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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.
| 201. |
A ring of radius `0.5 m` carries a total charge of `1.0xx 10^(-10)C` distributed non uniformly on its circum Ference, producting an electric field `vecE`. The value of `int_(r = oo)^(r = 0) -vecE.vecd r (r = 0` being the centre of the ring)isA. `+1.8 V`B. `-6 V`C. `-1.8 V`D. `+6 V` |
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Answer» Correct Answer - A `V = (9 xx 10^(9) xx 1xx 10^(-10))/(0.5) "volt" = 1.8 "volt"`. |
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| 202. |
The capcity of a parallel plate condenser is `10 muF`, when the distance between its plates is `8 cm`. If the distance between the plates is reduced to `4 cm`, then the capacity of this parallel plate condenser will beA. `5 muF`B. `10 muF`C. `20 muF`D. `40 muF` |
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Answer» Correct Answer - C `C prop (1)/(d) rArr (C_(1))/(C_(2)) = (d_(2))/(d_(1))`, so `(C_(2))/(10) = (8)/(4) rArr C_(2) = 20 muF` |
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| 203. |
Two capacitors `C_(1) = 2 muF` and `C_(2) = 6 muF` in series, are connected in parallel to a third capacitor `C_(3) = 4 muF`. This arrangement is then connected to a battery of e.m.f `= 2V`, as shown in the figure. How much energy is lost by the battery in charging the capacitors A. `22 xx 10^(-6) J`B. `11 xx 10^(-6) J`C. `((32)/(3)) xx 10^(-6) J`D. `((16)/(3)) xx 10^(-6) J` |
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Answer» Correct Answer - B `C_(eq) = (C_(1)C_(2))/(C_(1) + C_(2)) + C_(3) = (2 xx 6)/(2 + 6) + 4 = 5.5 muF` Energy supplied `(E) = QV = CV^(2) = 22 xx 10^(-6) J` `P.E.` stored `(U) = (1)/(2) C_(eq)V^(2) = (1)/(2) xx 5.5 xx (2)^(2) = 11 xx 10^(-6) J` `rArr` Energy lost `= E - U = 11 xx 10^(-6) J` |
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| 204. |
Assertion: All the charge in a conductor gets distributed on whole of its outer surface. Reason: In a dynamic system, charges try to keep their potential energy minimum.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. |
| Answer» Correct Answer - A | |
| 205. |
A point charge is surrounded symmetrically by six identical charges at distance `r` as shown in the figure How much work is done by the froces of electrostatic repulsion when the point charge at the centre is removed to infinity ? A. zeroB. `(6Q)/(4pi epsilon_(0)r)`C. `(6Q)/(4pi epsilon_(0)r^(2))`D. `(6Q^(2))/(4pi epsilon_(0)r)` |
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Answer» Correct Answer - D The potential of the point charge due to individual charges are identical, the resualtant potential of the point charge is `6U` or `(6Q^(2))/(4pi epsilon_(0)r)`. Hence , the work done on moving the point charge to infinity is `(6Q^(2))/(4pi epsilon_(0)r)`. |
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| 206. |
A bullet of mass `2 gm` is having a charge of `2 muc`. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of `10 m//s`A. `5 kV`B. `50 kV`C. `5 V`D. `50 V` |
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Answer» Correct Answer - B By using `(1)/(2) mv^(2) = QV` Given, `m = 2g = 2 xx 10^(-3) kg, v = 10 m//s` `q = 2muC = 2 xx 10^(-6) C` Substituting the value in relation for `V`, we obtain `V = (2 xx 10^(-3) xx (10)^(2))/(2 xx 2 xx 10^(-6))` `= 50 xx 10^(3) V = 50 kV` |
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| 207. |
A parallel plate air capacitor has capcity `C` distance of separtion between plates is `d` and potential difference `V` is applied between the plates force of attraction between the plates of the parallel plate air capacitor isA. `(C^(2)V^(2))/(2d^(2))`B. `(C^(2)V^(2))/(2d)`C. `(CV^(2))/(2d)`D. `(CV^(2))/(d)` |
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Answer» Correct Answer - C `F = (Q^(2))/(2 epsilon_(0)A)` `:. Q = CV` and `C = (epsilon_(0)A)/(d) rArr epsilon_(A) = Cd` So `F = (C^(2)V^(2))/(2Cd) = (CV^(2))/(2d)` |
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| 208. |
