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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.
| 251. |
Two charges, Q each, are fixed on a horizontal surface at separation 2a. Line OY is vertical and is perpendicular bisector of the line joining the two charges. Another particle of mass m and charge q has two equilibrium positions on the line OY, at A and B. The distances OA and OB are in the ratio `1:3sqrt(3)` (a) Find the distance of the point on the line OY where the particle will be in stable equilibrium.(b) Where will the particle experience maximum electric force – at a point above B or at a point between A and B or somewhere between O and A? Where is the acceleration of particle maximum on y axis from O to B? |
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Answer» Correct Answer - (a). `(3)/(2)a` (b). Maximum electric force is between A and B. Maximum acceleration is at O. |
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| 252. |
A nonumiform but spherically symmetric distribution of charge has charge density ` prop` given as follow - ` prop =prop_0 (1 - r //R)` for prop le R lt brgt ` prop =0 `for `r le R` where ` prop_0 = 3 Q // pi R^3` is a constant ( a) Show that the total charge contained in the charge distrubution is `Q`. (b) Show that , for the region defined by ` r le R`, the electric field is identical to that produced by a point charge Q. Obtain an expression for the electric field in the region ` r le R`. (d) Compare your results in part (b) and (c ) `r=R`. |
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Answer» Correct Answer - (c ) `(kQ r//R^2) (4 -3 r//R)` (a) For total charge `q = int rho d v rArr q = int rho_0 ( 1 - r/R ) ( 4 pi r^2 dr ) = rho_0^x` ` 4 pi int_0^R (r^2 - r^3/R) dr` `q =rho_0 4 pi [R^3 /3 - R^4 /(4R) ] = rho 4 pi [R^3 /(12)]` …(1) put `rho _0 = (3Q)/(piR^3)` in equation (1) `q = (3 Q)/(pi R^3) xx ( pi R^3)/3 = Q` (b) From gauss theorem for `r le R` `r le R oint vec E.s vec S = (sumq)/(varepsilon_0) E(4 pi r^2) = (sumq)/(varepsilon_0)` Since `sin q = Q` for `r le R` so `E = 1/( 4 pi in_0) Q/r^2` (c) `E = (sumq)/(4 pi varepsilon_0 r^2)` ...(1) `sum q = int rho d V = int_0^r rho 4 pi r der = rho_0 xx 4 pi int_0^r (r^2-r^3)/(R) dr` ` = (3Q)/(piR^2) xx 4 pi [ r^2 /3 - r^2 /(4R)]` ` q = ( 12Q)/R^2 (r^3 /3 - r^4 /(4R))` ...(2) Sub. (2) in (1) ` E = ( 12 Q)/( 4 pi varepsilon_0r^2 xx R^3) ( r^3 /3 - R^4 /(4 R) ) = (KQr)/R^3 ( 4 - (3r)/R)` (d) In (b) `r =R rArr E = (KQ)/R^2` in ( c) `r = R rArr E = (KQ)/R^2`. |
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| 253. |
A solid sphere of radius R has total charge Q. The charge is distributed in spherically symmetric manner in the sphere. The charge density is zero at the centre and increases linearly with distance from the centre. (a) Find the charge density at distance r from the centre of the sphere. (b) Find the magnitude of electric field at a point ‘P’ inside the sphere at distance `r_1` from the centre. |
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Answer» Correct Answer - (a). `rho(r)=(Q)/(piR^(4))r` (b) `(Qr_(1)^(2))/(4pi in_(0)R^(4))` |
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| 254. |
A disc of radius `a//4` having a uniformly distributed charge `6C` is placed in the x-y plane with its centre at `(-a//2, 0, 0)`. A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from `x=a//4` to `x=5a//4`. Two point charges `-7C` and `3C` are placed at `(a//4, -a//4, 0)` and `(-3a//4, 3a//4, 0)`, respectively. Conisder a cubical surface formed by isx surfaces `x=+-a//2`, `y=+-a//2`, `z=+-a//2`. The electric flux through this cubical surface is A. `(-2C)/(epsilon_(0))`B. `(2C)/(epsilon_(0))`C. `(10 C)/(epsilon_(0))`D. `(12 C)/(epsilon_(0))` |
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Answer» Correct Answer - 1 Total charge inside cube `q_("in") = 3 C + 2C - 7C = - 2 C` `phi = (q_("in"))/(in_(0)) = - (2 C)/(in_(0))` |
