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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.
| 51. |
Three plates of common surface area `A` are connected as shown. The effective capacitance will be A. `(epsilon_(0)A)/(d)`B. `(3epsilon_(0)A)/(d)`C. `(3)/(2)(epsilon_(0)A)/(d)`D. `(2epsilon_(0)A)/(d)` |
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Answer» Correct Answer - D The given circuit is equivalent to parallel combination of two identical capacitors, each having capacitance `C = (epsilon_(0)A)/(d)`. Hence `C_(eq) = 2C = (2epsilon_(0)A)/(d)` |
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| 52. |
The elctric potential in a region is given by, `v = 3x^(2) + 4x`. An electron is placed at the origin will experience:A. force along positive `x`B. force along negative `x`C. zero forceD. none of these |
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Answer» Correct Answer - A `E_(x) = - (dv)/(dx) = - (6x +4)` `E_(y) = 0 E_(s) = 0` At origin `E_(x) = - 4` Thus, electorn will experience force in the direction opposite to electirc field i.e., in `+x` direction. |
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| 53. |
Three charges `+q, 2 q and -4q` are placed on the three vertices of an equilateral triangle of each side `0*1m`. Calculate electrostatic potential energy of the system, take `q = 10^(-7) C` |
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Answer» Correct Answer - `9xx10^(-3)J` |
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| 54. |
How much work has to be done in assembling three charged particles at the vertices of an equilateral triangle as shown in figure |
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Answer» Correct Answer - 234J |
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| 55. |
What must be te magnitud eof an isolated positive charge so as to produce an electric potential of `3xx10^(5)V` at a distance of 3 m from it? |
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Answer» Correct Answer - `100muC` |
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| 56. |
The sides of rectangle ABCD are 15 cm and 5 cm, as shown in figure. Point cahrges of `-5muC` and `+2muC` are placed at the vertices B and D respectively. Calculate electric potential at the vertices A and C. Also calculate the work done in carrying a charge of `3muC` from A to C. |
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Answer» Correct Answer - 2.52J |
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| 57. |
Three particles, each having a charge of `10 muC`, are. placed at the vertices of an equilateral triabngle of side. 10 cm. Find the work done by a person in pulling them. apart to infinite separtions. |
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Answer» Correct Answer - `-27J` |
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| 58. |
A charge `q = + 1 muC` is held at O between two points A and B such that `AO = 2m and BO = 1m`, Fig . Calculate the value of potential differences `(V_(A) - V_(B))`. What will be the value of potential differences `(V_(A) - V_(B))` if position of B is charged as shown in Fig ? |
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Answer» Correct Answer - `-4500V,-4500V` |
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| 59. |
Two point charges one of `+100 muC` and another of `-400 muC`, are kept 30 cm apart. Find the point of zero potential on the line joining the two charges. |
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Answer» Correct Answer - 6 cm from `+100muC` charge |
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| 60. |
Eight charged drops of water, each of radius 1 m and having a charge of `10^(-10)` coulomb, combine to form a bigger drop. Determine the potential of the bigger drop. |
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Answer» Correct Answer - 3600 V |
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| 61. |
A point charge of `10^(-7)`C is situated at the centre of a cube of side 1m. Calculate the electric flux through its surface. |
