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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.
| 1301. |
The outer cylinder of a cylindrical capacitor of length 0.15 m and radial 1.61 cm and 1.5 cm is earthed while inner cylinder of this capacitor is given a charge of `8 muC`. Find the capacitance and potential of inner part of the capacitor. |
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Answer» Here, l = 0.15 m, `r_(b) = 1.61xx10^(-2)m, r_(a) = 1.50xx10^(-2) m` `q = 8.0xx10^(-6) C, C = ?, V = ?` For cylindrical capacitor, `C = (2pi in_(0) l)/(2.3031 log_(10) ((r_(b))/(r_(a))))` `= (2pi(8.854xx10^(-12)) (0.15))/(2.303 log_(10) ((1.61xx10^(-2))/(1.50xx10^(-2))))` `C = 1.18xx10^(-10) F`. `V = (q)/(C ) = (8.0xx10^(-6))/(1.18xx10^(-10)) = 6.78xx10^(4) V` |
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| 1302. |
Two capacitors of capacitance of `6 muF and 12 muF` are connected in series with a battery. The voltage across the `6 muF` capacitor is 2V. Compute the total battery voltage. |
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Answer» As capacitors are connected in series, charge on each capacitor must be same. Charge on `6 muF` capacitor = charge on `12 muC` capacitor `6xx10^(-6)xx2 = 12xx10^(-6)xx V_(2)` `V_(2) = (6xx2)/(12) = 1 vol t` Total battery voltage `= V_(1) + V_(2) = 2 + 1 = 3V` |
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| 1303. |
Seven capacitors each of capacitance `2muF` are to be connected in a configuration to obtain an effective capacitance of `(10/11)muF`. Which of the combination (s) shown in figure will achieve the desired result? |
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Answer» Suppose a parallel combition of n capacitors is connected in series with a series with a series combition of `(7-n)` capacitors. Capacitance of paralllel combination, `C_(1) = 2n muF` Capacitance at `(7-n)` series, and `C_(s) = (10)/(11) muF`, therefore, form `(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(2n) + (7-n)/(2) = (11)/(10)` Multiplying both sides by 10n, we get `5+5n (7-n) = 11n` `5+35n-5 n^(2) = 11n` or `5 n^(2) - 24n - 5 = 0` `(n-5) (5n + 1) = 0` Rejecting parallel combination of 5 capacitors must be connected in series with combination of 2 capacitors. |
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| 1304. |
If infinite parallel plane sheet of a metal is charged to charge density `sigma` coulomb per square metre in a medium of dielectric constant K. Intensity of electric field near the metallic surface will beA. `E=(sigma)/(epsilon_(0)K)`B. `E=(K)/(3 epsilon_(0))`C. `E=(sigma)/(2 epsilon_(0)K)`D. `E=(K)/(2epsilon_(0))` |
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Answer» Correct Answer - C We have, `E=(sigma)/(2epsilon)=(sigma)/(2epsilon_(0)K)" "(because(epsilon)/(epsilon_(0))=K)` |
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| 1305. |
A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field `vecE` at the centre O is A. `q/(4pi^(2) epsi_(0) r^(2)) hat(j)`B. `-q/(4pi^(2) epsi_(0) r^(2)) hat(j)`C. `-q/(2pi^(2) epsi_(0) r^(2)) hat(j)`D. `q/(2pi^(2) epsi_(0) r^(2)) hat(j)` |
| Answer» Correct Answer - C | |
| 1306. |
An infinite dielectric sheet having charge density `sigma` has a hole of radius `R` in it. An electron is released on the axis of the hole at a distance `sqrt(3)R` from the center. Find the speed with which it crosses the center of the hole. |