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| 255. |
A disc of radius `a//4` having a uniformly distributed charge `6C` is placed in the x-y plane with its centre at `(-a//2, 0, 0)`. A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from `x=a//4` to `x=5a//4`. Two point charges `-7C` and `3C` are placed at `(a//4, -a//4, 0)` and `(-3a//4, 3a//4, 0)`, respectively. Conisder a cubical surface formed by isx surfaces `x=+-a//2`, `y=+-a//2`, `z=+-a//2`. The electric flux through this cubical surface is A. `-2 C//in_(0)`B. `2 C//in_(0)`C. `10 C//in_(0)`D. `12 C//in_(0)` |
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Answer» Correct Answer - A Now charge on disc contributing to electric flux through cubical surface `= (6C)/(2) = 3C` Charge on rod contributing to electric flux through cubical surface `= (8C)/(4) = 2C` `-7 C` charge also contributes to electric flux but `3C` charge si outside the cubical surface,so it does not contribute to electric flux. Therefore `q_(enclosed) = 3C +2C - 7C = -2C` Hence electrc flux =, `phi_(C) = (q_(enclosed))/(in_(0)) = (-2C)/(in_(0))` |
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| 256. |
A particle A having charge of `2.0xx10^-6C` and a mass of 100 g is fixed at the bottomk of a smooth inclined plane of inclination `30^@`. Where should another particle B having same charge and mass, be placed o the inclined plane so that B may remain in equilibrium?A. 8 cm from the bottomB. 13 cm from the bottomkC. 21 cm from the bottomD. 27 cm from the bottom |
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Answer» Correct Answer - D `1/(4piepsilon_0)q^2/r^2=mg sintheta` `r=sqrt((1/(4piepsilon_0))(q^2/(mgsintheta)))` `sqrt(((9xx10^9)(2.0xx10^-6)^2)/((0.1)(9.8)sin30^@))` `=27xx10^-2m` `=27cm |
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| 257. |
Two concentric spheres of radii `R` and `2R` are charged. The inner sphere has a charge if `1muC` and the outer sphere has a charge of `2muC` of the same sigh. The potential is `9000 V` at a distance `3R` from the common centre. The value of R isA. `1m`B. `2m`C. `3m`D. `4m` |
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Answer» Correct Answer - A `V=1/(4piepsilon_0).q_1/(3R)+1/(4piepsilon_0).q_2/(3R)` `=q_"net"/((4piepsilon_0)(3R))` `:. R=(1/(4piepsilon))(q_"net")/(3V))` `=((9xx10^9)(3xx10^-6))/((3)(9000))` `=1m` |
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| 258. |
A proton is released from rest, 10 cm from a charged sheet carrying charged density of `-2.21xx10^-9C//m^2`. It will strike the sheet after the time (approximately)A. `4mus`B. `2mus`C. `2sqertmus`D. `4sqrt2mus` |
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Answer» Correct Answer - A `|a|=|(qE)/m|=|(qsigma)/(2epsilon_0m)|` `t=sqrt((2s)/a)=sqrt((4sepsilon_0m)/(|q||sigma|))` `=sqrt((4xx0.1xx8.86xx10^-12xx1.67xx10^-27)/(1.6xx10^-19xx2.21xx10^-9))` `=4xx10^-s` |
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| 259. |
1000 drops of same size are charged to a potential of 1 V each. If they coalesce to form in single drop, its potential would beA. `V`B. `10V`C. `100V`D. `1000V` |
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Answer» Correct Answer - C `1000(4/3pir^3)=4/3piR^3implies R=10r` i.e. radius has become 10 times. charge will become 1000 times `V=1/(4piepsilon_0) .(("Charge"))/(("Radius")) or Vprop ("Charge")/("Radius")` hence, potential will become 100 times. |
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| 260. |
If n drops, each charged to a potential V, coalesce to form a single drop. The potential of the big drop will beA. `n^(1//3) U`B. `n^(2//3) U`C. `n^(4//3) U`D. `n^(5//3) U` |
| Answer» Correct Answer - D | |
| 261. |
An uncharged insulated conductor A is brought near a charged insulated condutor B. what happens to charge and potential of B ? |
| Answer» The charge on conductor B remains same, but the potential of B gets lowered because it induces charge of opposite sign on conductor A. | |