Two equal charges `q` are placed at a distance of `2a` and a third charge `-2a` is placed at the midpoint. The potential energy of the system isA. `(q^(2))/(8pi epsilon_(0)a)`B. `(6q^(2))/(8pi epsilon_(0)a)`C. `-(7q^(2))/(8pi epsilon_(0)a)`D. `(9q^(2))/(8pi epsilon_(0)a)` |
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Answer» Correct Answer - C `U_(system) = (1)/(4pi epsilon_(0)) ((q)(-2q))/(a) + (1)/(4pi epsilon_(0)) ((-2q)(q))/(a) + (1)/(4pi epsilon_(0)) ((q)(q))/(2a)` `U_(system) = (7q^(2))/(8pi epsilon_(0)a)` |
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| 209. |
Four electric charges `+q, +q, -q` and `-q` are placed at the corners of a square of side 2L (see figure). The electric potential at point `A`, mid-way between the two charges `+q` and `+q`, is A. `(1)/(4pi epsilon_(0))(2L)/(L) (1+(1)/(sqrt(5)))`B. `(1)/(4pi epsilon_(0))(2L)/(L) (1-(1)/(sqrt(5)))`C. zeroD. `(1)/(4pi epsilon_(0))(2L)/(L) (1+sqrt(5))` |
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Answer» Correct Answer - B `V = (1)/(4pi epsilon_(0)) (q)/(r )` Here, the electric potential at point `A` `V = 2V_(+Ve) + 2V_(-Ve)` `V = (1)/(4pi epsilon_(0)) [(2q)/(L)-(2q)/(Lsqrt(5))]` `V = (2q)/(4pi epsilon_(0)L) (1-(1)/(sqrt(5)))` |
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| 210. |
Two charged particles having charge `1 muC` and `-muC` and of mass `50 gm` each are held ar rest while their separtion is `2m`. Find the relative velocity of the particles when their separation is `0.5m`.A. `(1)/(5)m//s`B. `(sqrt(3))/(5)m//s`C. `(3sqrt(3))/(5)m//s`D. `(3sqrt(3))/(10)m//s` |
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Answer» Correct Answer - C Speed of both will be same at any time. `(KE + PE)f = (KE + PE)i` `2((1)/(2)mv^(2)) - (9 xx 10^(9)xx10^(-12))/(0.5) = -(9 xx 10^(9) xx 10^(-12))/(2) + 0` `mv^(2) = 9 xx 10^(-3) [(1)/(0.5)-(1)/(2)]` Putting `m = 50//100, v = (3sqrt(3))/(10)m//s` |
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| 211. |
What is the potential energy of the equal positive point charges of `1 mu C` each held `1 m` apart in air ?A. `9 xx 10^(-3) J`B. `9 xx 10^(-3)eV`C. `2 eV//m`D. Zero |
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Answer» Correct Answer - A By using `U = 9 xx 10^(9) (Q_(1)Q_(2))/(r )` `rArr U = 9 xx 10^(9) xx (10^(-6) xx 10^(-6))/(1) = 9 xx 10^(-3) J` |
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| 212. |
A `2 muF` capacitor is charged to `100 V`, and then its plates are connected by a conducting Wire. The heat produced is .A. `1 J`B. `0.1J`C. `0.01 J`D. `0.001 J` |
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Answer» Correct Answer - C Heat produced = Energy of charged capacitor `= (1)/(2)CV^(2) = (1)/(2) xx (2 xx 10^(-6)) xx (100)^(2) = 0.01 J` |
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| 213. |
Two protons in a `U^(238)` nucleus are `6.0xx10^(-15)m` apart. What is their mutual electric potential energy? |
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Answer» Correct Answer - `2.4xx10^(5)eV` |
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| 214. |
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 The total energy before connection `= (1)/(2) xx 4 xx 10^(-6) xx (50)^(2) + (1)/(2) xx 2 xx 10^(-6) xx (100)^(2)` `= 1.5 xx 10^(-2) J` When connected in parallel `4 xx 50 + 2 xx 100 = 6 xx V rArr V = (200)/(3)` Total energy after connection `= (1)/(2) xx 6 xx 10^(-6) xx ((200)/(3))^(2) = 1.33 xx 10^(-2) J` |
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| 215. |
Two equal charges, `2.0xx10^(-7) C` each, are held fixed at a separation of 20cm. A third charge of equal magnitude is placed midway between the two both the charges. How much work is done by the electric field during the process? |
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Answer» Correct Answer - `-3.6xx10^(-3)J` |
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| 216. |
Two point charges are kept at a certain distance from one another. The graph represents the variartion of the potential along the straight line connecting the two charges. At what point is the electric field zero ? A. `1`B. `2`C. `3`D. none of these |
