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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. |
The centres of two identical small conducting sphere are 1 m apart. They carry charge of opposite kind and attract each other with a force F. when they connected by conducting thin wire they repel each other with a force `F//3`. The ratio of magnitude of charges carried by the spheres initially in `n : 1`. Find value of n |
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Answer» Correct Answer - 3 |
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| 52. |
IN fig, calculate the total flux of the electrostatic field through the spheres `S_(1) and S_(2)`. The wire AB shown here has a linear charge density `lambda` . Given by `lambda = kx`, where x is the distance measured along the wire from end A. |
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Answer» Charge on an element of length dx of wire AB `dq = lambda, dx = kx dx` Total charge on wire AB, `q = int dq = int_(0)^(1) kx dx = k((x^(2))/(2))_(0)^(1) = (1)/(2) kl^(2)` Total flux through `S_(1) = phi = (Q)/(in_(0))` Total flux through `S_(2)` `= phi_(2) = (Q+q)/(in_(0)) = (Q + (1)/(2) kl^(2))/(in_(0))` |
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| 53. |
Two identical conducting spheres, fixed in space, attract each other with an electrostatic force of `0.108 N` when separated by `50.0 cm`, centre-to-centre. A thin conducting wire then connects the spheres. When the wire is removed, the spheres repel each other with an electrostatic force of `0.0360 N`. What were the initial charges on the spheres?A. `pm 5 xx 10^(6) C` and `pm 15 xx 10^(-6) C`B. `pm 1.0 xx 10^(6) C `and `pm 3.0 xx 10^(-6) C`C. `pm2.0xx10^(-6) C` and `pm 6.0 xx 10^(6)C`D. `pm0.5 xx 10^(-6) C` and `pm1.5 xx 10^(-6) C` |
| Answer» Correct Answer - B | |
| 54. |
The diagram shows the arrangement of three small uniformally charged spheres, `A,B` and `C`. The arrow indicate the direction of the electrostatic forces acting between the spheres (for example, the left arrow on sphere `A` indicates the electronstatic force on sphere `A` due to sphere `B`). At least two of the spheres are positively charged. Which sphere, if any, could be negatively charged? A. sphere AB. sphere BC. sphere CD. no sphere |
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Answer» Correct Answer - A It can be seen from the diagram that only the sphere B and sphere C repel. Hence they both must be of same type. According to the fact that at least two spheres are positively charged, therefore both spheres should be positively charged . Since attraction occurs for two remaining pairs it can be concluded that the sphere A is negatively charged. |
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| 55. |
The electrostatic field due to a point charge depends on the distance r as `(l//r^(2))`. Similarly, indicates how each of the following quantities depends on r : (a) Intensity of light from a point source (b) Electrostatic potential due to a point source (c) Electrostatic potential due to a distance r from the center of a charged metallic sphere of radius R `(r lt R)`. |
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Answer» (a) For a point source, light intensity obeys inverse square law, i..e, `1 prop (1)/(r^(2))`. (b) In case of a point charge, `V = (q)/(4pi in_(0) r) :. V prop r^(-1)` (c ) In case of a charged conducting sphere for an internal point `(r lt R)`. `V_(in) = V_(surface) = (q)/(4pi in R)` = constant. `:. V prop r^(0),` i.e, V does not depend on r. |
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| 56. |
Two charges of magnitudes –2Q and +Q are located at points (a,0) and (4a,0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin? |
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Answer» Electric flux ϕ = qinside/ε0 = -2Q/ε0. |
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| 57. |
Define the term electric dipole moment of a dipole. State its S.I. unit. |
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Answer» Strength of an electric dipole is measured by its electric dipole moment, whose magnitude is equal to product of magnitude of either charge and separation between the two charges i.e., vector (p = q.2a) and is directed from negative to positive charge, along the line joining the two charges. Its SI unit is Cm. |
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| 58. |