| 262. |
A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor, but has a thickness `3d//4`. Find the ratio of the capacitanace with dielectric inside it to its capacitance without the dielectric. |
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Answer» Without the dielectric, capacity of the capcitor `= C_(0) = (in_(0) A)/(d)` With dielectric of thickness `t = 3d//4`, the capacity of condenser becomes `C = (in_(0) A)/(d -t (1- (1)/(K)))` `C = (in_(0) A)/(d - (3d)/(4) (1- (1)/(K))) = (in_(0) A)/((d)/(4) (1 + (3)/(K)))` `:. (C )/(C_(0)) = (4)/(1 + (3)/(K)) =(4K)/((K + 3))` |
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| 263. |
`n` small drops of same size are charged to `V` volts each .If they coalesce to from a single large drop, then its potential will be - |
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Answer» Let r be the radius of each small drop and R be the radius of bigger drop. As volume of bigger drop = volume of n small drops `:. (4)/(3) pi R^(3) = n xx (4)/(3) pi r^(3)` As capacity varies directly as the radius, `:.` Capacity of bigger drop becomes `n^(1//3)` times the capacity of each small drop. Also, potential of bigger drop `= ("total charge")/("capacity")` `:. V = (nq)/(4 pi in_(0) R) = (nq)/(4pi in_(0) n^(1//3) r) = n^(2//3) (q)/(4pi in_(0) r)` `= n^(2//3)` times the potential of each small drops. |
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| 264. |
If dielectric is inserted in charged capacitor ( battery removed ), then quantity that remains constant isA. capacitanceB. potentialC. intensityD. charge |
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Answer» Correct Answer - D If a dielectric is inserted in charge capacitor or the battery is removed, then all the given quantities of parallel electric field and total energy of capacitor, will change except charge. |
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| 265. |
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. . |
| Answer» As `C = Q/V` and from the figure, for same value of `Q : A` has lower value of V and B has higher value of V. Therefore . A has higher capacitance. | |
| 266. |
Three capacitors `3muF, 6muF and 6muF` are connected in series to a source of 120 V. The potential difference, energy stored and charge of a parallel plate capacitor respectively. The quantities that increase when a dielectric slab is introduced between the plates without disconnecting the battery areA. 24B. 30C. 40D. 60 |
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Answer» Correct Answer - D The combination of three charges in series `(1)/(C)=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))=(1)/(3)+(1)/(6)+(1)/(6) rArr C=(6)/(4)=1.5muF` The charge of this circuit, `q=CV=1.5xx120=180muC` The potential difference across the `3muF` `q=CVrArr V=(q)/(C)=(180)/(3)=60V` |
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| 267. |
In the circuit shown, the potential difference across the `3muF` capacitor is `V` and the equivalent capacitance between `A` and `B` is `C` Then: A. `C(AB)=4muF`B. `C_(AB)=18/11 muF`C. `V=20V`D. `V=40V` |
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Answer» Correct Answer - A::D |
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| 268. |
An electric line of forces in the x-y plane is given by the equation `x^2+y^2=1`. A particle with unit poistive charge, initially at rest at the point `x=1`, `y=0` in the x-y plane, will move along the circular line of force. |
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Answer» For a particle to move in circular motion, we need a centripetal force which is not available. The statement is false. |
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| 269. |
A parallel plate capacitor is charged. If the plates are pulled apartA. the potential difference increasesB. the capacitance increasesC. the total charge increasesD. the charge and the potential difference remain the same |