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Answer» Potential function is not defined for infinite conducting sheet and hence to solve this either calculate potential difference or use force equations Electric field due to infinite dielectric sheet, `E_(1)=(sigma)/(2epsilon_(0))` Electric field at the axis of a disc of radius `R`. `E_(2)=(sigma)/(2epsilon_(0))[1-(x)/(sqrt(x^(2)-R^(2)))]` Resultant electric field `E=E_(1)-E_(2)=(sigma)/(2epsilon_(0))-(x)/(sqrt(x^(2)+R^(2)))` Force on the direction `F=-(sigmaex)/(2epsilon_(0)sqrt(x^(2)+R^(2)))` `mv(dv)/(dx)=-(sigmaex)/(2epsilon_(0)sqrt(x^(2)+R^(2)))` `m int_(0)^(v)vdv=-(sigmae)/(2epsilon_(0))int_(sqrt(3R))^(0)=(x)/(sqrt(x^(2)+R^(2)))dx` `m(v^(2))/(2)=-(sigmae)/(2epsilon_(0))[sqrt(x^(2)+R^(2))]_(sqrt(3R))^(0)` `v=sqrt((sigmaeR)/(mepsilon_(0)))` |
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| 1307. |
In an electron gun, the electrons are accelerated by the potential `V`. If the `e` is the charge and `m` is the mass of the electron, then the maximum velocity of these electrons will beA. `sqrt((eV)/(m))`B. `sqrt((e)/(Vm))`C. `sqrt((2eV)/(m))`D. none of these |
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Answer» Correct Answer - C `W=qV" ""i.e. "1/2mv^(2)=qV` `therefore" "v=sqrt((2qV)/m)=sqrt((2eV)/m)" "(thereforeq=e)` |
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| 1308. |
If a conductor has a potential `V != 0` and there are no charges anywhere else outside, thenA. Their must be charges on the surface or inside itselfB. there cannot be any charge in the body of the conductorC. there must be charges only on the surfaceD. there must be charges inside the surface |
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Answer» Correct Answer - A::B From the knowledge of theory, if a conductor has a potential `V != 0` and there are no charges anywhere elese outside, then there must be charges on the surface of conductor of inside the conductor. However, there cannot be any charge in the body of the conductor Choices (a) and (b) are correct. |
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| 1309. |
The electric charges of +9C, -10C and -8C are enclosed anywhere by a closed surface. T.N.E.I. through that surface isA. `-9 C`B. `+9 C`C. `5 C`D. `4 C` |
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Answer» Correct Answer - A `q_(1)=9C,q_(2)=-10C,q_(3)=-8C, T.N.E.I.=?` `T.N.E.I. = q_(1)+q_(2)+q_(3)` `=+9-10-8=-9C` |
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| 1310. |
What is electric field intensity at a point at a distance r meter from q coulomb of a charge in free space ? |
| Answer» `E = (1)/(4pi in_(0)) (q)/(r^(2))` , where `in_(0)` is absoulute electrical permittively of free space. | |
| 1311. |
Charges `Q_(1)` and `Q_(2)` lie inside and outside respectively of an uncharged conducting shell. Their separation is `r`. (i) The force on `Q_(1)` is zero (ii) The force on `Q_(1)` is `k((Q_(1) Q_(2))/(r^(2)))` (iii) The force on `Q_(2)` is `k((Q_(1) Q_(2))/(r^(2)))` (iv) The force on `Q_(2)` is zeroA. The force on `Q_(1)` is zeroB. The force on `Q_(1)` is `k(Q_(1)Q_(2))/(r^(2))`C. The force on `Q_(2)` is `k(Q_(1)Q_(2))/(r^(2))`D. The force on `Q_(2)` is zero. |
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Answer» Correct Answer - A::B::C::D |
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| 1312. |
The difference between the radill of the two spheres of a spherical capacitor is increased. Will the capacitance increase or decrease ? |
| Answer» Thecapacitance C decreases as `C = (4pi in_(0) ab)/(b-a)` , where a,b are radil of two spheres. So if `b-a` increases , capacitance C will decreases. | |
| 1313. |
Two charged conducting spheres of radill a and b are connected to eachother by a wire. What is the ratio of electric fields at the surface of two spheres ? Use the result obtained to explain why charge density on the sharp and pointed ends of a conducter is higher than on its fatter portions ? |
