InterviewSolution
Saved Bookmarks
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.
| 1451. |
The SI unit of electric field intensity isA. netweon/coulombB. joule/coulombC. volt-metreD. newton/metre |
|
Answer» Correct Answer - A |
|
| 1452. |
Figure shows the field lines on a positive charge. Is the work done by the field in moving a small positive charge from Q to P positive or negative? Give reason. |
|
Answer» The work done by the field is negative. This is because the charge is moved against the force exerted by the field. |
|
| 1453. |
Why do the electric field lines never cross each other? |
|
Answer» If the field lines cross each other, then at the point of intersection, there will be two directions for the same electric field which is not possible. |
|
| 1454. |
When an electric dipole of moment `| vec(p) | = q xx 2a` is held at an angle `theta`, with the direction of uniform external electric field `vec(E)`, a torque `tau = pE sin theta` acts on the dipole. This torque tries to align the electric dipole in the direction of the field. When `vec(p)` is along `vec(E), theta^(@) , tau = pE sin 0^(@) = zero`. The dipole is in stabel equilibrium. The energy possessed by the dipole by virtue of its particular position in the electric field is called potential energy of dipole. `U = W = -pE (cos theta_(2) - cos theta_(1))` `theta_(1) = 90^(@)` is the position of zero potential energy. ` :. U = W = -pE (cos theta - cos 90^(@)) = -pE cos theta`. For stable equillibrium, `theta 0^(@), :. U = -pE` = minimum. Read the above passage and answer the following questions : (i) What is the direction of torque acting on electric dipole held at an angle with uniform external electric field ? (ii) An electric dipole of length 10cm having charges `+- 6xx10^(-3) C`, placed at `30^(@)` with respect to a uniform electric field experiences a torque of magnitude `6 sqrt(3) N-m`. Calculate. (a) magnitude of electric field. (b) potential energy of dipole. (iii) What is the physical significance of this concept in our day to day life ? |
|
Answer» (i) `tau = pE sin theta` can be rewritten in vector form as : `vec(tau) = vec(p) xx vec(E)` Therefore, the direction of torque `vec(tau)` is perpendicular to both `vec(p) and vec(E)`. It is determined by right handed screw rule. (ii) Here, `2a = 10 cm = 10^(-1)m, q = +- 6xx10^(-3) C, theta = 30^(@) tau = 6 sqrt(3) N-m = ?, U = ?` From `tau = pF sin theta = q (2a) E sin theta` `6 sqrt(3) = 6xx10^(-3) (10^(-1)) E sin 30^(@) = 6xx10^(-4) E xx (1)/(2) = 3xx10^(-4) E` `E = (6 sqrt(3))/(3xx10^(-4)) = 2 sqrt(3) xx10^(4) NC^(-1)` `U = -pE cos theta = -q (2a) E cos theta = -6xx10^(-3) (10^(-1)) (2 sqrt(3) xx10^(4)) cos 30^(@)` `= -6xx10^(-4)xx2 sqrt(3) xx10^(4) xx (sqrt(3))/(2) = -18J` (iii) From teh given paragraph, we find that electric dipole held in an external field will be in stable equilibrium only when it is aligned along the field and it possesses minimum potential energy. The same is true in day to day life. At your work place, your boss tries always to align you along his/her plan. Your job/position is secured/stable after you are perfectly aligned and your persnal whims are at lowest ebb. |
|
| 1455. |
Explain why two field lines never cross each other at any point? |
|
Answer» If two field lines cross each other, then we can draw two tangents at the point of intersection which indicates that (as tangent drawn at any point on electric line of force gives the direction of electric field at that point) there are two directions of electric field at a particular point, which is not possible at the same instant. Thus, two field lines never cross each other at any point. |