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Answer» Correct Answer - `1883.2Nm^(2)C^(-1)` |
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| 62. |
A charge of 1500 `muC` is distributed uniformly over a very large sheet of surface area `300m^(2)`. Calculate the electric field at a distance of 25 cm. |
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Answer» Correct Answer - `5.6xx10^(5)NC^(-1)` |
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| 63. |
A uniformly charged sphere has a total charge of `300muC` and radius of 8 cm. find the electric field (i) at a point 16 cm from the centre of the sphere (ii) at a point on the surface of the sphere (iii) at a point inside the sphere. |
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Answer» Correct Answer - `1.05xx10^(8)NC^(-1),1.17xx10^(8)NC^(-1)`, zero |
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| 64. |
The electric field I a region is given gy `vecE=(E_(0)x)/(b)hati`. Find the charge contained in the ubical volume bounded by the surfaces x=0, x=a, y=0, y=0,z=0 and z=a. take `E_(0)=5xx10^(3)NC^(-1),a=1cm and b=2cm`. |
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Answer» Correct Answer - `2.2xx10^(-12)C` |
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| 65. |
If unifrom electric filed `vecE = E_(0)hati + 2E_(0) hatj`, where `E_(0)` is a constant, exists in a region of space and at `(0, 0)` the electric potential `V` is zero, then the potential at `(x_(0),0)` will be.A. ZeroB. `-E_(0)x_(0)`C. `-2E_(0)x_(0)`D. `-sqrt(5) E_(0)x_(0)` |
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Answer» Correct Answer - B |
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| 66. |
A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants `k_1`, `k_2` and `k_3` as shown. If a isngle dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectic constant k is given by A. `(1)/(k) = (1)/(k_(1)) + (1)/(k_(2)) + (1)/(2k_(3))`B. `(1)/(k) = (1)/(k_(1) + k_(2)) + (1)/(2k_(3))`C. `k = (k_(1)k_(2))/(k_(1) + k_(2)) + 2k_(3)`D. `k = k_(1) + k_(2) + 2k_(3)` |
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Answer» Correct Answer - B `C_(1) = (K_(1)epsilon_(0)(A)/(2))/(((d)/(2))) = (K_(1)epsilon_(0)A)/(d)` `C_(2) = (K_(2)epsilon_(0)(A)/(2))/(((d)/(2))) = (K_(2) epsilon_(0) A)/(d)` and `C_(3) = (K_(3)epsilon_(0)(A)/(2))/(((d)/(2))) = (2K_(3)epsilon_(0)A)/(d)` `= (1)/(C_(eq)) = (1)/(C_(1) + C_(2)) + (1)/(3) = (1)/((epsilon_(0)A)/(d)(K_(1) + K_(2))) + (1)/((epsilon_(0))/(d) xx2K_(3))` `(1)/(C_(eq) = (d)/(epsilon_(0)A) [(1)/(K_(1) + K_(2)) + (1)/(2K_(3))]` `C_(eq) = [(1)/(K_(1) + K_(2)) + (1)/(2K_(3))]^(-1). (epsilon_(0)A)/(d)` So `K_(eq) = [(1)/(K_(1) + K_(2)) + (1)/(2K_(3))]^(-1)` |
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| 67. |
The capacitance of an air capacitor is `15 muF` te separation between the parallel plates is `6 mm`. A copper plate of `3 mm` thickness is introduced symmetrically between the plates. The capacitance now becomesA. `5 muF`B. `7.5 muF`C. `22.5 muF`D. `30 muF` |
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Answer» Correct Answer - D By using `C_(air) = (epsilon_(0)A)/(d), C_(medium) = (epsilon_(0)A)/(d - t+(t)/(K))` For `K = oo C_(medium) = (epsilon_(0)A)/(d - t)` `rArr (C_(m))/(C_(a)) = (d)/(d - t) rArr (C_(m))/(15) = (6)/(6 - 3) rArr C_(m) = 30 muC` |
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| 68. |
The capacitor of capacitance `4 muF` and `6 muF` are connected in series. A potential difference of `500` volts applied to the outer plates of the two capacitor system. Then the charge on each capacitor is numericallyA. `6000 C`B. `1200 C`C. `1200 muC`D. `6000 muC` |
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Answer» Correct Answer - C `C_(eq) = (C_(1)C_(2))/(C_(1) + C_(2)) = 2.4 muF`. Charge flown `= 2.4 xx 500 xx 10^(-6) C = 1200 muC`. |
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| 69. |
Three capacitors `2, 3` and `6 mF` are joined in series with each other. What is the minimum effective capacitanceA. `(1)/(2)muF`B. `1 muF`C. `2 muF`D. `3 muF` |