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Answer» Correct Answer - C `E_(x) = - (dV)/(dx) = 0 rarr` at point `3`. |
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| 217. |
A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed ny a spherical conducting shell B of radius b and the two are connected by a wire ? |
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Answer» If the charge on the sphere of radius a is q, then `V = (1)/(4 pi epsilon _0) q/a` `i.e, q = (4 pi epsilon_0 a) V` Now, when sphere A is enclosed by spherical conductor B and the two are connected by a wire, charge will reside on the outer surface of B and so the potential of B will be `V_B= (1)/(4 pi epsilon _0) q/b = (1)/(4 pi epsilon_0)(4pi epsilon_(0)a)/(b)V = a/b V` Now as sphere A is inside B, so its potential is `V_A = V_B = a/b (V)" " [ V "as a" lt b]`. (##BMS_V03_C03_S01_031_S01##). |
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| 218. |
A solid conducting sphere of radius `10 cm` is enclosed by a thin metallic shell of radius `20 cm`. A charge `q = 20 mu C` is given to the inner sphere is connected to the shell by a conducting wire.A. 12 JB. 9 JC. 24 JD. zero |
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Answer» Correct Answer - b On connecting, the entire amount of charge will shift to the outer sphere. Heat generated is `U_i - U_f = (q^2)/(8 pi epsilon_0 R_1) - (q^2)/(8 pi epsilon_0 R_2)` =`((20 xx 10^-6) xx 9 xx 10^9)/(2) [(1)/(0.01) - (1)/(0.20)] = 9 J`. |
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| 219. |
If the electric potential of the inner metal sphere is `10` volt & that of the outer shell is `5` volt then the potential at the centre will be A. 10 voltB. 5 voltC. 15 voltD. 0 |
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Answer» Correct Answer - A |
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| 220. |
Assertion: A positively charged rod is held near a neutral conducting solid sphere as illustrated below. The sphere lies on a insulated stand. The potential of ground (or earth) is zero. The potential at point `A` (point `A` need not be centre of the sphere) is higher compared to potential of gound (earth). Reason: In this situation of assertion, the potential at the centre of conducting sphere is positive. the solid sphere being conducting, potential at each point in the sphere is same.A. If both assertion and reason are true and reson is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A The potential at cnetre of sphere is only due to the charged rod and hence positive. The potential at centre of sphere due to induced charges on its surface is zero. Hence the net potential at centre is positive. The solid sphere being conducting, potential at each point in the sphere is same. Hence Assertion is true, Reason is true, Reason is a correct explanation for Assertion. |
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| 221. |
Inside a conducting hollow sphere of inner radius `R_(1)` and outer radius `R_(2)`, a point charge q is placed at a distance `x` from the center as shown in (Fig. 3.70.) Find. . (i) electric potential at C (ii) electric field and potential at a distance r from the center outside the shell. |
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Answer» The electric potential at center hue to this system is due to q, induced charge -q on inner surface, and induced charge +q on outer surface. Thus, `V_C = (K q)/(x) -(K q)/R_(1) + (K q)/R_(2)` Electric field and potential at a distance r from the center outside the shell will only be due to the charge on outer surface because outside cavity the field due all the cavity charges is always zero. As induced charge on inner surface of cavity always nullifies the effect of point charge inside it. `E_(out) = (K q)/(r^2) and V_(out) = (K q)/r`. |
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| 222. |