|
| 1456. |
The electrostaic potential of a charged body represents the degree of electrification of the body. It detemines the direction of flow of charge between two charged bodies placed in contact with eachother. Charege always flows a body at higher potential to another body at lower potential. The flow of charge stops as soon as the potentials of the two bodies become equal. Electrostatic potential in electrically corresponds to level in case fo liquids , pressure in case of gases and temperature in case of heat. Due to a point charge q in air, electrostatic potentials at a distance r from the charge is `V = (q)/(4pi in_(0) r)` The SI unit of potential is volt. Read the above passage and answer the follwing questions : (i) The capacity of a body A is 100 times the capacity of body B and charge on A is 10 times the charge on B. When A and B are put in contact with eachother, will charge flow from A to B to A ? Why ? (ii) Calcualte the potential in air at a point 1 meter away from charge of `1muC`. (iii) What values of life do yo+-earn from the concept of electric potential ? |
|
Answer» (i) `V_(A) = (q_(A))/(C_(A)) and V_(B) = (q_(B))/(C_(B)) :. (V_(B))/(V_(A)) = (q_(B))/(C_(B)) xx (C_(A))/(q_(A)) = (C_(A))/(C_(B)) xx (C_(A))/(q_(A)) = 100xx (1)/(10) = 10` `V_(B) gt V_(A)` `:.` Charge would flow from B to A, as potential of B is greater than potential of A. (ii) From `V = (q)/(4pi in_(0) r) = ((10^(-6))xx9xx10^(9))/(1) = 9xx10^(3)` volt. (iii) The concept of potential emphasises taht charges flows from a body at higher potential to a body at lower potential . In day to day life, knowledge flows from a person of higher learning level to the the person of lower learning level. If we happen to have some health problem, we must always seek the opinion of a super specialist in the field concerned. The knowledge and information the super specialist have is of much higher level. The line of treatment adopted by him is going to cure us. The quacks who might have more knowledge of a variety of other fields shall be of no use. |
|
| 1457. |
An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? |
|
Answer» An electrostatic field line represents the actual path travelled by a unit positive charge in an electric field. If the line have sudden breaks it means the unit positive test charge jumps from one place to another which is not possible. It also means that electric field becomes zero suddenly at the breaks which is not possible. So, the field line cannot have any sudden breaks. |
|
| 1458. |
(a) An electrostaic field line is a continous curve. That is a field line cannot have sudden breaks. Why not ? (b) explain why two filed lines never cross each other at any point. |
| Answer» In an electrostaic field, a charge experiences a continuous force and it moves continuosly. As the charge does not jump from one point to another, the field line cannot have sudden breaks. | |
| 1459. |
Figure shown a section through two long thin concentric cylinders of radii a & b with `a gt b`. The cylinders have equal and opposite per unit length `lambda`. Find the electric field at a distance r from the axis for (i) `r lt a` (ii) `a lt r lt b` (iii) `r gt b` |
|
Answer» Correct Answer - `[ 0 , (2K lambda)/(r) , 0]` |
|
| 1460. |