| Answer» Correct Answer - A | |
| 270. |
Two protons A and B are placed in between the two plates of a parallel plate capacitor charged to a potential difference V as shown in the figure. The forces on the two protons are identical. |
| Answer» True is because two parallel plates is uniform `(E)`, charge on each proton is same `(q)`. Therefore, force on each proton `F = qE` is same. | |
| 271. |
Two protons A and B are placed in between the two plates of a parallel plate capacitor charged to a potential difference V as shown in the figure. The forces on the two protons are identical. |
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Answer» The electric field produced between the parallel plate capacitor is uniform. The force acting on charged particle placed in an electric field is given by `F=qE`. In the case of two protons, q and E are equal and hence force will be equal. |
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| 272. |
A positive charge `q` is placed in a spherical cavity made in a positively charged sphere . The centres of sphere and cavity ar displaced by a small distance vec `l`. Force on charge `q` si :A. in the direction parallel to vectro ` vec l`B. in radial directionC. in a direction which depends on the magitued of charge density in sphereD. direction can not be delermined . |
| Answer» Correct Answer - A | |
| 273. |
Distance between centres of two spheres A andB, each of radius R is r as shown in figure`-1.450`. SphereB has a spherical cavity of radius `R//2` such that distance of centre of cavity is (r-R/2) from the centre of sphere A and `R//2` from the centre of sphere B. Di-electric constant of material of each sphere is K = I and material of each sphere has a uniform charge density· p per unit volume. Calculate interaction energy of the two spheres : |
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Answer» Correct Answer - `(pirho^(2)R^(5)(7r-4R))/(9epsi_(0)r(2r-R))` |
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| 274. |
A rod lies along the X-axis with one end at the origin and the other at `x rarr oo`. It carries a uniform charge `lambda Cm^(-1)`. The electric field at the point `x=-a` on the axis will beA. `E=lambda/(4pi epsi_(0)a) (-hat(i))`B. `E=lambda/(4pi epsi_(0)a) (hat(i))`C. `E=lambda/(2pi epsi_(0)a) (-hat(i))`D. `E=lambda/(2pi epsi_(0)a) (hat(i))` |
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Answer» Correct Answer - C |
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| 275. |
The electric field in a region is given by `vecE=E_0x/iota veci`. Find the charge contained inside a cubical volume bounded by the curfaced `x=0, x=alpha, y=0, y=alpha, z=0 and z= alpha`. Take `E_0= 5 xx 10^3 N C^(-1)`, `l= 2cm and alpha=1 cm`. |
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Answer» Correct Answer - `2.2xx10^(-12)C` |
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| 276. |
Explain why a neutral object can be attracted to a charged objec. Why can this neutal object not be repelled by a charged object ? |
| Answer» This is because a neutral object consists of equal amount of positive and negative charge. When a charged object is brought closer to a neutral object, then opposite charges develop on account of charging by induction which results in attractive forces. | |
| 277. |
A capacitor charged from a 50 V d.c. supply is found to have charge of `10 muC`. What is the capacitance of the capacitor and how much energy is stored in it? |
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Answer» As `C = (q)/(V) = (10xx10^(-6))/(50) = 0.2 muF` Energy stored , `U = (1)/(2) CV^(2)` `= (1)/(2) xx 0.2 xx10^(-6)xx(50)^(2) = 2.5 xx10^(-4) J`. |
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| 278. |
A capacitor of capacitance `6 muF` is charged to a potential of 150V. Its potential falls to 90V, when another uncharged capacitor is connected to it. Find the capacitance of the second capacitor and the amount of energy lost due to the connection. |
