednt ? Establish a relation between electric field and potential gradient.A. Electric field is in the direction in which the potential decreases steepestB. Magnitude of electric field is given by the charge in the magnitude of potential per unit displacement jnormal to the equipotential surface at the point.C. In the region of strong electrric field, equipotential surfaces are far apart.D. Both the statements (a) and (b) are correct. |
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Answer» Correct Answer - D In the region of strong electric fields equipotential surfaces are close together and in the region of weak electric fields , equipotential surfaces are far apart. |
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| 30. |
A capacitor is charged through a potential difference of 200 V, when 0.1 charge is stored in it. The amount of energt released by it, when it is discharged isA. 5 JB. 10 JC. 20 JD. 2.5 J |
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Answer» Correct Answer - B Here , `V = 200 V , q = 0 .1 C`. Energy stored , `U = (1)/(2)qV = (1)/(2) xx 0.1 xx 200 = 10 J` When the capacitor is discharged it relaease the same amount of enegry i.e., 10 J. |
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| 31. |
Energies stored in capacitor and dissipated during charging a capacitor bear a ratioA. `1:1`B. `1:2`C. `2:1`D. `1:3` |
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Answer» Correct Answer - C Half of the energy is dissipated during charging a capactitor . `therefore " " ("Energy stored in a capacitor")/("Energy dissipated during charging a capacitor") = (2)/(1)` |
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| 32. |
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(½) QE`, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor `½`. |
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Answer» Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of `Deltax`. Hence , work done by the force to do so =`FDeltaX` ltBrgt As a result, the potential energy of the capacitros increases by an amount given as uA`Deltax`. Where, u=energy density A=area of each plate d=distance between the plates V=potential difference across the plates the work done will be equal to the increases in the potential energy i.e,. `F Detlax=uA Deltax ` `F=uA=(1/2 in_(0)E^(2))A` Electric intensity is given by `E=V/d` `:. F=1/2 in_(0)(V/d)EA=1/2 (in_(0)AV/d)E` however, capacitance, `C=(in_(0)A)/d` `:. F=1/2 (CV)E` charge on the capacitor is given by `Q=CV` `:. F=1/2 QE` The physical origin of the factors , `1/2` , in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. it is the avearge value, `F/2`, of the field that contributes to the force. |
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| 33. |
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports. Show that the capacitance of this spherical capacitor is given by `C = (4pi in_(0) r_(1) r_(2))/(r_(1) - r_(2))`, Where `r_(1) and r_(2)` are radial of outer and inner spheres respectively. |
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Answer» Radius of the outer shell =`r_(1)` radius of the inner shell =`r_(2)` The inner surface of the outer shell has charge +Q the outer surface of the inner shell induced charge -Q. potential difference between the two shells is given by, `V=Q/(4piin_(0)r_(2))-Q/(4piin_(0)r_(1))` where, `in_(0)`=permitivity of free space `V=Q/(4pi in_(0))[1/(r_(2))-1/(r_(1))]` `=(Q(r_(1)-r_(2)))/(4piin_(0)r_(1)r_(2))` capacitance of the given system is given by `C=("Charge (Q)")/("Potential difference (V)")` `=(4pi in_(0)r_(1)r_(2))/(r_(1)-r_(2))` Henced proved. |
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| 34. |
In a region of constant potentialA. the electric field is uniform.B. the electric field is zero.C. there can be no charge inside the region.D. both (b) and (c ) are correct. |
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Answer» Correct Answer - D In a region of constant potential `E=-"dV"/"dr"=0 " " (because "V=constant")` i.e., electric field is zero , so there can be no charge inside the region. |
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| 35. |
Equipotential surfacesA. are closer in regions of large electric fields compared to regions of lower electric fieldsB. will be more crowded near sharp edges of a conductorC. will always be equally spacedD. both (a) and (b) are correct |
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Answer» Correct Answer - D Electric field,`E=-"dV"/"dr" "or" dr prop 1/E` i.e., equipotential surfaces are closer in regions of large electric fields compared to regions of lower electric field. At sharp edges of a conductor, charge density is more . Therefore electric field is stronger. Hence equipotential surfaces are more crowded. |
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| 36. |
Describe schematically the equipotential surfaces corresponding to (a) a constant electric field in the z-direction, (b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction, (c) a single positive charge at the origin, and (d) a uniform grid consisting of long equally spaced parallel charged wires in a plane. |
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Answer» (a) Equidistance planes parallel to the x-y plane are the equipotential surfaces (b) planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer , the field incraeses. (c) concentric spheres centred at the origin are equipotential surface (d) a periodically varying shape near the given grid is the equipotential surface. this shape gradually reaches the shape of planes parallel to the grid at a larger distance. |