If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change? |
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Answer» The electric flux remains the same, as the charge enclosed remains the same. |
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| 59. |
The product of small area element and normal component of electric intensity isA. electric fluxB. flux densityC. electric potentialD. charge density |
| Answer» Correct Answer - A | |
| 60. |
What is the electric flux through a cube of side 1 cm which encloses an electric dipole ? |
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Answer» Zero (as net charge enclosed by the surface is zero.) |
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| 61. |
If the radius of the Gaussion surface enclosing a charge q is halved, how does the electric flux through the Gaussion surface change? |
| Answer» As `phi = (q)/(in_(0))`, so flux does not depend upon the radius of Gaussion surface, it will remain unchanged. | |
| 62. |
An arbitrary surface encloses a dipole. What is the electric flux through this surface ?A. zeroB. positiveC. negativeD. infinite |
| Answer» Correct Answer - A | |
| 63. |
A charge q is placed at the centre of a cube of side l what is the electric flux passing through two opposite faces of the cube ? |
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Answer» Flux through each face `= (q)/(6 in_(0))` `:.` flux through two opposite faces `= (q)/(6 in_(0)) + (q)/(6 in_(0)) = (2q)/(6 in_(0)) = (q)/(3 in_(0))` |
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| 64. |
A charge q is placed at the centre of a cube of side `l` what is the electric flux passing through two opposite faces of the cube ?A. zeroB. `(q)/(8 epsilon_(0))`C. `(q)/(6 epsilon_(0))`D. `(q)/(6 epsilon_(0))` |
| Answer» Correct Answer - 2 | |
| 65. |
What is the net flux of the uniform electric field of Q.15 thorugh a cube of side 20cm oriented so that its faces are parallel to the co-ordinate planes ? |
| Answer» Net flux over the cube is zero, because the number of lines entering the cube is the same as the number of lines leaving the cube. | |
| 66. |
A charged particle of radius `5xx10^(-7)m` is located in a horizontal electric field of intensity `6.28xx10^(5)Vm^(-1)`. The surrounding medium has the coefficient of viscosity `eta=1.6xx10^(5)Nsm^(-2)`. The particle starts moving under the effect of electric field and finally attains a uniform horizontal speed of `0.02 ms^(-1)`. Find the number of electrons on it. Assume gravity free space. |
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Answer» Correct Answer - [30] |
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| 67. |
A thin uniform rod of mass M and length 2L is hinged at its centre O so that it can rotate freely in horizontal plane about the vertical axis through O. at its ends the insulating rod has two point charges 2q and q (see figure). An electrif field E is switched on making and angle `theta_(0)=60^(@)` with the initial position of the rod. The field is uniform and horizontal. (a) Calculate the maximum angular velocity of the rod during subsequent motion. (b) Find the maximum angular acceleration of the rod. |
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Answer» Correct Answer - (a). `sqrt((3qR)/(ML))` (b). `(3sqrt(3))/(2)(qE)/(ML^(2))` |
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| 68. |
The total charge of an isolated system is always conserved. Explain with an example. |
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Answer» 1. When a glass rod is rubbed with silk, it becomes positively charged and silk becomes negatively charged. 2. The amount of positive charge on glass rod is found to be exactly the same as negative charge on silk. 3. Thus, the systems of glass rod and silk together possesses zero net charge after rubbing. Hence, the total charge of an isolated system is always conserved. |
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| 69. |
Is the total charge of the universe conserved ? |
| Answer» Yes, charge conservation is a global phenomeon. | |
| 70. |
what do mean by conservation of electric charge ? |
| Answer» Convervation of electric charge means that the total charge on an isolated system remains uncharged with time. | |
| 71. |
What do you mean by additivity of electric charge ? |
| Answer» Additivity of charge means the total charge on a system is the algebraic sum (with proper signs) of all individual charges in the system. | |