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Answer» Here, `C_(1) = 6 muF, V_(1) = 150 V, V_(2) = 0` `V = 90 V, C_(2) = ?` Common potential `V = (C_(1) V_(1) + C_(2) V_(2))/(C_(1) + C_(2))` `90 = (6xx10^(-6)xx150+0)/(6xx10^(-6) + C_(2))` `C_(2) = 4xx10^(-6) F = 4 muF` Initial energy, `U_(1) = (1)/(2) C_(1) V_(1)^(2)` `= (1)/(2)xx6xx10^(-6)xx(150)^(2) = 6.75xx10^(-2) J` Final energy, `U_(2) = (1)/(2) (C_(1) + C_(2)) V^(2)` `= (1)/(2) (6+4) xx 10^(-6) xx (90)^(2) = 4.05xx10^(-2) J` Loss of energy on connecting two capacitors, `DeltaU = U_(2) - U_(1) = (6.75xx10^(-2) - 4.05 xx10^(-2))` `= 0.027 J` |
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| 279. |
A `10 muF` capacitor is charged to a potential difference of `50 V` and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes `20` volt. The capacitance of second capacitor isA. `10muF`B. `20muF`C. `30muF`D. `15muF` |
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Answer» Correct Answer - D Charge of first capacitor, `q=CV = 10xx50=500muC` Common potentail `=("Total charge")/("Total capacitance")` `20=(500)/(C_(1)+C_(2)) rArr 20 =(500)/(10+C_(2))` `C_(2)=15muF` |
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| 280. |
Two charges `2muC and -2muC` are placed at points A and B 6 cm apart. (a) Identify an equipotenital surface of the system. (b) What is the direction of the electric field at every point on this surface ? |
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Answer» (a) The plane normal to AB and passing through its middle point has zero potential everywhere. (b) The direction of electric field at every point on this surface is along normal to the plane in the direction AB. |
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| 281. |
Determine the force between two free electrons spaced 1 angstrom `(10^(-10)m)` apart. |
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Answer» `F = (kq_(1) q_(2))/(r^(2)) = (9xx10^(9)xx(1.6xx10^(-19))^(2))/((10^(-10))^(2))` `= 2.304xx10^(-8)N` |
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| 282. |
Potential energy of two equal negative point charges `2muC` each held 1 m apart in air isA. 2 JB. 2 eVC. 4 JD. 0.036 J |
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Answer» Correct Answer - D `"As, PE"=(1)/(4pi epsilon_(0)).(q_(1)q_(2))/(r)=(9xx10^(9)(2xx10^(-6))^(2))/(1)=0.036J` |
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| 283. |
An attractive force of `5 N` is acting between two charges of `+ 2muC and - 2 mu C` placed at some distance. If the charges are mutually touched and placed again at the same distance , what will be the new force, between them ? |
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Answer» On touching, charges neutralise. Therefore, `F = 0` |
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| 284. |
The electric potential in a region is represented as `V=2x+3y-z` obtain expression for electric field strength. |
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Answer» Correct Answer - A::B::C `E=-[(delV)/(delx)hati=(delV)/(dely)hatj+(delV)/(delz)hatk]` `(delV)/(delx)=del/(delx)(2x+3y-z)=2` `(delV)/(dely)=del/(dely)(2x_3y-z)=3` `(delV)/(delz)=del/(delz)(2x+3y-z)=-1` `:.E=-2hati-3hatj+hatk` |
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| 285. |
`1 mu C` charge is shifted from A to B, and it is found that work done by an external force is `40 mu J`. In doing so against electrostatic forces, find potential difference `V_A - V_B`. |
| Answer» `(W_(AB))_("ext")=q(V_(B)-V_(A))" "implies" "40 muJ=1 muC (V_(B)-V_(A))" "implies" "V_(A)-V_(B)=-40 V` | |
| 286. |
An electric field of `10^5 N//C` points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of `+2muC and -5muC` at this spot? |
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Answer» Correct Answer - A::B::D Force on `+2muC=qE=(2xx10^-6)(10^5)` `=0.2N` (due west) Force on `-5muC=(5xx10^-6)(10^5)` `=0.5N` (due east) |