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Answer» The charge flows from the sphere at higher potential to the other at lower potential, till their potentials become equal. After sharing, the charges on two spheres would be `(Q_(1))/(Q_(2)) = (C_(1) V)/(C_(2) V)` where `C_(1), C_(2)` are the capaciteres of two spheres. But `(C_(1))/(C_(2)) = (a)/(b) :. (Q_(1))/(Q_(2)) = (a)/(b)` Ratio of surface density of charge on the two spheres: `(sigma_(1))/(sigma_(2)) = (Q_(1))/(4pi a^(2)) . (4pi b^(2))/(Q^(2)) = (Q_(1))/(Q_(2)) . (b^(2))/(a^(2)) = (a)/(b) (b^(2))/(a^(2)) = (b)/(a)` Hence ratio of electric fields at the surfaces of two spheres `(E_(1))/(E_(2)) = (sigma_(1))/(sigma_(2)) = (b)/(a)` A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius. Therefore, charge density on sharp and pointed ends of conductor is much higher than on its flatter portions. |
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| 1314. |
A solid sphere of radius R has a volume charge density `rho=rho_(0) r^(2)` (where `rho_(0)` is a constant ans r is the distance from centre). At a distance x from its centre (for `x lt R`), the electric field is directly proportional to :A. `1//x^(2)`B. `1//x`C. `x^(3)`D. `x^(2)` |
| Answer» Correct Answer - C | |
| 1315. |
Two charges, each equal to q, aer kept at `x=-a` and `x=a` on the x-axis. A particle of mass m and charge `q_0=q/2` is placed at the origin. If charge `q_0` is given a small displacement (ylt lt a) along the y-axis, the net force acting on the particle is proportional toA. yB. `-y`C. `1/y`D. `-1/y` |
| Answer» Correct Answer - A | |
| 1316. |
A solid sphere of radius `R` is charged uniformly. The electrostatic potential `V` is plotted as a function of distance `r` from the centre of th sphere. Which of the following best represents the resulting curve?A. B. C. D. |
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Answer» Correct Answer - C |
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| 1317. |
a point charge `q` is situated at a distance `r` from one end of a thin conduction rod of length `L` having a charge `Q` (uniformly distributed a long its length).find the magnitudes of electric force between the two.A. `(Kqq)/r^(2)`B. `(2KQ)/(r(r+L))`C. `(KQq)/(r(r-L))`D. `(KQq)/(r(r+L))` |
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Answer» Correct Answer - D |
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| 1318. |
In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as function of distance from the centre, The graph which would correspond to the above will be:A. B. C. D. |
| Answer» Correct Answer - C | |
| 1319. |
In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as function of distance from the centre, The graph which would correspond to the above will be:A. B. C. D. |
| Answer» Correct Answer - C | |
| 1320. |
In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as function of distance from the centre, The graph which would correspond to the above will be:A. B. C. D. |
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Answer» Correct Answer - C |
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| 1321. |
A solid non conducting sphere of radius R and uniform volume charge density `rho` has its centre at origin. Find out electric field intensity in vector form at following positions : (i) `(R//2, 0, 0)" "`(ii) `(R/sqrt(2), R/sqrt(2), 0)" "`(iii) `(R, R, 0)` |