| 72. |
An isolated conducting sphere id given s positive charge. Does its mass increase, decrease or remain the same ? |
| Answer» Its mass decrease slighly as it loses some electrons. | |
| 73. |
What is meant by quantisation of charge ? |
| Answer» Charge on any body or particle can be integral multiple of cahrge on an electron `(-e)`, i.e., `q = +- n e`, when `n = 1,2,3,….` | |
| 74. |
An ebonite rod is rubbed with fur or wool. What type of charges do they acquire ? |
| Answer» The ebonite rod acquires negative charge and fur/wool acquires an equal positive charge. | |
| 75. |
When ebonite rod is rubbed with fur, ebonite rod and fur acquires ……… |
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Answer» When ebonite rod is rubbed with fur, ebonite rod and fur acquires negative and positive charge respectively |
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| 76. |
An ebonite rod is rubbed with fur or wool. What type of charges do they acquire ?A. positive chargeB. negative chargeC. can not be chargedD. none of these |
| Answer» Correct Answer - B | |
| 77. |
A glass rod rubbed with silk acquires a charge `+ 1.6xx10^(-12)C`. What is tha charge on the silk? |
| Answer» Charge on silk is equal and opposite to charge on glass rod, `i.e., q = -1.6xx10^(-12)C`. | |
| 78. |
When ebonite rod is rubbed with fur, ebonite rod and fur acquires ………… |
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Answer» When ebonite rod is rubbed with fur, ebonite rod and fur acquires Negative and positive charge respectively |
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| 79. |
Who observed that a material named Amber is rubbed with wool, acquires the property of attracting light bodies ?A. EdisonB. FranklinC. CoulombD. Thales |
| Answer» Correct Answer - D | |
| 80. |
Show that the normal component of electrostatic field has a discontinuly form one side of a charged. Surface to another given by `(vec(E_(2)) - vec(E_(1))). hat(n) = (sigma)/(in_(0))` where `hat(n)` is a unit vector normal to the surface at a point and `sigma` at a point and `sigma` is the surface charge density at that point. (The direction of `hat(n)` is from side 1 to side 2). Hence show that justy outside a conductor, the electric field `sigma hat(n)//in_(0)`. (b) Show that the tangential componet of electrostatic field is contionous from one side fo a charged surface to another. |
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Answer» Proceeding as in Art, normal of electric field intensity due to thin infinitie plane sheet of charge, on left side (side 1) `vec(E)_(1) = - (sigma)/(2in_(0)) hat(n)` and on right side (side 2), `vec(E_(2)) = (sigma)/(2in_(0)) hat(n)` Discotinuity is the normal component from one side to the other is `vec(E_(2)) - vec(E_(1)) = (sigma)/(2 in_(0)) hat(n) + (sigma)/(2 in_(0)) hat(n) = (sigma)/(in_(0)) hat(n) or (vec(E_(2)) - vec(E_(1))) hat(n) = (sigma)/(in_(0)) hat(n). hat(n) = (sigma)/(in_(0))` Inside a closed conductor, `vec(E)_(1) = 0` `:. E = vec(E_(2)) = (sigma)/(in_(0)) hat(n)` (b) To show that the tangential component of electrostatic field is continous from one side of a charged surface to another, we use the fact that work done by electrostatic field on a closed loop is zero. |
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| 81. |
A charged particle having some mass is resting in equilibrium at a height `H` above the centre of a uniformly charged non-conducting horizontal ring of radius ` R`. The equilibrium of the particle will be stable .A. for all values of HB. only if ` H gt (R)/(sqrt2)`C. only if ` H lt (R)/(sqrt2)`D. only if ` H = (R)/(sqrt2)` |
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Answer» Correct Answer - B Since electric field due to ring will be maximum at ` H = R/(sqrt 20`. |
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| 82. |
An infinite line cahrge is perpendicular to the plane of the figure having linear charge density `lamda`, A partcle having charge Q and mass m is projected in the field of the line cahrge fromo point P. the point P is at a distance R from the line cahrge and velocity given tot he particle is perpendicular tot he radial line at P (see figure) (i). Find the speed of the particle when its distance from the line cahrge grows the `etaR(eta gt 1)` (ii). Find the velocity component of the particle along the radial line (joining the line charge to the particle) at the instant its distance becomes `etaR`. |