Pick out the statement which is incorrect?A. A negative test charge experiences a force opposite to the direction of the fieldB. The tangent drawn to a line of force represents the direction of electric fieldC. Field lines never intersectD. The electric field lines form closed loop |
|
Answer» Correct Answer - D |
|
| 1461. |
Two identical balls of charge `q_(1)` & `q_(2)` initially have equal of the same magnitude and direction. After a ubiform electric field is applied for some time, the direction of the velocity of the first ball charges by `60^(@)` and the magnitude is reduced by half. The direction of the velocity of the second ball charges there by `90^(@)`. In what proportion will the velocity of the second ball changes? |
|
Answer» Correct Answer - `(v)/(sqrt(3))` |
|
| 1462. |
A positive charge `+Q` is fixed at a poibt A. Another positively charged particle of mass m and charge +q is projected from a point B with velocity u as shown in (Fig. 3.103). Point B is at a large distance from A and at distance d from the line A C. The initial velocity is parallel to the line A C. The point C is at a very large distance from A. Find the minimum distance (in meter) of +q from +Q during the motion. Take `Q q = 4 pi epsilon_0 m u^2 d` and `d (sqrt(2) - 1) m`. . |
|
Answer» Correct Answer - 1m |
|
| 1463. |
A charged particle of charge `2 muC` and mass `10` milligram, moving with a velocity of `1000 m//s` enters a uniform electric field of strength `10^(3) N//C` directed perpendicular to its direction of motion. Find the velocity and displacement of the particle after `10s`. |
|
Answer» Here, `q = 2 muC = 2xx10^(-6) C`, `m = 10 mg = 10^(-5) kg` `v_(x) = 1000 m//s E_(y) = 10^(3) N//C ` `v_(y) = (u_(y) + a_(y)t = 0 + (qE_(y) t)/(m))` `(2xx10^(-6)xx10^(-3)xx10)/(10^(-5)) = 2000 m//s` As ` v_(x)` remains constant, therefore net velocity after `10s` `= sqrt(V_(x)^(2) + v_(y)^(2)) = sqrt((1000)^(2) + (2000)^(2))` `= 1000 sqrt(5) m//s` Displacement along x-axis after `10s` `x = v_(x) t = 1000xx10 = 10000 m` Displacement along y-axis after `10s` `y = u_(y) xx t+ (1)/(2) (q E_(y))/(m) t^(2)` `= 0+ (1)/(2) xx (2xx10^(-6)xx10^(3))/(10xx10^(-6)) (10^(2)) = 10000m` Net displacement `= sqrt(x^(2) + y^(2))` `= sqrt((10000)^(2) + (10000)^(2)) = 10000 sqrt(2) m` |
|
| 1464. |
Flux coming out from a positive unit charge placed in air, isA. `epsi_(0)`B. `epsi_(0)^(-1)`C. `(4 pi epsi_(0))^(-1)`D. `4 pi epsi_(0)` |
|
Answer» Correct Answer - B |
|
| 1465. |
find the electric field strength due to a point charge of `5 muC` at a distance of 80 cm from the charge.A. `8 xx 10^(4) NC^(-1)`B. `7 xx 10^(4) NC^(-1)`C. `5 xx 10^(4) NC^(-1)`D. `4 xx 10^(4) NC^(-1)` |
|
Answer» Correct Answer - B |
|
| 1466. |
An electric field `E = (20hati + 30 hatj)` N/C exists in the space. If thepotential at the origin is taken be zero, find the potential at `(2 m, 2 m)`. |
|
Answer» `V_(1,2)-V_(0 ,0) = - int _(0,0)^(1,2) vec(E ) . d vec( r)` `= - int_(0,0)^(1,2) (E_(x)dx + E_(y) dy)` `= -[int_(0)^(1) E_(x) dx + int_(0)^(2) E_(y) dy]` `= [ int_(0)^(1) 10 dx + int _(0)^(2) 20 dy]` `= [ 10|x|_(0)^(1) + 20|y|_(0)^(2) ]` `= -[10 (1 - 0) + 20(2 - 0)] = -30` `V_(1,2) - 0 = -30 ` `V_(1,2) = -30` volt |
|
| 1467. |
An electric field `vecE = vec(i20 + vecj30 ) NC^(-1)` exists in the space. If the potential at the origin is taken to be zero find the potential at (2m, 2m). |
|
Answer» `V_(0 ,0) - V_(1 ,2) = -int_(1 , 2)^(0,0) vec(E ) . D vec( r) = - int_(1)^(0) E_(x) dx` `= - int_(1)^(0) 20 dx = - 20|(x^(2))/(2)|_(0)^(t)` `= -20( 0 -(1^(2))/(2)) = 10` `V_(0,0) - 0 = 10` `V_(0,0) = 10` volt |
|
| 1468. |