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Answer» (i) At `(R//2, 0, 0)` : Distance of point from centre `=sqrt((R//2)^(2)+0^(2)+0^(2))=R//2 lt R`, so point lies inside the sphere, so `vec(E)=(p vec(r))/(3 epsi_(0))=rho/(3 epsi_(0)) [R/2 hat(i)]` (ii) At `(R/sqrt(2), R/sqrt(2), 0)` distance of point from centre `=sqrt((R//sqrt(2))^(2)+(R//sqrt(2))^(2)+0^(2))=R=R`, so point lies at the surface of sphere, therefore `vec(E)=(KQ)/R^(3) vec(r) =(k 4/3 pi R^(2) rho)/R^(3) =[R/sqrt(2) hat(i) +R/sqrt(2)hat(j)]=rho/(3 epsi_(0))[R/sqrt(2) hat(i) +R/sqrt(2)hat(j)]` (ii) The point is outside the sphere So, `vec(E)=(KQ)/r^(3) vec(r) =(K4/3 pi R^(3) rho)/((sqrt(2) R)^(3)) [Rhat(i)+Rhat(j)]=rho/(6sqrt(2) epsi_(0)) [Rhat(i)+Rhat(j)]` |
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| 1322. |
A table tennis ball covered with a conducting paint is suspended by a silk thread so that it hangs between tow metal plates. One plate is earthed, when the other plate is connected to a high voltage generator, what will happen to the ball. A. is attached to the high voltageB. hangs without movingC. swings backward and forward hitting the platesD. is repelled to the earthed plate and stays there |
| Answer» Correct Answer - 3 | |
| 1323. |
Figure shows a uniformly charged sphere of radius R and total charge Q. A point charge q is situated outside the sphere at a distance r from centre of sphere. Find out the following : (i) Force acting on the point charge q due to the sphere. (ii) Force acting on the sphere due to the point charge. |
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Answer» (i) Electric field at the position of point charge `vec(E)=(KQ)/r^(2) hat(r)` So, `vec(F)=(KqQ)/r^(2) hat(r)" "|vec(F)|=(K qQ)/r^(2)` (ii) Since we know that every action has equal and opposite reaction so `vec(F)_("sphere")=(K qQ)/r^(2) hat(r)` `vec(F)_("sphere")=(K qQ)/r^(2)` |
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| 1324. |
Two concentric uniformly charge spherical shells of radius `R_(1)` and `R_(2) (R_(2) gt R_(1))` have total charges `Q_(1)` and `Q_(2)` respectively. Derive an expression of electric field as a function of r for following positions. (i) `r lt R_(1)` (ii) `R_(1) le r lt R_(2)` (iii) `r ge R_(2)` |
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Answer» (i) For `r lt R_(1)`, therefore, point lies inside both the spheres `E_("net")=E_("inner")+E_("outer")=0+0` (ii) For `R_(1) le r lt R_(2)`, point lies outside inner sphere but inside outer sphere : `:. E_("net")=E_("inner")+E_("outer")` `(KQ_(1))/r^(2) hat(r)+0=(KQ_(1))/r^(2) hat(r)` (iii) For `r ge R_(2)` point lies outside inner as well as outer sphere. Therefore, `E_("Net")=E_("inner")+E_("outer")=(KQ_(1))/r^(2) hat(r) + (KQ_(2))/r^(2) hat(r) =(K (Q_(1)+Q_(2)))/r^(2) hat(r)` |
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| 1325. |
Two concentric conducting thin spherical shells A andB having radii rA and `r8 (r_(8) gt r_(A))` are charged to `Q_(A)` and `- Q_(B)(|Q_(B)| gt |Q_(A)|)`. The electric field strength along a line passing through the centre varies with the distance x as :A. B. C. D. |
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Answer» Correct Answer - A |
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| 1326. |
Two large , parallel conducting plates `X` and `Y` , kept close to each other , are given `Q_(1)` and `Q_(2) (Q_(1) gt Q_(2))`. The four surfaces of the plates are `A,B,C` and `D` , as shown (i) The charge on `A` is `(1)/(2) (Q_(1) + Q_(2))` (ii) The charge on `B` is `(1)/(2) (Q_(1) - Q_(2))` (iii) The charge on `C` is `(1)/(2) (Q_(2) - Q_(1))` (iv) The charge on `D` is `(1)/(2) (Q_(1) + Q_(2))`A. The charge on `A` is `1/2 (Q_(1)+Q_(2))`B. The charge on `B` is `1/2 (Q_(1)-Q_(2))`C. The charge on `C` is `1/2(Q_(2)-Q_(1))`D. The charge on `D` is `1/2(Q_(1)+Q_(2))` |
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Answer» Correct Answer - A::B::C::D |
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| 1327. |