Three charge `+q, -q`, and `+2q` are placed at the vertices of a right angled triangle (isosceles triangle)as shown. The net electrostatic energy of the configuration is: A. `-(kq^(2))/(a) (sqrt(2 + 1))`B. `(Kq^(2))/(a)(sqrt(2) + 1)`C. `-(Kq^(2))/(a) (sqrt(2) -1)`D. None of these |
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Answer» Correct Answer - C `U = (Kq^(2))/(a) (-1+2-(2)/(sqrt(2)))` `= (Kq^(2))/(a) (1-sqrt(2)) rArr U = - (kq^(2))/(a)(sqrt(2)-1)` |
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| 223. |
A charge of `5 mu C` is placed at the center of a square `ABCD` of side `10 cm`. Find the work done `("in" mu J)` in moving a charge of `1 mu C` from A to B. |
| Answer» All the four corners will be at the same potential. | |
| 224. |
(Figure 3.139) shows three circular arcs, each of radius `R` and total charge as indicated. The net electric potential at the center of curvature. .A. `q/(2 pi epsilon_0 R)`B. `(5 Q)/(12 pi epsilon_0 R)`C. `(3 Q)/(32 pi epsilon_0 R)`D. none of these. |
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Answer» Correct Answer - a Charge can be considered as located at a distance `R` from the center. Total charge is `(Q - 2 Q+ 3Q) = 2 Q` Hence, `V = (1)/(4 pi epsilon_0) (2 Q)/r = Q/(2 pi epsilon_0 R)`. |
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| 225. |
Three charges `2q, -q, and -q` are located at the vertices of an equilateral triangle. At the center of the triangle,A. the field is zero but potential is non-zeroB. the field is non-zero but potential is zeroC. both field and potential are zeroD. both field and potential are non-zero |
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Answer» Correct Answer - B |
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| 226. |
Three charges `2q, -q, and -q` are located at the vertices of an equilateral triangle. At the center of the triangle,A. the field is zero but potential is nonzero.B. the field is nonzero but potential is zero.C. both field and potential are zero.D. both field and potential are nonzero. |
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Answer» Correct Answer - b E is a vector quantity, and `V` is a scalar quantity. |
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| 227. |
At a point in space, the eletric field points toward north. In the region surrounding this point, the rate of change of potential will be zero along.A. northB. southC. north - southD. east - west |
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Answer» Correct Answer - d Equipotential surfaces are perpendicular to the electric lines of forces. |
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| 228. |
a. A circle is drawn with center as a charge +q. What is the work done in moving a charge +q from B to C along the circumferebce of the circle ? B In the above question, if the charge +q is first taken from B to A and then from A to C, on which path is the magnitude of work greater ? . |
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Answer» a. Points B and C lie on equipotential surface. So work done is zero. b. `V_B - V_A = V_C - V_A`, so magnitude of change in potential is same, hence, magnitude of work done will be the same. |
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| 229. |
A nonconducting sphere of radius `R = 5 cm` has its center at the origin O of the coordinate system as shown in (Fig. 3.112). It has two spherical cavities of radius `r = 1 cm`, whose centers are at `0,3 cm` and `0,-3 cm`, respectively, and solid material of the sphere has uniform positive charge density `rho = 1 // pi mu Cm^-3`. Calculate the electric potential at point `P (4 cm, 0)`. . |
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Answer» Correct Answer - `(43)/(45 pi epsilon_0) xx 10^-10` Volt Charge on the sphere (including cavities) `Q = (4)/(3) pi R^3 rho = (4)/(3) pi (5 xx 10^-2)^3 (1)/(pi) xx 10^-6 = (500)/(3) xx 10^-12 C` Charge in a volume equal to that of cavity `q = (4)/(3) pi R^3 rho = (4)/(3) pi (1 xx 10^-2)^3 (1)/(pi) xx 10^-6 = (4)/(3) xx 10^-12 C` Potential at P `V_p = V_("whole sphere") - 2 V_("cavity")` =`(k Q)/(2 R)[3 - (x^2)/(R^2)] - 2(k q)/(5 xx 10^-2)` =`(9 xx 10^9 xx (500//3) xx 10^-12)/(2 xx 5 xx 10^-2)[ 3 - ((4)/(5))^2]` `-(2 xx 9 xx 10^9 xx (4//3) xx 10^-12)/(5 xx 10^-2)` =`(873)/(25) = 34.92 V`. |