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Answer» Correct Answer - B `(1)/(C_(eq)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3))` `= (1)/(2)+(1)/(3)+(1)/(6)` `= (3+2+1)/(6) = (6)/(6) = 1 muF` |
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| 70. |
Two capacitors of capacitance `2 muF` and `3 muF` are joined in series. Outer plate first capacitor is at `1000` volt and outer plate of second capacitor is earthed (grounded). Now the potential on inner plate of each capacitor will beA. `700` VoltB. `200` VoltC. `600` VoltD. `400` Volt |
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Answer» Correct Answer - D Equivalent capacitance `= (2 xx 3)/(2 + 3) = (6)/(5) muF` Total charge by `Q = CV = (6)/(5) xx 100 = 1200 muC` Potential `(V)` across `2 muF` is `V = (Q)/(C ) = (1200)/(2) = 600 V` `:.` Potential on internal plates `= 1000 - 600 = 400 V` |
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| 71. |
Three capacitors of `2 muF, 3 muF` and `6 muF` are joined in series and the combination is charged by means of a `24` volt battery. The potential difference between the plates of the `6 muF` capacitor isA. `4` voltB. `6` voltC. `8` voltD. `10` volt |
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Answer» Correct Answer - A `(1)/(C_(eq)) = (1)/(2) + (1)/(3) + (1)/(6) rArr C_(eq) = 1muF` Total charge `Q = C_(eq).V = 1 xx 24 = 24 muC` Potential diff. across `6muF` capacitor `= (24)/(6) = 4` volt |
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| 72. |
Three capacitors each of capacitance `C` and of breakdown voltage `V` are joined in series. The capacitance and breakdown voltage of the combination will beA. `(C )/(3), (V)/(3)`B. `3C, (V)/(3)`C. `(C )/(3), 3V`D. `3C, 3V` |
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Answer» Correct Answer - C In series combination charge on each plate of each capacitor has the same magnitude. The potential difference is distributed inversely in the ratio of capacitors, i.e., `V = V_(1) + V_(2) + V_(3)`.. Here, `V = 3V` The equivalent capacitance `C_(s)` is given by `(1)/(C_(s)) = (1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))+..` `C_(s) = (C )/(3)` |
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| 73. |
Two infinite parallel,non- conducting sheets carry equal positive charge density `sigma` ,One is placed in the `yz` plane other at distance `x=a` Take potential `V=0` at `x=0` thenA. For `0 le x le a`, potential `V_(x)=0`B. For `x ge a`, potential `V_(x) = - (sigma)/(epsilon_(0))(x-a)`C. For `x ge a` potential `V_(x)=(sigma)/(epsilon_(0))(x-a)(X-a)`D. For `xle0` potential `V_(X)=(sigma)/(epsilon_(0))x` |
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Answer» Correct Answer - A::B::D |
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| 74. |
Two infinite parallel,non- conducting sheets carry equal positive charge density `sigma` ,One is placed in the `yz` plane other at distance `x=a` Take potential `V=0` at `x=0` thenA. for `0 le x le a`, potential `V = 0`B. for `x ge a`, potential `V = (sigma)/(epsilon_0)(x - a)`C. for `x ge a`, potential `V = (sigma)/(epsilon_0)(x - a)`D. for `x le 0`, potential `V = (sigma)/(epsilon_0) x` |
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Answer» Correct Answer - a.,c.,d. `0 le x le a, V_x = [- underset(0)overset(x)(int) E_x d x]+ V_((0)) = 0` `("as" E_x = 0)` `x ge a, V = - underset(a)overset(x)(int) E_x d x + V_((a))` =`[- underset(a)overset(x)(int) (sigma)/(epsilon_0) dx] + V_((a)) = -(sigma)/(epsilon_0)(x - a) = (- sigma)/(epsilon_0) (x - a)` `x le 0, V = - underset(a)overset(x)(int) E_x d x + V_((0)) [ because E_x = (-sigma)/(epsilon_0)]` =`- (-(sigma)/(epsilon_0) x) + V_((0)) = (sigma)/(epsilon_0) x`. |
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| 75. |
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 Common potential `V = (6xx20+3xx0)/((6+3)) = (120)/(9)`volt lt So, charge on `3muF` capacitor `Q_(2)= 3 xx 10^(-6) xx (120)/(9) = 40muC` |
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| 76. |
Two thin dielectric slabs of dielectric constants `K_(1)` and `K_(2) (K_(1) lt K_(2))` are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field `E` between the plates with distance `d` as measured from plate `P` is correctly shown by A. B. C. D. |