Ten positively charged particles are kept fixed on the x-axis at points x=10cm, 20cm, 30cm, …, 100cm. The first particle has a charge `1.0xx 10^(-8)C`, the second `8xx10^(-8)` C, the third `27xx10(-8)`C and so on. The tenth particle has a charge `1000xx10^(-8)C.` find the magnitude of the electric force acting on a 1 C charge placed at the origin. |
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Answer» By superposition principle, force on charge 1C placed at origin `F_(0) = F_(01) + F_(02) +…. + F_(010)` = `(q_(0))/(4pi in_(0)) [(q_(1))/(r_(1)^(2)) + (q_(2))/(r_(2)^(2)) + (q_(10))/(r_(n)^(2))]` = `9xx10^(9)xx1 [(1.0xx10^(-8))/((0.10)^(2)) + (8xx10^(-8))/((0.20)^(2)) + (27xx10^(-8))/((0.30)^(2)) +(1000xx10^(-8))/((1.00)^(2))]` `= 9xx10^(9)xx10^(-6) [1+2+3+...+10]` = `9xx10^(3)xx55 = 4.95xx10^(5)N` |
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| 1328. |
A uniform electric field of 20 N/C exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 40 cm. |
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Answer» `E=-(dv)/(dr) implies dv=-vec(E). Vec(dr)` for `vec(E)=` constant `implies" "DeltaV=- vec(E). Vec(Delta r)` `DeltaV=-20 (-hat(j)). (40xx10^(-2)) hat(j)=8` volts. |
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| 1329. |
A thick conducting spherical shell of inner radius a and outer radius b is shown in figure. It is observed that the inner face of the shell carries a uniform charge density `-sigma`. The outer surface also carries a uniform surface charge density `+sigma`. (a) Can you confidently say that there must be a charge inside the shell? Find the net charge present on the shell. (b) Find the potential of the shell. |
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Answer» Correct Answer - (a). `Yes` `4pisigma(b^(2)-a^(2))` (b) `(sigmab)/(in_(0))` |
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| 1330. |
An electron (charge = e, mass = m) is projected horizontally into a uniform electric field produced between two oppositely charged parallel plates, as shown in figure. The charge density on both plates is `+- sigma C//m^2` and separation between them is d. You have to assume that only electric force acts on the electron and there is no field outside the plates. Initial velocity of the electron is u, parallel to the plates along the line bisecting the gap between the plates. Length of plates is 2L and there is a screen perpendicular to them at a distance L. (i) Find s if the electron hits the screen at a point that is at same height as the upper plate. (ii) Final the angle q that the velocity of the electron makes with the screen while it strikes it. |
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Answer» Correct Answer - (i). `sigma=(epsilon_(0)m du^(2))/(8eL^(2))` (ii). `theta=tan^(-1)((epsilon_(0)m u^(2))/(2Le.sigma))` |
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| 1331. |
A uniform electric fied exists in x-y plane. The potential of ponts `A(2m,2m),B(-2m,2m)` and `C(2m,3m)` are `4V, 16V` and `12V` respectively. The electric field isA. `(4hati+5hatj)V//m`B. `(3hati+4hatj)V//m`C. `-(3hati+4hatj)V//m`D. `(3hati-4hatj)V//m` |
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Answer» Correct Answer - D `E=E_xhati+E_yhatj` Now we can use `intdV=-intE.dr` two times and can find vales of `E_x and E_y` |
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| 1332. |
Three point charges ` Q, 4 Q` and ` 16 Q` are placed on a straight line ` 9 cm` long. Charges are placed in such a wat that the system has minimum potential energy . Then .A. `4Q` and `16 Q` must be at the ends and `Q` at a distance of `3` cm from the `16Q`B. `4Q` and `16 Q` must be at the ends and `Q` at a distance of `3` cm from the `16Q`C. Electric field at the position of `Q` at a distance of `6 cm` from the `16 Q`.D. Electric field at the position of `Q` at a distance of `6 cm` from the `16 Q`. |