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| 230. |
The electric field in a region surrounding the origin and along the x- axis is uniform. A small circle is drawn with the center at the origin cutting the axes at points `A, B, C, and D` having coordinates `(a,0),(0.a), (-a,0), and (0, -a)`, respectively, as shown in (Fig. 3.127). Then the potential is minimum at. .A. `A`B. `B`C. `C`D. `D` |
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Answer» Correct Answer - a Potential decreases in the direction of electric field. |
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| 231. |
Three charges `q_1 = 1 mu C, q_2 = -2 mu C, and q_3 = -1 mu C` are placed at `A (0,0,0), B(-1 , 2,3,) and C (2 , -1, 1)`. Find the potential of the system of three charges at `P( 1, - 2, -1)`. |
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Answer» If `vec r_(P i)` is the position of P from the charge, its potential at P is `V_i = (q_i)/(4 pi epsilon _0 |vec r _P - vec r _i |)` Then, potential at P due to charge at A is `V_1 = (q _1)/(4 pi epsilon_0 |vec r _P - vec r _A|)` `= (10^-6 xx 10^9 xx 9)/(|(hat i -2 hat j - hat k) - (0 hat i + 0 hat j + 0 hat k)| )= (9 xx 10^3)/(sqrt (6)) V` similarly, `V_2 = (q _2)/(4 pi epsilon _0 | vec r _P - vec r _B|)` `= (-2 xx 10^-6 xx 9 xx 10^9)/(|(hat i - 2 hat j - hat k) -(- hat i + 2 hat j + 3 hat k)|) = -3 xx 10^3 V` `V_3= (q_3)/(4 pi epsilon _0 | vec r _P - vec r _C|)` `= (-10^-6 xx 9 xx 10^9)/(|(hat i - 2 hat j - hat k)-(2 hat i - hat j + hat k)|) = -(9)/(sqrt(6)) xx 10^3 V` Then, `V_P = Sigma V_i = V_1 + V_2 +V_3 = -3 xx 10^3 V`. |
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| 232. |
(Fig. 3.95) shows four orientations of an electric dipole in an external electric field. Rank the orientations acording to the a. magnitude of the torque on the dipole, and b. potential energy of the dipole, greatest first. . |
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Answer» For 2 and 4, `tau = p E sin theta, U = -p E cos theta` For 1 and 3,` tau = p E sin (180 - theta) = p E sin theta` `U = -p E cos (180 - theta) = p E cos theta` Hence, torque is same for all, and U is greater for 1 and 3 than 2 and 4. |
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| 233. |
Determine the electric field strength vector if the potential of this field depends upon x - and y - coordinates as : (i) `V = a(x^2 - y^2)` (ii) `V = a x y`. |
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Answer» (i) As `V = a(x^2 - y^2)`, `E_x = - (del V)/(del x) = - (del)/(del x)a(x^2 - y^2) = -2 a x` and `E_y = -(del V)/(del y) = + 2 a y` `:. vec E = E_x hat i + E_y hat j = - 2 a x hat i + 2a y hat j = - 2 a(x hat i - y hat j)` (ii) As `V = a x y`, `E_x = (del V)/(del x) = - a y` and `E_y = -(del V)/(del y) = - a x` `:. vec E = E_x hat i + E_y hat j= - a y hat i - a x hat j = - a (y hat i + x hat j)`. |
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| 234. |
Two capacitances of capacity `C_(1)`and `C_(2)` are connected in series and potential difference `V` is applied across it. Then the potential difference across `C_(1)` will beA. `V(C_(2))/(C_(1))`B. `V(C_(1) + C_(2))/(C_(1))`C. `V (C_(2))/(C_(1) + C_(2))`D. `V(C_(1))/(C_(1) + C_(2))` |
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Answer» Correct Answer - C Charge flowing ` = (C_(1)C_(2))/(C_(1) + C_(2))V` Potential diff.across `C_(1) = (C_(1)C_(2)V)/(C_(1) + C_(2)) xx (1)/(C_(1)) = (C_(2)V)/(C_(1) + C_(2))` |
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| 235. |
The potential energy of a particle in a force field is: `U = (A)/(r^(2)) - (B)/(r )`,. Where `A` and `B` are positive constants and `r` is the distance of particle from the centre of the field. For stable equilibrium the distance of the particle isA. `B//A`B. `B//2A`C. `2A//B`D. `A//B` |