Assume that an electric field `vecE=30x^2hatj` exists in space. Then the potential difference `V_A-V_O`, where `V_O` is the potential at the origin and `V_A` the potential at `x=2m` is:A. (a) `120J//C`B. (b) `-120J//C`C. (c) `-80J//C`D. (d) `80J//C` |
|
Answer» Correct Answer - C Potential difference between any two points in an electric field is given by, `dV=-vecE.vecdx` `int_(VO)^(VA)dV=-int_0^(2)30x^2dx` `V_A-V_O=[10x^3]_0^2=-80J//C` |
|
| 1469. |
Let `u_(a)` and `u_(d)` represent the energy density (energy per unit volume) in air and in a dielectric respectively, for the same field in both. Let `K=` dielectric constant. ThenA. `u_(a)=u_(d)`B. `u_(a)=Ku_(d)`C. `u_(d)=Ku_(a)`D. `u_(a)=(K-1)u_(d)` |
|
Answer» Correct Answer - C |
|
| 1470. |
In a parallel-plate capacitor, the region between the plates is filled by a dielectric slab. The capacitor is connected to a cell and the slab is taken out. ThenA. Some charge is drawn form the cellB. Some charge is returned to the cell.C. The potential differences across the capacitor is reduced.D. No work is done by an external agent in taking the slab out. |
|
Answer» Correct Answer - B |
|
| 1471. |
Three charged particle sare in equilibrium under their electrostatic forces only. ThenA. The particles must be collinear.B. All the charges cannot have the same magnitudeC. All the charges cannot have the same sign.D. The equilibrium is unstable. |
|
Answer» Correct Answer - B |
|
| 1472. |
Four identical charges are placed at the points `(1,0,0), (0,1,0),(-1,0,0)`, and `(0,-1,0)`. Then,A. The potential at the origin is zero.B. The field at the origin is zeroC. The potential at all ponts on the `z`-axis, othehr than the origin, is zero.D. The field at all points on the `z`- axis, other than the origin, acts along the `z`-axis. |
|
Answer» Correct Answer - A::B::C::D |
|
| 1473. |
For charges A,B,C and D are plasced at the four corners of a squre of a side a. The energy required to take the charges C and D to infinity (they are also infinitely separated from each other )is A. `q^2/(4piepsilon_0a)`B. `(2q^2)/(piepsilon_0a)`C. `q^2/(4piepsilon_0a)(sqrt2+1)`D. `q^2/(4piepsilon_0a)(sqrt2-1)` |
|
Answer» Correct Answer - C Energy, required `=/_U=U_f-U_i` `=1/(4piepsilon_0)[(q^2/a)-(q^2/a-q/(sqrt2a)-q^2/a-q^2/a-q^1/(sqrt2a)+q^2/a)]` `q^2=(4piepsilon_0)[sqrt2+1]` |
|
| 1474. |
Four charges , all of the same magnitude , are placed at the four corners of a square . At the centre of the square , the potential is `V` and the field is `E`. By suitable choice of the signs of the four charges ,which of the following can be obtained? (i) `V = 0 , E = 0` (ii) `V = 0 , E != 0` (iii) `V != 0 , E = 0` (iv) `V != 0 , E != 0`A. `V=0,E=0`B. `V=0,E!=0`C. `V!=0,E=0`D. `V!=0,E!=0` |
|
Answer» Correct Answer - B::D |
|
| 1475. |
Assertion. A metallic shield in the form of a hollow shell, can be built to block an electric field. Reason. In a hollow spherical shell, the electric field inside is not zero at every point.A. both, Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. both, Assertion and Reason are true, but Reason is not the correct explanation of the Asserrtion.C. Assertion is true, but the Reason is false.D. both, Assertion and Reason are false. |
|
Answer» Correct Answer - c The assertion correct but the reason is false. |
|
| 1476. |
The electric field in the space between the plates of a discharge tube is `3.24 xx 10^(4)NC^(-1)`. If mass of proton is `1.67xx10^(-27)` kg and its charge is `1.6xx10^(-19)C`, the force often the proton in the field isA. `10.4xx10^(-15)N`B. `2.0xx10^(-23)`C. `5.40xx10^(-15)N`D. `5.20xx10^(-15)N` |