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Answer» Correct Answer - (a) `V=sqrt(V_(0)^(2)+(lamdaQ)/(pi epsi_(0)m)ln eta)` (b). `V_(r)=sqrt(V_(0)^(2)(1-(1)/(eta^(2)))+(lamdaQ)/(pi epsi_(0)m)ln eta)` |
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| 83. |
If two conducting spheres are separately charged and then brought in contactA. the total energy of two spheres is conservedB. the total charge on two spheres is conservedC. both the total energy and charge are conservedD. the final potential is always the mean of the original potentials of the two spheres |
| Answer» Correct Answer - B | |
| 84. |
The capacitance of a parallel plate condenser does not depend uponA. area of the platesB. metal of the platesC. medium between the platesD. distance between the plates |
| Answer» Correct Answer - B | |
| 85. |
If two charged conductors are brought in contact, then they showA. gain in energyB. loss of some energyC. gain in chargeD. loss of same charge |
| Answer» Correct Answer - B | |
| 86. |
The charge and energy stored in the capacitor of capacity `32muF`, when it is charged to a potential difference of 0.6 kV are respectivelyA. `1.92xx10^(-2)C,5.76J`B. `2.92xx10^(-2)C,5.76J`C. `1.92xx10^(2)C,5.76J`D. `1.92xx10^(-2)C,4.76J` |
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Answer» Correct Answer - A Q = CV `=32xx10^(-6)xx0.6xx10^(3)=1.92xx10^(-2)C` `E=1/2CV^(2)=(32xx10^(-6)xx(0.6xx10^(3))^(2))/2=5.76J.` |
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| 87. |
The energy stored in a capacitor of capacity C and potential V is given byA. `0.5C^(2)V`B. `0.5CV^(2)`C. 0.5 CVD. `0.5C^(2)V^(2)` |
| Answer» Correct Answer - B | |
| 88. |
Capacity of a capacitor is `48mu F`. When it is charged from `0.1` C to `0.5` C , change in the energy stored isA. 2500 JB. `2.5xx10^(-3)J`C. `2.5xx10^(6) J`D. `2.42xx10^(-2)` |
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Answer» Correct Answer - A `DeltaE=E_(2)-E_(1)=Q_(2)^(2)/(2C)-Q_(1)^(2)/(2C)` `=1/(2C)[Q_(2)^(2)-Q_(2)^(2)]=((0.25-0.01)/(2xx48xx10^(-6)))` = 2500 J |
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| 89. |
Ever heard a crackling sound while taking out your sweater in winter? |
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Answer» Yes, sometimes while removing our sweater in winter, some crackling sound is heard and the sweater appears to stick to body. |
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| 90. |
Birds perched on electrical transmission wires do not suffer electric shock, but if a person touches both the wires at once receives a tremendous shock. Why? |
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Answer» The danger of electric shock arises not from mere contact with a live wire but rather from simultaneous contact with a live wire and another body or wire at a different potential so that our body provides a conducting path between the two and a current passes through our body. Touching a single wire by the birds does not result in a current through their bodies because then the electric circuit is not complete. But if a person touches two wires at different potentials at once, or if a barefooted person touches the live wire only, the electric circuit is complete and the person receives an electric shock. In the latter case, the current from the wire passes to the Earth through the body. [Notes : (1) We must not touch any electric appliance. when bare-footed or with wet hands. When a bare-footed person touches a short-circuited electric appliance, the current from such an appliance goes to the Earth through his body, thus completing the circuit. When our skin is dry, the electrical resistance of our body is about 50 kΩ, a wet skin lowers the resistance to 10 kΩ. It needs a minimum of 1 mA of electric current to pass through our body for us to experience a shock. Thus, when dry, it needs at least 50 V potential difference to get a shock, but only 10 V is enough when wet. (2) The Earth often serves as a charge reservoir known as a ground. A ground can accept or provide electrons freely, and it is so large that the addition or subtraction of electrons has a negligible effect on it. So, the ground remains essentially neutral at all times. When something is connected to the ground by a conductor, we say that it is earthed or grounded.] |
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| 91. |