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Answer» Correct Answer - C Force experienced by the particle in field, `F = - ((dU)/(dr)) = - ((-2A)/(r^(3))+(B)/(r^(2)))` At equilibrium `F = 0 F = - ((-2A)/(r^(3))+(B)/(r^(2))) = 0` `rArr r = (2A)/(B)` |
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| 236. |
When charge `10 mu C` is shifted from inifinity to a point in an electric field, it is found that work done by eletrotatic forces is `10 muJ`. If the charge is doubled and taken again from inifinity to the same point without accelerating it, then find the amount of work done by electric field and against electric field. |
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Answer» `(W_("ext"))_(oo P) = - (W_(e l))_(oo P) = (W_(e l)) _(p oo) = 10 mu J` Because `Delta KE = 0` `V_P = ((W_("ext"))_(oo p))/q = (10 mu J)/(10 mu C)= 1 V` So if now the charge is doubled and taken from infinity, then `1 = ((W_("ext"))_( oo P))/(20 mu C)` or `(W_(ext))_( oo P) = 20 mu J` `(W_(e l))_( ooP) = -20 mu J`. |
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| 237. |
A charge `3 mu C` is released at rest from a point P where electric potential is (20 V). Find its kinetic energy when it reaches inifinity. |
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Answer» `W_(e l) = Delta K = K_f - 0` `(W_(e l))_(P rarr oo) = q V_P = 60 mu J` So `K_f = 60 mu J`. |
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| 238. |
A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge.A. remains a constant because the electric field is uniformB. increases because the charge moves along the electric fieldC. decreases because the charge moves along electric fieldD. decreases because the charge moves opposite to the electric field. |
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Answer» Correct Answer - C |
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| 239. |
A point charge `q` is located at the centre `O` of a spherical uncharged coducting layer provided with small orifice. The inside and outside radii of the layer are equal to a and `b` respectively. The amount of work that has to be performed to slowly transfer the charge `q` from teh point `O` through the orifice and into infinity is |
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Answer» Correct Answer - `(q^2)/(8 pi epsilon_0) ((1)/a - (1)/b)` Potential at center O due to induced charges is `V_1 = - q/(4 pi epsilon_0 a) + q/(4 pi epsilon_0 b)` `V_(oo)=0` small work done to transfer a small charge dq from center to infinity is `d W = d q (V_oo - V_1) = (1)/(4 pi epsilon_0) [(1)/a - (1)/b] q d q` or `W = int d W = (q^2)/(8 pi epsilon_0) [(1)/a - (1)/b]`. |
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| 240. |
A series combination of `n_(1)` capacitors, each of value `C_(1)`, is charged by a source of potential difference `4 V`. When another parallel combination of `n_(2)` capacitors, each of value `C_(2)`, is charged by a source of potential difference `V`, it has same (total) energy stored in it, as the first combination has. the value of `C_(2)`, in terms of `C_(1)`, is thenA. `(2C_(1))/(n_(1)n_(2))`B. `16 (n_(2))/(n_(1))C_(1)`C. `2(n_(2))/(n_(1))C_(1)`D. `(16 C_(1))/(n_(1)n_(2))` |
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Answer» Correct Answer - D Case I. When the capacitors are connected in series `U_("series") = (1)/(2) (C_(1))/(n_(1))(4V)^(2)` Case II. When the capacitors are connected in parallel `U_("parallel") = (1)/(2) (n_(2)C_(2)) V^(2)` According to question, `U_("series") = U_("parallel")` or `= (1)/(2)(C_(1))/(n_(1)) (4V)^(2) = (1)/(2) (n_(2)C_(2)) V^(2)` `rArr C_(2) = (16C_(1))/(n_(2)n_(1))` |
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| 241. |
The capacities of two conductors are `C_(1)` and `C_(2)` and their respectively potentials are `V_(1)` and `V_(2)`. If they are connected by a thin wire, then the loss of energy will be given byA. `(C_(1)C_(2)(V_(1) +V_(2)))/(2(C_(1) + C_(2)))`B. `(C_(1)C_(2)(V_(1) -V_(2)))/(2(C_(1) + C_(2)))`C. `(C_(1)C_(2)(V_(1) -V_(2)^(2)))/(2(C_(1) + C_(2)))`D. `((C_(1) + C_(2))(V_(1) - V_(2)))/(C_(1)C_(2))` |