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| 287. |
There are two charge `+1mu c` and `+2muc` kept at certain separation ,the ratio of electrostatic forces acting on them will be in the ratioA. `1 : 5`B. `1 : 1`C. `5 : 1`D. `1 : 25` |
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Answer» Correct Answer - B |
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| 288. |
The electric potential at point `A` is `20 V` and at `B` is `-40 V`. Find the work done by an external and electrostatic force in moving an electron slowly from B to A. |
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Answer» Correct Answer - A::B Here, the test charge is an electron, i.e. `q_0=1.6xx10^-19C` `V_A=20V` `and V_B=-40V` work done by external force `(W_(B-A))_("external force" =q_0(V_A-V_B)` `=(-1.6xx10^-19)[(20)-(-40)]` `=-9.6xx10^-18J` work done electric force `(W_(B-A))_("electric force")=-(W_(B-A))_("external force")` `=-(-9.6xx10^-18J)` `=9.6xx10^-18J` |
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| 289. |
Find the work done by some external force in moving a charge `q=2muC` from infinity to a point where electric potential is `10^4` V. |
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Answer» Correct Answer - A::B Using the relation `(W_(oo-a))_("external force")=qV_a` We have, `(W_(oo-a))_("external force") =(2xx10^-6)(10^4)` `=2xx10^-2J` |
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| 290. |
Which of the following is/are incorrect statement ?A. Electric field is always conservativeB. Electric field die to a varying magnetic field is non-conservativeC. Electri field due to stationary charge is conservativeD. Electric field lines are always closet loops |
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Answer» Correct Answer - A::D |
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| 291. |
A capacitor stores `60muC` charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of `120muC` flows through the battery. The dielectric constant of the material inserted is:A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C |
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| 292. |
Which combination of four identical capacitors has the maximum capacitance? Which combination of these capacitors will store minimum energy when a constant p.d. is applied across it ? |
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Answer» The maximum capacitance of four identical capacitors, each of capacitance C, is obtained for their parallel combination : Cp = 4C. Their series combination has the minimum capacitance. The charge stored in their parallel combination is four times that in their series combination. For the same constant p.d. V, the energy stored in the parallel combination is 1/2 (4Q)V and that in the series combination is 1/2 QV. Thus, the series combination will store minimum energy. |
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| 293. |
Find the capacitance of the combination shown in Fig. |
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Answer» Starting from the left side of network, two `2 muF` capacitors are in series. Therefore, `(1)/(C_(s)) = (1)/(2) + (1)/(2) = 1, C_(s) = 1 muF`. This 1 `muF` and third 1 `muF` capacitor are in parallel. Therefore,`C_(p) = 1 + = 2 muF` and so on. Proceeding in this way, finally two `2 muF` capacitors are in series. Therefore, capacitanace of the combination between A and B is `1 mu F`. |
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| 294. |
It is required to construct a `10 mu` F capacitor which can be connected across a 200V battery . Capacitance `10 mu F` are available but they can withstand only 50V ,Design a combination which can yield the desired result . |