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Answer» Correct Answer - B::C For potential enrgy to be minmuum charge ` 11 Q` and ` 16Q` to be placed at the ebd of line ` U = (Kq_1q_2)/r_(12) + (Kq_1q_3)/r_(13) + (Kq_2q_3)/r_(23)` `U= (K. 4Q.Q)/((9-x) xx 10^(-12)) + (K. (4 Q) 16Q)/((9) xx 10^(-2)) + (KQ.16Q)/x` ` U= 4/((9-x)) + 100KA^2 . (64)/9 + 100 KQ^2` for `U` to be minimum `(dU)/(dx) =0` `1/((9-0x)^2) = 4/x^2` `(0-x)^2 x^2` ` x= 6 cm or 18 cm` since ` F=- (dU)/(dx) rArr F0.`, when `U` is min. |
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| 1333. |
Two infinite sheets of uniform charge debsity ` + sigma` and ` - sigma` are parallel to each other as shown in the gugrue . Elctric field at the .A. points to the left or to the right of the sheets is zero .B. midpoint between the sheets is zero.C. midpoint of the sheets is ` sigma //varepsilon_(0)` and is directed towards right .D. midpoint of the sheet is `2 sigma // varepsilon_(0)` and is directed towards right . |
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Answer» Correct Answer - A::C Since out side the plate field due to sheets are equal and opposite so zero and at mid point = `(isgma)/(2 in_0) + (sigma)/(2 in_0) = (sigma)/(in_0)`. |
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| 1334. |
Two particles (free to move) with charges `+q` and `+4q` are a distance L apart. A third charge is placed so that the entire system is in equilibrium. (a) Find the location, magnitude and sign of the third charge.(b) Show that the equilibrium is unstable.A. `Q=(4)/(9)q "(negative)at" (l)/(3)`B. `Q=(4)/(9) q"(positive) at" (l)/(3)`C. `Q=q "(positive) at" (l)/(3)`D. `Q = q"(negative) at" (l)/(3)` |
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Answer» Correct Answer - A |
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| 1335. |
The figure shows thrednfinite non-conducting plates of charge perpendicular to the plane of the paper with charge per unit area `+sigma, + 2sigma" and " -sigma ` . Find the ratio of the net electric field atthat point A to that at point B. The points A and Bare located midway between the plates : |
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Answer» Correct Answer - zero |
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| 1336. |
Find the electric field intensity due to a positively charged conducting cylinder with radius 1 cm and surface charge density `10^(-9)C//m^(2)`, at a point at a distance of 2 m from the axis of cylinder. The dielectric constant of the medium surrounding the cylinder is 2.A. `0.28V//m`B. `1.28V//m`C. `2.28V//m`D. `1.00V//m` |
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Answer» Correct Answer - A `E=(sigmaR)/(epsi_(0)kr)=(10^(-9)xx1xx10^(-12))/(8.85xx10^(-12)xx2xx2)=0.28V//m.` |
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| 1337. |
Two conducting spheres of radii `r_(1)` and `r_(2)` are equally charged. The ratio of their potentral is-A. `r_(2)/r_(1)`B. `r_(1)/r_(2)`C. `(r_(1)//r_(2))^(2)`D. `(r_(2)//r_(1))^(2)` |
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Answer» Correct Answer - A `V=1/(4piin_(0)k)Q/r" "thereforeVprop1/r" "thereforeV_(1)/V_(2)=r_(2)/r_(1)` |
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| 1338. |
The capacitance between the points `A` and `B` in the given circuit will be A. `3.5muF`B. `3muF`C. `2muF`D. `1muF` |
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Answer» Correct Answer - D `C_(1)=3muF,C_(2)=C_(3)=1.5muF,C_(4)=3muF,C_(PQ)=?` Now `C_(P)=C_(2)+C_(3)=1.5+1.5=3muF` Now `C_(1), C_(P) and C_(4)` are in series `1/C_(s)=1/C_(1)+1/C_(p)+1/C_(4)=1/3+1/3+1/3` `C_(s)=C_(PQ)=1muF` |