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Answer» Correct Answer - C Initial energy `U_(i) = (1)/(2)C_(1)V_(1)^(2) + (1)/(2)C_(2)V_(2)^(2)`. Final energy `U_(f) = (1)/(2)(C_(1) + C_(2))V^(2)` (where `V = (C_(1)V_(1) + C_(2)V_(2))/(C_(1)C_(2)))` Hence energy loss `DeltaU = U_(i) - U_(f) = (C_(1)C_(2))/(2(C_(1) + C_(2))) (V_(1) - V_(2))^(2)` |
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| 242. |
Two condensers `C_(1)` and `C_(2)` in a circuit are jhioned as shown in . The potential fo point `A` is ` V_1` and that of `B` is ` V_2` . The potential of point `D` will be ` A. `(1)/(2) (V_(1) + V_(2))`B. `(C_(2)V_(1) + C_(1)V_(2))/(C_(1) + C_(2))`C. `(C_(1)V_(1) + C_(2)V_(2))/(C_(1) + C_(2))`D. `(C_(2)V_(1) - C_(1)V_(2))/(C_(1) + C_(2))` |
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Answer» Correct Answer - C Charge on `C_(1) =` charge on `C_(2)` `rArr C_(1) (V_(A - V_(D)) = C_(2) (V_(D) - V_(B))` `rArr C_(1) (V_(1) - V_(D)) = C_(2) (V_(D) - V_(2)) rArr V_(D) = (C_(1)V_(1) + C_(2)V_(2))/(C_(1) + C_(2))` |
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| 243. |
Two condensers `C_(1)` and `C_(2)` in a circuit are jhioned as shown in . The potential fo point `A` is ` V_1` and that of `B` is ` V_2` . The potential of point `D` will be ` A. `(1)/(2) (V_(1) + V_(2))`B. `(C_(2)V_(1) + C_(1)V_(2))/(C_(1) + C_(2))`C. `(C_(1)V_(1) + C_(2)V_(2))/(C_(1) + C_(2))`D. `(C_(2)V_(1) - C_(1)V_(2))/(C_(1) + C_(2))` |
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Answer» Correct Answer - C Charge on `C_(1) =` charge on `C_(2)` `rArr C_(1) (V_(A - V_(D)) = C_(2) (V_(D) - V_(B))` `rArr C_(1) (V_(1) - V_(D)) = C_(2) (V_(D) - V_(2)) rArr V_(D) = (C_(1)V_(1) + C_(2)V_(2))/(C_(1) + C_(2))` |
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| 244. |
A conducting sphere `S_1` of radius `r` is attached to an insulating handle. Another conduction sphere `S_2` of radius `R` is mounted on an insulating stand. `S_2` is initially uncharged. `S_1` is given a charge `Q` brought into contact with `S_2` and removed. `S_1` is recharge such that the charge on it is again `Q` and it is again brought into contact with `S_2` and removed. This procedure is repeated `n` times. a. Find the electrostatic energy of `S_2` after `n` such contacts with `S_1`. b. What is the limiting value of this energy as `nrarroo` ? |
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Answer» (i) When `S_1` and `S_2` come in contact, there is transfer of charges till the potentials of the two spheres become equal. During first contact, `V_1 = V_2 (q_1 " charge shifts from " S_1 " to " S_2)` `rArr (K(Q - q_1))/(r) = (Kq_1)/rR " or " q_1 = Q (R/(R + r))` During second contact, again `V_1 = V_2` `rArr (K[Q - (q_2 - q_1)]]/r = (K q_2)/R` `[(q_2 - q_1) "charge shifts from" S_1 "to" S_2]` `:. q_2 = Q[(R)/(R + r)+ ((R)/(R + r))^2]` On third contact, again `V_1 = V_2` `rArr (K[Q-(q_3 - q_2)])/(r) = (K q_3)/(R)` `[(q_3 - q_2) "charge shifts from" S_1 "to" S_2]` `:. q_3 = Q[(R)/(R + r)+ ((R)/(R + r))^2 + ((R)/(R + r))^3]` On nth contact, by symmertry `V_1 = V_2` `rArr (K[Q - (q_n - q_(n-1))])/(r) = (K q_n)/(R)` `[(q_n - q_(n-1) "charge shift from" S_1 "to" S_2]` `q_n = Q[(R)/(R + r) + ((R)/(R + r))^2 +...+ ((R)/(R + r))^n]` =`(Q R)/(r) [1 -((R)/(R + r))^n]` The eletrostatic energy of `S_2` after n contacts is `U_n = (1)/(2) (q_n^2)/(C) = (1)/(2) xx (1)/(4 pi epsilon_0 R) xx {(Q R)/(r) [1 -((R)/(R + r))^n]}^2` (ii) The limiting value is ` underset(n rarr oo)(Lt) U_n = underset(n rarr oo)(Lt) [(1)/(2) xx (1)/(4 pi epsilon_0 R){(Q R)/(r)[1 -((R)/(R + r))^n]}^2]` =`(Q^2 R)/(2( 4 pi epsilon_0)r^2)`. |
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| 245. |
A small metal sphere, carrying a net charge `q_1 = -2 mu C`, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of `q_2 = -8 mu C` and mass `1.50 g`, is projected toward `q_1`. When the two spheres are `0.800 m` apart, `q_2` is moving toward `q_1` with speed `20 ms^-1` as shown in (Fig. 3.151). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. . The speed of `q_2` when the spheres are `0.400 m` apart is.A. `2 sqrt(10) ms^-1`B. `2 sqrt(6) ms^-1`C. `4 sqrt(10) ms^-1`D. `4 sqrt(6) ms^-1` |