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| 37. |
A parallel plate capacitor is to be designed with a voltage rating 1 KV using a material of dielectric constant 3 and dielectric strength about `10^(7) Vm^(-1)`. [Dielectric strength is the maximum electric field a material can tolerate without break down, i.e, without starting to conduct electrically through partial ionization. For safety, we should like the field never to exceed say `10%` of the dielectric strength]. What minimum area of the plates is required to have a capacitance of 50 pF ? |
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Answer» Potential rating of a parallel plate capacitor , V=1kv =1000 V dielectric constant of a meterial `in_(r)=3` dielectric strength `=10^(7)V//m` For safety, the field intensity never exceeds 10 % of the dielectric strength. Hence, electric field intensity, `E=10 % ` of `10^(7) =10^(6) V//m` Capactiance of the parallel plate capacitor , C=50 pF =`50xx10^(-12)F` distance between the plates in given by , `d=V/E` `=1000/(10^(6))=10^(-3) m` Capacitance is given by the relation, `C=(in_(o)in_(r)A)/d` where, A= area of each plate `in_(0)`=permitivity of free space =`8.85xx10^(-12) N^(-1)C^(2)m^(-2)` `:. A=(Cd)/(in_(0)in_(r))` `=(50xx10^(-12)xx10^(-3))/(8.85xx10^(-17)xx3) ~~19cm^(2)` Hence, the area of each plate is about `19 cm^(2)`. |
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| 38. |
Two charges of magnitude `5 nC and -2 nC` are placed at points (2cm,0,0) and (x cm,0,0) in a region of space. Where there is no other external field. If the electrostatic potential energy of the system is `-0.5 muJ`. What is the value of x ?A. 20 cmB. 80 cmC. 4 cmD. 16 cm |
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Answer» Correct Answer - A Potential energy of system `U=1/(4piepsilon_0)(q_1q_2)/r` `=0.5xx10^(-6)=(9xx10^(9)xx5xx10^(-9)xx(-2)xx10^(-9))/((x-2)xx10^(-2))` `rArr` x =20 cm |
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| 39. |
Consider two conductinbg spheres of radill `R_(1) and R_(2)` with `R_(1)gt R_(2).` If the two are at the same potential, and the larger sphere has more charge than the smaller sphere, thenA. the charge density of smaller sphere is less then that of larger sphere,B. the charge density of smaller sphere is more than tahat of larger sphere.C. both spheres may have same charge density.D. none of these |
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Answer» Correct Answer - B Here ` V_(1) = V_(2) or ( q_(1))/(4pi epsi_(0)R_(1)) = (q_(2))/(4piepsi_(0)R_(2))` `therefore " "(q_(1))/(q_(2)) = (R_(1))/(R_(2)) " "......(i)` Given `R_(1) gt R_(2)` `therefore " "q_(1) gt q_(2)` `therefore` Larger sphere has more charge than the smaller sphere. Now charge densities. `sigma = (q_(1))/(4piR_(1)^(2)) and sigma_(2) = (q_(2))/(4piR_(2)^(2))` `therefore (sigma_(2))/(sigma_(1)) = (q_(2))/(q_(1)) (R_(1)^(2))/(R_(2)^(2)) or (sigma_(2))/(sigma_(1)) = (R_(2))/(R_(1)) (R_(1)^(2))/(R_(2)^(2)) = (R_(1))/(R_(2)) " "("using (i)")` As `R_(1) gt R_(2) ` therefore `sigma_(2) gt sigma_(1)` Charge density of smaller sphere is more thatn the charge density of large sphere. |
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| 40. |
Two metal spheres, one fo radius R and the other of radius 2R, both have same surface charge density s. They are brought in contact and seprated. What will be new surface charge densitites on them ?A. `5/2sigma, 5/4sigma`B. `5/3sigma, 5/6sigma`C. `3/5sigma, 6/5sigma`D. `2/3sigma,1/2sigma` |
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Answer» Correct Answer - B Before contant, changes of each sphere, ` q_(1) = sigma4piR^(2) ` and `q_(2) = sigma 4pi(2R)^(2) = 4 q_(1)` When the two sphere are brought it contact ,their charges are shared till their potentials become equal i.e., `V_(1) + V_(2)` `therefore " "(q_(1))/(4piepsi_(0)R) =(q_(2))/(4piepsi_(0)(2R))` `therefore " " q_(2) = 2q_(1)" "...(i)` As there is no loss of charge in the process `therefore " " q_(1) + q_(2) = q_(1) + q_(2) = q_(1) + 4q_(1) = 5q_(1) = 5(sigma 4piR^(2))` or `q_(1) + 2q_(1) = 5 sigma 4 piR^(2) " "("using (i)")` `q_(1) = (5)/(3) sigma4piR^(2) and q_(2)= 2q_(1) = (10)/(3)(sigma 4 piR^(2))` Hence `sigma_(1) = (q_(1))/(4piR^(2)) = (5)/(3) sigma, sigma_(2) = ( q_(2))/(4pi(2R)^(2)) = (5)/(6) sigma ` |
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| 41. |
Minimum number of capacitors each of `8 muF` and 250 V used to make a composite capacitor of `16 muF` and 1000 V areA. 8B. 32C. 16D. 24 |
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Answer» Correct Answer - B Minimun number of capacitors in each row ` =(1000)/(250)= 4` Therefore , 4 capacitors connected in series. If there are m such rows, that total cacpacity ` = m xx 2 = 16 " "therefore m = 16//2 8` `therefore` minimum number of capcitors ` = 4 xx 8 = 32` |
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| 42. |
Two idential capacitors are joined in parallel, charged to a potential `V` and then separated and then connected in series i.e. the positive plate of one is connected to negative of the otherA. The charges on the free plated connected together are destoyed.B. The energy stored in the system increases.C. The potential difference between the free plates is 2V.D. The potential difference remains constant. |
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Answer» Correct Answer - C When the two capacitor charged to same potenital are connected in series, then total potential difference `V = V_(1) +V_(2) = V + V =2V` |
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