|
Answer» Correct Answer - D As, the force on the proton `F=qE=1.6xx10^(-19)xx3.25xx10^(4)=5.20xx10^(-15)N` |
|
| 1477. |
Electric charge is uniformly distributed along a along straight wire of radius `1 mm`. The charge per centimeter length of the wire is `Q` coulomb. Another cyclindrical surface of radius `50 cm` and length `1 m` symmetrically enclose the wire ask shown in figure. The total electric flux passing through the cyclindrical surface is A. `(Q)/(epsi_(0))`B. `(100 Q)/(epsi_(0))`C. `(10 Q)/((pi epsi_(0)))`D. `(100 Q)/((pi epsi_(0)))` |
|
Answer» Correct Answer - B |
|
| 1478. |
Four coulomb charge is uniformly distributed on 2 km long wire. Its linear charge density is(A) 2 C/m (B) 4 C/m(C) 4 × 103 C/m(D) 2 × 10-3 C/m |
|
Answer» Correct option is: (D) 2 × 10-3 C/m |
|
| 1479. |
Assume that an electric field `vec(E) = 30 x^(2) hat(i)` exists in space. Then the potential differences `V_(A) - V_(0)` where `V_(0)` is the potential at the origin and `V_(A)`, the potential at `x = 2m` isA. 120 JB. `-120 J`C. `-80` JD. 80 J |
| Answer» Correct Answer - C | |
| 1480. |
A `5C` charge experience a force of `2000 N` when moved between two points separated by a distance of `2 cm` in a uniform electric field. The potential difference between the two points isA. 8 VB. 200 VC. 800 VD. 20, 000 V |
| Answer» Correct Answer - A | |
| 1481. |
An electron of mass `m_(e )` initially at rest moves through a certain distance in a uniform electric field in time `t_(1)`. A proton of mass `m_(p)` also initially at rest takes time `t_(2)` to move through an equal distance in this uniform electric field.Neglecting the effect of gravity, the ratio of `t_(2)//t_(1)` is nearly equal toA. `m_(p)/m_(e)`B. `(m_(p)/m_(e))^(2)`C. `(m_(p)/m_(e))^(1//2)`D. `(m_(p)/m_(e))^(3//2)` |
|
Answer» Correct Answer - C `S=1/2at^(2)=((eE)/m)t^(2)` Now, `t^(2)/m` = Constant for same distance Thus, `mpropt^(2)` `therefore" "m_(e)/m_(p)=t_(e)^(2)/t_(p)^(2)" "thereforet_(p)/t_(e)=sqrt(m_(p)/m_(e))` `i.e" "t_(2)/t_(1)=(m_(p)/m_(e))^(1//2)` |
|
| 1482. |
Consider a uniformly charged ring of radius R. Find the pint on the axis where the electrie field is maximum.A. `1/(4pi epsi_(0)) q/(3sqrt(3) R^(2))`B. `1/(4pi epsi_(0)) (2q)/(3R^(2))`C. `1/(4pi epsi_(0)) (2q)/(3sqrt(3) R^(2))`D. `1/(4pi epsi_(0)) (3q)/(2sqrt(3)R^(2))` |
| Answer» Correct Answer - C | |
| 1483. |
A flat circular fixed disc has charge +Q uniformly distributed on the disc. A charge +q is thrown with kinetic energy K, towards the disc along its axis. The charge q :A. may hit the disc at the centreB. may return back along its path after touching the discC. may return back along its path without touching the discD. any of the above three situations is possible depending on the magnitude of K |
| Answer» Correct Answer - D | |
| 1484. |
A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest?A. `(2m u^(2))/(qE)`B. `(m u^(2))/(qE)`C. `(m u^(2))/(2qE)`D. `(m u^(2))/(2 qE)` |
|
Answer» Correct Answer - 4 Retardation `a = (qE)/(m)` `v^(2) = u^(2) - 2as` `0 = u^(2) - (2qE)/(m) s` `s = (mu^(2))/(2 qE)` |
|
| 1485. |
A charged particle of mass `m` and charge `q` is released from rest in an electric field of constant magnitude `E`. The kinetic energy of the particle after time `t` isA. `(Eqm)/t`B. `(E^(2)q^(2)t^(2))/(2m)`C. `(2E^(2) t^(2))/(mq)`D. `(Eq^(2) m)/(2t^(2))` |
| Answer» Correct Answer - B | |
| 1486. |