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Answer» As each capacitor of `10 muF` can withstand only 50V, therefore to be connected across a 200V battery, four capacitors must be connected in series in a row. Capacity `C_(1)` of each row of four capacitors is `(1)/(C_(1)) = (1)/(10) + (1)/(10) +(1)/(10) +(1)/(10) = (4)/(10)` `C_(1) = (10)/(4) = 2.5 mu F` For a total capacity of `10 muF`, four such rows of capacitors must be connected in parallel so that `C_(p) = 4 C_(1) = 4xx2.5 = 10 mu F`. Hence, we need 16 capacitors in series in each row and 4 such rows in parallel. |
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| 295. |
Find equivalent capacity between A & B, Fig. |
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Answer» In fig. If `C_(1)` is capacity of LMNO, `(1)/(C_(1)) = (1)/(3) + (1)/(3) + (1)/(3) = 1, C_(1) = 1 muF` This is an parallel with `2 muf` capacitor in LO. `:. C_(2) = 1 + 2 = 3mu F` Similarly, capacity across KP, `C_(3) = 1 + 2 = 3mu F`. `:.` Capacity between A and B = `C_(4) = 1 muF` |
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| 296. |
Five identical capacitor plates, each of area A, are arranged such that adjacent plates are at a distance d apart, the plates are connected to a source of emf V as shown in the figure The charge on plate 1 is ………..and on plate 4 is............. |
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Answer» The five plates make up four identical capacitors A,B,C,D, each of capacity `(in_(0) A//d)`, connected in parallel. As plate 1 is connected to `+` of battery and is a part of one capacitor only, therefore, charge on it is `q_(1) = +((in_(0) A)/(d)) V` However, plate 4 is connected to negaative of battery, and is common to two identical capacitors C and D is parallel, therefore, `q_(4) = -(q_(C) + q_(D)) = -2q_(1) = (-2 in_(0) A)/(d)` V |
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| 297. |
A water drop of mass 11.0 mg and having a charge of 1.6 × 10-6 C stays suspended in a room. What will be the magnitude and direction of electric Held in the room? |
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Answer» As the drop is suspended, Force (F) due to electric field balances the weight of the drop. ∴ F = mg ………….. (1) Here, m = 11.0 mg = 11 × 10-6 kg, q = 1.6 × 10-6 C Electric field is given by, E = \(\frac{F}{q}\) = \(\frac{mg}{q}\) = \(\frac{11\times10^{-6}\times9.8}{1.6\times10^{-6}}\) E = 67.4 N/C As upward force balances the weight, hence direction of electric field must be vertically upwards. |
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| 298. |
A proton is accelerated from rest through a potential difference of 500 volts. Find its final momentum. [mp = 1.67 × 10-27 kg, e = 1.6 × 10-19 C] |
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Answer» Data : mp = 1.67 × 10-27 kg, e = 1.6 × 10-19 C, u = 0, V = 500 V Initial KE, KE; = \(\cfrac12\) mpu2 = 0 ∴ ∆KE = KEf – KEi = KEf ∆KE = qV ∴ KEf = qV KEf = \(\cfrac12\) mpv2 = \(\cfrac{p^2_1}{2m_p}\) where pf = mpv ≡ the magnitude of the final momentum of the proton. ∴ Pf2 = 2 mpqV ∴ Pf = \(\sqrt{2m_pqV}\) = \(\sqrt{2(1.67\times10^{-27}\,kg)(1.6\times10^{-19}\,C)(500\,V)}\) = 5.169 × 10-22 kg∙m/s The momentum is directed along the applied electric field. |
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| 299. |
The formation of a dipole is due to two equal and dissimilar point charges placed at aA. short distanceB. long distanceC. above each otherD. None of these |
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Answer» Correct Answer - A |
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| 300. |
A point `Q` lies on the perpendicular bisector of an electrical dipole of dipole moment `p`, If the distance of `Q` from the dipole is `r` (much larger than the size of the dipole), then electric field at `Q` is proportional toA. `p^(-1)` and `r^(2)`B. `p` and `r^(-2)`C. `p^(2)` and `r^(-3)`D. p and `r^(-3)` |
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Answer» Correct Answer - D |
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