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Answer» Correct Answer - C Electric field inside parallel plate capacitor having charge `Q` at place where dielectric is absent `= (Q)/(A epsilon_(0))` where dielectric is present `= (Q)/(KA epsilon_(0))` As `K_(1) lt K_(2)` so `E_(1) gt E_(2)` Hence graph `(c )` correctly depicts the variation of electric field `E` with distance `d`. |
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| 77. |
Three capacitors of capacitances `3 muF, 9muF` and `18 muF` are connected one in series and another time in parallel. The ratio of equivalent capacitance in the two cases `((C_(s))/(C_(p)))` will beA. `1 : 15`B. `15 : 1`C. `1 : 1`D. `1 : 3` |
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Answer» Correct Answer - A `(1)/(C_(s)) = (1)/(3) + (1)/(9) + (1)/(18) = (1)/(2) rArr C_(s) = 2muF` `C_(p) = 3 + 9 + 18 = 30 muF rArr (C_(s))/(C_(p)) = (2)/(30) = (1)/(5)` |
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| 78. |
A `2muF` capacitor is charged as shown in the figure. The percentage of its stored energy disispated after the switch S is turned to poistion 2 is A. `80%`B. `0%`C. `20%`D. `75%` |
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Answer» Correct Answer - A Initial energy stored in capacitor `2 muF` `U_(i) = (1)/(2)2(V)^(2) = V^(2)` Final voltage after switch `2` is `ON` `V_(f) = (C_(1)V_(1))/(C_(1)+C_(2)) = (2V)/(10) = 0.2 V` Final energy in both the capacitors `U_(f) = (1)/(2) (C_(1) + C_(2)) V_(f)^(2) = (1)/(2) 10 ((2V)/(10))^(2) = 0.2 V^(2)` So energy dissipated `= (V^(2) - 0.2 V^(2))/(V^(2)) xx 100 = 80%` |
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| 79. |
Three equal charges `Q` are placed at the three vertices of an equilateral triangle. What should be the va,ue of a charge, that when placed at the centroid, reduces the interaction energy of the system to zero ?A. `(-Q)/(2)`B. `(-Q)/(3)`C. `(-Q)/(2 sqrt(3))`D. `(-Q)/(sqrt(3))` |
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Answer» Correct Answer - D |
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| 80. |
A uniform electric field exists in x y plane as shown in Fig. 3.45. Find the potential difference. Between origin O and `A(d,d,0)`. . |
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Answer» Here `vec E = E cos theta hat i + E sin theta hat j` `vec l = vec(OA) = d hat i + d hat j` `- Delta V = vec E . Vec l` =`E d cos theta + E d sin theta = E d(cos theta + sin theta)` `:. -(V_A - V_0) = E d (cos theta + sin theta)` or `V_0 - V_A = E d (cos theta + sin theta)`. |
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| 81. |
If identical charges `(-q)` are placed at each corner of a cube of side `b`, then electric potential energy of charge `(+q)` which is palced at centre of the cube will be |
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Answer» Correct Answer - `(-4 q^2)/(pi epsilon_0 sqrt(3) b)` Length of the cube diagonal is `sqrt(3) b`. Distance of center of the cube from its each corner is `r = (sqrt(3))/(2) b` Total PE of charge q at the center is `(-8 q(q))/(4 pi epsilon_0 r) = (-8 q^2)/(4 pi epsilon_0 sqrt(3)(b //2))= (-4q^2)/(pi epsilon_0 sqrt(3) b)`. |
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| 82. |
If identical charges `(-q)` are placed at each corner of a cube of side `b`, then electric potential energy of charge `(+q)` which is palced at centre of the cube will beA. `(8 sqrt(2)q^(2))/(4pi epsilon_(0)b)`B. `(-8sqrt(2)q^(2))/(pi epsilon_(0)b)`C. `(-4sqrt(2)q^(2))/(piepsilon_(0)b)`D. `(-4 q^(2))/(sqrt(3)pi epsilon_(0)b)` |
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Answer» Correct Answer - D |
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| 83. |
If identical charges `(-q)` are placed at each corner of a cube of side `b`, then electric potential energy of charge `(+q)` which is palced at centre of the cube will beA. `(8sqrt(2)q^(2))/(4pi epsilon_(0)b)`B. `(-8sqrt(2)q^(2))/(pi epsilon_(0)b)`C. `(-4sqrt(2)q^(2))/(pi epsilon_(0)b)`D. `(-4sqrt(2)q^(2))/(sqrt(3)pi epsilon_(0)b)` |