A charged particle of mass `m` and charge `q` is released from rest in an electric field of constant magnitude `E`. The kinetic energy of the particle after time `t` isA. `(2 E^(2) t^(2))/(mq)`B. `(Et^(2) m)/(2t^(2))`C. `(E^(2) q^(2) t^(2))/(2m)`D. `(E q m)/(2t)` |
|
Answer» Correct Answer - 3 `a = (F)/(m) = (qE)/(m)` `v = u + at = 0 + (qEt)/(m)` `K = (1)/(2) mv^(2) = (1)/(2) m((qEt)/(m))^(2) = (q^(2) E^(2) t^(2))/(2m)` |
|
| 1487. |
A conducting sphere of radius R having charge Q is placed in a uniform external field E. O is the centre of the sphere and A is a point on the sphere of the sphere such that AO makes an angle of `theta_(0)=60^(@)` with the opposite direction fo external field. calculate the potential at point A due to charge on the sphere only. |
|
Answer» Correct Answer - `(1)/(4piepsilon_(0))(Q)/(R)-(ER)/(2)` |
|
| 1488. |
A conducting shell having no charge has radius R. A point charge Q is placed in front of it at a distance `r_0` from its centre. Find potential due to charge induced on the surface of the shell at a point P inside the shell. Distance of point P from point charge Q is r. |
|
Answer» Correct Answer - `(kQ)/(r_(0))-(KQ)/(r)` |
|
| 1489. |
A hemispherical bowl of radius R carries a uniform surface charge density of `sigma`. Find potential at a point P located just outside the rim of the bowl (see figure). Also calculate the potential at a point A located at a distance R/2 from the centre on the equatorial plane. |
|
Answer» Correct Answer - At both P and A the potential is `(sigmaR)/(2epsilon_(0))` |
|
| 1490. |
Plate A of parallel plate air filled capacitor is connected to a spring having force constant k and plate B is fixed. They are held on a frictionless table top as shown in Fig. If charge `+q` is placed on plate A and a charge `-q` on plate A and a charge `-q` on plate B, by how much does the spring expand ? |
|
Answer» In a capacitor, plates carry equal and opposite charges. Therefore, there is a force of attraction between the plates. In a parallel plate capacitor, energy stored, `U = (1)/(2) (q^(2))/(C ) and C = (in_(0) A)/(x)` `:. U = (q^(2) x)/(2 in_(0) A)` As the electric field is conservation, therefore, As the electric field is conservative , therefore, `F = - (dU)/(dx) = - (d)/(dx) ((q^(2) x)/(2in_(0) A)) = - (q^(2))/(2 in_(0) A)` Negative sign imples that force is attractive. If k is force constant of the spring and it expands through a distance x, then `-F = kx` or `x = - (F)/(k) = (q^(2))/(2 in_(0) Ak)` |
|
| 1491. |
Plot 1 in Fig gives the charge q that can be stored on capacitor `C_(1)` versus electric potential V set up across it. Plots 2 and 3 are simillar plots for capacitor `C_(2) and C_(3)` respectively. The three capacitors are connected to 6.0 V battery as shown here. Calculate charge stored in capacitor `C_(2)`. |
|
Answer» As is clear from Fig. `C_(1) = (q_(1))/(V) = (12)/(2) = 6muF` `C_(2) = (q_(2))/(V) = (8)/(2) = 4 muF` `C_(3) = (q_(2))/(V) = (4)/(2) = 2muF` As `C_(2) and C_(3)` are in parallel, therefore `C_(p) = C_(2) + C_(3) = 4 + 2 = 6muF` The combination is in seires with `C_(1)` `:. (1)/(C_(123)) = (1)/(C_(1)) + (1)/(C_(p)) = (1)/(6) + (1)/(6) = (1)/(3)` `:. C_(123) = 3muF` Charge on capacitor `C_(1) is q_(1) = 3xx6 = 18 muC` Voltage across `C_(1) = (q_(1))/(C_(1)) = (18)/(6) = 3 vol t`. Voltage across `C_(2) = 6-3 = 3 vol t` `:.` Charge on `C_(2) = 4xx3 = 12 muC` |
|
| 1492. |
(i) Find potential at point A and B sue to the small charge - system fixed near origin. (Distance between the charges is negligible). (ii) Find work done to bring a test charge `q_(0)` from point A to point B, slowly. All parameters are in S.I. units. |