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| 1339. |
A capacitor of `20 muF` is charged to `500` volts and connected in parallel ith another capacitor of `10 muF` and charged to `200` volts. The common potential isA. 500VB. 400VC. 300VD. 200V |
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Answer» Correct Answer - B `C_(1)=20mF,V_(1)=500V,` `C_(2)=10mF,V_(2)=200V` V = ? `V=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))` `=(20xx10^(-6)xx500+10xx10^(-6)xx200)/(30xx10^(-6))` = 400 V |
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| 1340. |
A capacitor of `20 muF` is charged to `500` volts and connected in parallel ith another capacitor of `10 muF` and charged to `200` volts. The common potential isA. 400 VB. 200 VC. 100 VD. 50 V |
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Answer» Correct Answer - A Suppose `C_(1) and C_(2)` are capacities of two condensers charged to potentials `V_(1) and V_(2)`. Respectively. Total charge before sharing `=C_(1)V_(1)+C_(2)V_(2)` If V is the common poetnial on sharing charges, then total charge after sharing `=C_(1)V+C_(2)V=(C_(1)+C_(2))V` As mo charge is lost in the process of sharing, therefore `(C_(1)+C_(2))V=C_(1)V_(1)+C_(1)V_(1)+C_(2)V_(2)` `therefore" "V=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))` `=((20xx10^(-6)xx500)+(10xx10^(-6)xx200))/((20xx10^(-6))+(10xx10^(-6)))` = 400 V |
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| 1341. |
If `4 xx 10^(20)eV` energy is required to move a charge of `0.25` coulomb between two points. Then what will be the potential difference between them ?A. 100 VB. 256 VC. 200 VD. 128 V |
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Answer» Correct Answer - B `q=0.25C,U=4xx10^(20)eV,V=?` U = qV `therefore" "V=U/q=(4xx10^(20)xx1.6xx10^(-19))/0.25=256V` |
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| 1342. |
Two electrons are separated by a distance of `1Å`. What is the Coulomb force between them?A. `2.3xx10^(-8)N`B. `4.6xx10^(-8)N`C. `1.5xx10^(-8)N`D. None of these |
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Answer» Correct Answer - A Change on electron `16xx10^(-19)C` Using the relation, `F=(1)/(4pi epsilon_(0)).(q_(1)q_(2))/(r^(2))` `"Here, "q_(1)=q_(2)=1.6xx10^(-19)C` `r=1Å=10^(-10)m` So, Force between the electrons `F=(9xx10^(9)xx(1.6xx10^(-19))^(2))/((10^(-10))^(2))=2.3xx10^(-8)N` |
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| 1343. |
There is infinitely long straight thread carrying a charge with linear density `lambda`.Calculate the potential difference `A` and `B`. |
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Answer» The electric field at perpendicular `x` from wire `vec(E ) = (lambda)/(2 pi in_(0) x) hat(i) , dx hat(i) + dyhat(j) + dz hat(k)` `vec(E ). d vec(r ) = (lambda)/(2 pi in_(0) x) dx` `V_(A) - V_(B) = - int_(eta d)^(d) vec(E) . dvec(r ) = -int_(eta d)^(d) |log_(e ) x|_(eta d)^(d)` ` = - (lambda)/(2 pi in_(0)) log_(e ) ((1)/(eta))` `= (lambda)/(2 pi in_(0)) log_(e ) (eta)` |
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| 1344. |
Find electric field at point A, B, C, D due to infinitely long uniformly charged wire with linear charge density `lambda` and kept along z-axis (as shown in figure). Assume that all the parameters are in S.l. units. |
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Answer» `E_(A)=(2 K lambda)/(3) (hat(i)) implies E_(B)=(2 K lambda)/4 (hat(j))` `E_(c)=(2 K lambda)/5 overset(^^)(OC)=(2 K lambda)/5((3hat(i)+4hat(j))/5)" "E_(D)=(2 k lambda)/5((3 hat(i)+4hat(j))/5)" "implies E_(D)=E_(C)` |
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| 1345. |