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Answer» Correct Answer - D Length of the diagonal of a cube having each side `b` is `sqrt(3)b`. So distance of centre of cube from each vertex is `(sqrt(3) b)/(2)`. Hence potential, energy of the given system of change is `U = 8 xx {(1)/(4pi epsilon_(0)).((-q)(q))/(sqrt(3)b//2)} = (-4q^(2))/(sqrt(3) pi epsilon_(0)b)` |
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| 84. |
Marke the correct statements :A. a given conducting sphere can be charged to any extentB. a given conducting sphere cannot be charged to a potential greater than a certain valueC. a given conducting sphere cannot be charged to a potential less than a certain minimum valueD. none of the above |
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Answer» Correct Answer - B::C |
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| 85. |
A conducting sphere of radius r has a charge . Then .A. the charge is uniformly distributed over its surface. If there is no external electric fieldB. distribution of charge over its surface will be non-uniform, if an external electric field exists in the space.C. the electric field strength inside the space will be equal to zero only when no external electric field existsD. potential at every point of the sphere must be the same |
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Answer» Correct Answer - A::D |
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| 86. |
The electric potential decreases unifromly from 120 V to 80 V as one moves on the x-axis from x=-1 cm to x=+1 cm. The electric field at the originA. must be equal to `20 V cm^-1`B. must be equal to `20 Vm^-1`C. may be greater than `20 Vm^-1`D. may be less than `20 V cm^-1` |
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Answer» Correct Answer - c `E_x = (d V)/(d x) = (120 - 80)/(2) = 20 Vm^-1` There may be the y - and z - components of field also. |
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| 87. |
A spherical conductor of radius `2 m` is charged to a potential of `120 V`. It is now placed inside another hollow spherical conductor of radius `6 m`. Calculate the potential to which the bigger sphere would be raisedA. 20 VB. 60 VC. 80 VD. 40 V |
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Answer» Correct Answer - D |
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| 88. |
If yoy know ( E) at a given point, can yoy calculate V at that point ? If not, what further information do you need ? |
| Answer» If we know E in certain region, then we can find potential difference between any two points in that region. To find potential at one point, we should know the potential at some other points. | |
| 89. |
An arc of radius r carries charge. The linear density of charge is `lamda` and the arc subtends an angle `(pi)/(3)` at the centre.A. `(lamda)/(4 epsilon_(0))`B. `(lamda)/(8epsilon_(0))`C. `(lamda)/(12 epsilon_(0))`D. `(lamda)/(16 epsilon_(0))` |
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Answer» Correct Answer - C |
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| 90. |
The electric field in a region is given by `E = (4 axy sqrt(z))hat i + (2 ax^2 sqrt(z)) hat j + (ax^2 y// sqrt(z)) hat k` where A is a positive constant. The equation of an equipotential surface will be of the form. |
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Answer» At equipotential, surface potential is constant. Therefore, `E = (- d v)/(d r) - int d v = int vec E . d vec r` `vec r = x hat i + y hat j + z hat k` or `d vec r = d x hat i + d y hat j + d z hat k` `int vecE (dx hat i + d y hat j + d z hat k) = constant` or `int [(4 a x y sqrt(z)) hat i + (2 a x^2 sqrt(z)) hat j + (a x^2 y// sqrt(z)) hat k]` `(d x hat i + d y hat j + d z hat k) = constant` or `int 4 a x y sqrt(z) d x + int2 ax^2 sqrt(z) d y + int (a x^2 y d z)/sqrt(z) = constant` or `(4 a x^2 y sqrt(z))/(2) + 2a x^2 y sqrt(z) + 2 a x^2 y sqrt (z) = constant` or `6 a x^2y sqrt(z) = constant` or `sqrt(z) = (constant)/(6 a x^2 y)` or `z = (constant)/(x^4 y^2)`. |
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| 91. |