|
Answer» (i) Dipole moment of the system is `vec(P)=(qa) hat(i)+(qa) hat(j)` Potential at point a due to the dipole `V_(A)=K ((vec(P).vec(r)))/r^(3)=(K[(qa)hat(i)+(qa)hat(j)]. (4hat(i)+3hat(j)))/5^(3)=(k(qa))/125 (7)` `implies V_(B)=(K[(qa)hat(i)+(qa)hat(j)](3hat(i)-4hat(j)))/((5)^(3))=(-K (qa))/125` (ii) `W_(A rarr B)=U_(B)-U_(A)=q_(0) (V_(B)-V_(A))=q_(0) [-(K(qa))/125-((K(qa)(7))/125)]` `implies W_(A rarr B)=(-Kq q_(0)a)/125 (8)` |
|
| 1493. |
An isolated conducting sphere of diameter 20 cm placed in air carries a charge of `9muC`. The electric intensity at a point at a distance of 10 cm from the surface of the charged sphere isA. `10.25xx10^(5)N//C`B. `15.25xx10^(5)N//C`C. `25.25xx15^(5)N//C`D. `20.25xx10^(5)N//C` |
|
Answer» Correct Answer - D `E=1/(4piepsi_(0))q/r^(2)=(9xx10^(9)xx9xx10^(-6))/((10xx10^(-2))^(2))` `=20.25xx10^(5)N//C.` |
|
| 1494. |
If number of electric lines of force fro charge q are 10, then find out number of electric lines of force from 2q charge. |
|
Answer» No. of ELOF `prop` charge `10 prop q" "implies" "20 prop 2q` So, number of ELOF will be 20. |
|
| 1495. |
Some electric lines of force are shown in figure. For points a and B A. `E_(A) gt E_(B)`B. `E_(B) gt E_(A)`C. `V_(A) gt V_(B)`D. `V_(B) gt V_(A)` |
|
Answer» Correct Answer - C Lines are more dense at A, so `E_(A) gt E_(B)` in the direction of Electric field, potential decreases so `V_(A) gt V_(B)` |
|
| 1496. |
Five thousand lines of force enter a certain volume of space and three thousand lines emerge from it. What is the total charge in coulomb within this volume ? |
|
Answer» Correct Answer - `-1.77xx10^(8) C` Here, number of lines of force emerging from the volume `= phi = 3000 = 8.85xx10^(-12) (-2000)` `= -1.77xx10^(-8)C` |
|
| 1497. |
In the above question, what is the electric flux passing throguh a face of the given cube ? |
|
Answer» Correct Answer - `9.42 NC^(-1) m^(2)` The given charge is located at one corner of the cube. So it is being directly shared by three faces of the cube. Electric flux throgh any one of these faces `phi = (1)/(3) xx ((1)/(8) (q)/(in_(0))) = (1)/(3) xx28.25 NC^(-1) m^(2)` |
|
| 1498. |
what is the nature of sysmmetery of the electric field due to (i) point charge and (ii) electric dipole ? |
| Answer» The electric field due to a point charge has spherical symmetry with point charge at the center, It is so because, the electric field due to dipole will be same at every point on the surface of a right circular cylinder with electric dipole as sthe axis. | |
| 1499. |
when an electric dipole is suspended in a uniform electric field, then under what conditions the dipole is in (i) stable equilibrium and (ii) unstable equaliibrium. |
|
Answer» The dipole in an electric field will be in stable equalibrium if the following conditions are satisfied : (i) The resultant force on dipole is zero, i..e, there is no translatory motion of dipole. (ii) The torque on dipole is zero, i.e, there is no ratatroy motion pf dipole is minimum. it will be so when dipole will be so when dipole is aligned along lthe direction lof electric field. The dipole will be in unstable equalibrium if (i) the resultant force on diople is zero. (ii) the torque on the diople is zero. (iii) the potential energy of dipole is maximum. it will be so when diple is aligned opposite to the direction of electric field. |
|
| 1500. |
At what position is the electric dipole in uniform electric field in its most stable equilibrium position? |
|
Answer» When θ= 0° between vector (P and E) |
|