A very long charged wire (lying in the xy plane ) which is having a linear charge density `lambda` is having one of its end at a point P as shown in figure . What is electric field intensity at point Q : |
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Answer» Correct Answer - `[|E| = (sqrt2 k lambda)/(r) vecE = (sqrt2klambda)/(r) (-hatj)]` |
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| 1346. |
A capacitor c1=4microfarad in series with another capacitor c2=1microfarad, the combination is connected across d.c source of 200v. The ratio of potential across c2 to c1 is. a)2:1 b) 4:1 c) 8:1 d) 16:1 |
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Answer» Option (b) is correct as explained below Given the capacitances of two capacitors ,C1=4 μF.and C2 = 1 μF. As they are connected in series they will have same charge Q. So Q=C1V1=C2V2, where V1 and V2 are respective P.D across the capacitors So V1/V2= C2/C! = > V2/V1=4/1 |
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| 1347. |
A parallel plate air capacitor of capacitance `C` is connected to a cell of `emF V` and then disconnected from it. A dielectric slab of dielectric constant `K`, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?A. The potential difference between the plates decreases K timesB. The energy stored in the capacitor decreases K timesC. The change in energy is `(1)/(2)C_(0)V^(2)(K-1)`D. The change is energy is `(1)/(2)C_(0)V^(2)((1)/(K)-1)` |
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Answer» Correct Answer - D `"Initial energy, "U_(i)=(1)/(2)C_(0)V^(2)` `"Final energy, "U_(f)=(1)/(2)(KC_(0))((V)/(K))^(2)=(1)/(K)((1)/(2)C_(0)V^(2))` `"Change in energy "=U_(f)-U_(i)=(1)/(2)C_(0)V^(2)((1)/(K)-1)` Note that this change of energy is negative, i.e. there is a decrease of energy. |
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| 1348. |
A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively areA. constant, decrease, increaseB. constant, decrease, decreaseC. constant, increase, decreaseD. increase, decrease, decrease |
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Answer» Correct Answer - C After separation charge is constant. Potential, `V=E.d` Capacity, `C=(epsilon_(0)A)/(d)` |
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| 1349. |
A parallel plate air capacitor of capacitance `C` is connected to a cell of `emF V` and then disconnected from it. A dielectric slab of dielectric constant `K`, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?A. The potential difference between the plates decreases `K` timesB. The enerty stored in the capacitor decreased `K` timesC. The change in energy is `1/2C_(0)varepsilon^(2)(K-1)`D. The change in energy is `1/2C_(0)varepsilon^(2)(1- 1/K)` |
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Answer» Correct Answer - A::B::D |
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| 1350. |
State the expression for the capacitance of a parallel-plate capacitor filled with a dielectric. Explain how its capacitance can be increased. |
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Answer» The capacitance of a parallel-plate capacitor filled with a dielectric is C = \(\cfrac{Akɛ_o}d\) where A is the area of each plate, k is the relative permittivity (dielectric constant) of the medium between the plates, ε0 is the permittivity of free space and d is the uniform plate separation. The capacitance of a parallel-plate capacitor can be increased by 1. increasing the area of each plate 2. decreasing the distance between the two plates 3. filling the space between the two plates by a medium of greater relative permittivity. |
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