Assertion: If a dielectric is placed in external field then field inside dielectric will be less than applied field. Reason: Electric field will induce dipole moment opposite to field direction.A. If both assertion and reason are true and reson is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If the assertion is true but the reason are false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - C Dipole moment will be in the same direction as the external field. The collective effect of dipole moments produces a field that oppose the external field and hence, the net electric field inside the dielectric is less than the external electric field. |
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| 92. |
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 `Q_(1) = Q_(2) + Q_(3)` because in series combination charge same on both condenser and `V = V_(1) + V_(2)` because in parallel combination `V_(2) = V_(3)`. Hence `V = V_(1) + V_(2)` |
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| 93. |
A uniform surface charge of density `sigma` is given to a quarter of a disc extending up to infinity in the first quadrant of `x-y` plane. The centre of the disc is at the origin `O`. Find the z-component of the electric field at the point `(0,0,z)` and the potential difference between the point `(0,0,d)` and `(0,0,2d)` |
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Answer» Correct Answer - `(sigma)/(8 epsilon_(0))` |
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| 94. |
Potential difference beween centre and surface of the sphere of radius R and uniorm volume charge density `rho` within it will beA. `(rho R^(2))/(6 epsilon_(0))`B. `(rho R^(2))/(4 epsilon_(0))`C. 0D. `(rhoR^(2))/(2 epsilon_(0))` |
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Answer» Correct Answer - A |
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| 95. |
Electric lines of force are as shown in the figure. Then potential at point `P` A. is zeroB. is not zeroC. may be zero alsoD. is not defined |
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Answer» Correct Answer - B |
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| 96. |
Four similar charges each of charge `+q` are placed at four corners of a square of side a. Find the value of integral `-int_(infty)^(a//sqrt(2)) vec(E).vec(d)r.` (in volts) if value of integral `-int_(a//sqrt(2))^(0) vec(E).vec(dr)=(sqrt(2)Kq)/(a)("where" K=(1)/(4 pi epsilon_(0)))`A. `(4Kq)/(a)`B. `(3Kq)/(a)`C. `(3sqrt(2)Kq)/(a)`D. `(4sqrt(2)Kq)/(a)` |
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Answer» Correct Answer - C |
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| 97. |
If you carry out the integral of the electric field `int vec E. vec d l` for a closed path like that as shown in (Fig. 3.40), the integral will always be equal tom zero, independent of the shape of the path and independent of where charges may be located relative to the path. Explain why. . |
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Answer» `V_f - V_i = int _i^f vec E. d l` In a closed path, `V_i = V_f`. Hence, `int_i ^f vec E. d l` in a close path will be zero. |
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| 98. |
Two electric charges `12 mu C` and `-6 mu C` are placed `20 cm` apart in air. There will be a point `P` on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of `P` from `6 mu C` chrage isA. 0.10 mB. 0.15 mC. 0.20 mD. 0.25 m |
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Answer» Correct Answer - C |
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| 99. |
How can you check whether the electric potential in a given region of space has constant value ? |
| Answer» Electric field intensity E in that region should be zero if potential is constant in a given region. This we can check by putting a charged particle in that region. If it experiences no force, E should be zero. | |
| 100. |
The electric field inside a hollow charged conductor is zero. Is this true or false ? |
| Answer» True, because induced charges cancel the external electric field. | |