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Answer» Correct Answer - C `E_i = K_i + U_i = (1)/(2) (0.0015 kg)(20.0 ms^-1)^2+(k (2.00 xx 10^-6 C)(8.00 xx 10^-6 C))/(0.800 m)` =`0.48 J` `E_i = E_f = (1)/(2) mv_f^2 + (k q_1 q_2)/(r_f)` `v_f = sqrt((2(0.48 J - 0.3t6 J))/(0.0015 kg)) = 4 sqrt(10) ms^-1` At the closest point, the velocity is zero. `0.48 J = (kq_1 q_2)/(r)` or `r = (k(2.00 xx 10^-6 C)(8.00 xx 10^-6 C))/(0.48 J) =0.30 m`. |
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| 246. |
A small sphere with mass `1.2 g` hangs by a thread between two parallel vertical plates `5.00 cm` apart. The plates are insulating and have uniform surface charge densities `+ sigma and - sigma`. The charge on the sphere is `q = 9 xx 10^-6 C`. What potential difference between the plates will cause the thread to assume an angle of `37^(@)` with the vertical as shown in (Fig. 3.134) ? .A. 30 VB. 12 VC. 50 VD. 25 V |
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Answer» Correct Answer - c `F_e = m g tan theta` =`(1.20 xx 10^-3 kg)(10 ms^-2) tan (37^(@)) = 0.0090 N` `F_e = E q = (V q)/D` `:. V = (F_e d)/q = ((0.009 N)(0.0500 m))/(9.0 xx 10^-6 C) = 50.0 V`. |
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| 247. |
An insulating rod having linear charge density `lambda = 40.0 mu Cm^-1` and linear mass dinsity `mu = 0.100 kg m^-1` is released from rest in a uniform electric field `E = 100 Vm^(-1)` directed perpendicular to the rod. (a) Determine the speed of the rod after it has travelled 2.00 m (b) How does your answer to part (a) change if the electric field is not perpendicular to the rod ? |
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Answer» Arbitrarily take `V = 0` at the initial point. Then at distance d downfield, where L is the rod length, `V = -E d` and `U_e = lambda L E d` a. `(K + U)_i = (K + U)_f` or `0 + 0 = (1)/(2) mu L v^2 - lambda L E d` or `v = sqrt((2 lambda E d)/mu) = 0.400 ms^(-1)` b. The same. |
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| 248. |
A small metal sphere, carrying a net charge `q_1 = -2 mu C`, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of `q_2 = -8 mu C` and mass `1.50 g`, is projected toward `q_1`. When the two spheres are `0.800 m` apart, `q_2` is moving toward `q_1` with speed `20 ms^-1` as shown in (Fig. 3.151). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. . How close does `q_2` get to `q_1` ?A. 0.20 mB. 0.30 mC. 0.10 mD. 0.15 m |
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Answer» Correct Answer - B `E_i = K_i + U_i = (1)/(2) (0.0015 kg)(20.0 ms^-1)^2+(k (2.00 xx 10^-6 C)(8.00 xx 10^-6 C))/(0.800 m)` =`0.48 J` `E_i = E_f = (1)/(2) mv_f^2 + (k q_1 q_2)/(r_f)` `v_f = sqrt((2(0.48 J - 0.3t6 J))/(0.0015 kg)) = 4 sqrt(10) ms^-1` At the closest point, the velocity is zero. `0.48 J = (kq_1 q_2)/(r)` or `r = (k(2.00 xx 10^-6 C)(8.00 xx 10^-6 C))/(0.48 J) =0.30 m`. |
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| 249. |
A point charge `q_(1)=+6e` fixed at the origin of a coordinate system, and another point charge `q_(2)=-10e` is fixed at `x=8nm`, `y=0` the locus of all points in the xy plane for which potential `V=0` (other man inifinity) is a circle centered on the x-axis as shown. Q. The potential at the centre of the circle isA. 0.32 VB. 0.77 VC. 1.2 VD. `-1.2 V` |
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Answer» Correct Answer - B |
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| 250. |
A point charge `q_(1)=+6e` fixed at the origin of a conducting system, and another point charge `q_(2) = -10e` is fixed at `x=8` nm, `y=0`. The locus of all points in the `xy` plane for which potential `V=0` (other than infinity) is a circle contered on the x-axis, as shown x-coordinate of the centre of the circle isA. `-2` nmB. `-3` nmC. `-4.5` nmD. `-7.5` nm |
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Answer» Correct Answer - C |
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