Have you experienced a shock while getting up from a plastic chair and shaking hand with your friend? |
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Answer» Yes, sometimes a shock while getting up from a plastic chair and shaking hand with friend is experienced. |
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| 92. |
An electric dipole has two point charges of 1.6 × 10-19 C and -1.6 × 10-19 C separated by 2 A. If the dipole is placed in a uniform electric field of 10 N/C, making an angle of 300 with the dipole moment, find (i) the magnitude of the torque acting on the dipole due to the field (ii) the potential energy of the dipole. |
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Answer» Data: q = 1.6 × 10-19 C, 2l = 2Å = 2 × 10-10 m, E = 10 N/C, θ = 30° Electric dipole moment, p = 2ql = q (2l) = (1.6 × 10-19 C)(2 × 10-10 m) = 3.2 × 10-29A∙m2 (j) The magnitude of the torque, τ = pE sin θ = (3.2 × 10-29 Am2 )(10 N/C) sin 30° = 3.2 × 10-28 × 0.5 = 1.6 × 10-28 N∙m (ii) The potential energy, U = -pE cosθ = -(3.2 × 10-29 A∙m2 )(10 N/C) cos30° = -3.2 × 10-28 × 0.866 = -2.771 × 10-28 J |
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| 93. |
What constitutes an electric shock? |
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Answer» Living organisms are electrical conductors. Electric shock is the result of the passage of electric current through our body. During electric shock we experience an extreme stimulation of nerves and muscles. It needs a minimum of 1 mA of electric current to pass through our body for us to experience a shock. |
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| 94. |
At time t=0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them beocme 4V? `[Take : In 5 = 1.6, In3 = 1.1]`. . |
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Answer» Here, supply voltage, `V_(s) = 10 vol t`. Effective resistance, `R = (2xx2)/(2+2) = 1 M Omega = 10^(6) Omega` Effective capacitanace, `C = 2+2 = 4 muF = 4xx10^(-6)F` While charging the capacitor with voltage applied, voltage across the capacitor is given by `V_(c) = V_(s) [1-e^(-t//RC )]=10 [1-e^(-t//10^(6))xx4xx10^(-6)]` `4 = 10 [1-e^(-t//4)]` `1-e^(-t//4) = (4)/(10) = 0.4` `e^(-t//4) = 1 - 0.4 = 0.6` `- (t)/(4) log_(e) e = log_(e) 0.6 = log_(e) ((3)/(5)) = log_(e) 3 - log_(e) 5` `- (t)/(4) = 1.1 - 1.6 = -0.5` `t = 4xx0.5 = 2s` |
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| 95. |
A particle carrying charge + q is held at the center of a square of each side arranged on the square as shown in Fig. If q = 2 muC, what is the net force on the particle? |
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Answer» As in clear from Fig. I(a)., forces on the particle at O due to `(-2 q, -2 q) , (-3 q, -3 q) and (+4 q, + 4 q)` are equal and opposite They cancel out in pairs. However, forces due to +7 q add up. There, net force on the particle at O is ` F = (1)/(4pi in_(o)) xx ((7q) (q) + 7q(q))/((1//2)^(2))` = `(9xx10^(9)xx14(2xx10^(-6))^(2))/(1//4)` ` = 36xx14xx4xx10^(-3) N` |
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| 96. |
An electric dipole has opposite charges of magnitude 2 × 10-15 C separated by 0.2 mm. It is placed in a uniform electric field of 103 N/C. (i) Find the magnitude of the dipole moment. (ii) What is the torque on the dipole when the dipole moment is at 60° with respect to the field? |
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Answer» Data : q = 2 × 10-15 C, 2l = 0.2 mm = 2 × 10-4 m, E = 103 N/C, θ = 60° (i) The magnitude of the dipole moment is V = q (2l) = (2 × 10-15 C) (2 × 10-4 m) = 4 × 10-19 C∙m (ii) The torque on the dipole is τ = pE sin θ = (4 × 10-19 C∙m) (103 N/C) sin 60° = 4 × 10-16 × 0.866 = 3.46 × 10-16 N∙m |
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| 97. |
The smilling face of Fig, consists of three parts , (i) a thin rod of charge `-3.0 muC` that forms a full circle of radius 6.0 cm. (ii) a thin rod of charge `2.0 muC` that forms a circular are of radius 4.0 cm, subtending an angle of `90^(@)` about the centre of full circle, and (iii) and electric diipole with dipole moment `= 1.28xx10^(-21)` Cm perpendicular to a radial line as shown in Fig. What is the net electric potentail as the centre ? |
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Answer» Here, `q_(1) = -3.0 muC, r_(1) = 6.0 cm` `V_(1) = (q_(1))/(4pi in_(0) r_(1)) = (k-(-3.0))/(6.0) = - (k)/(2) muC//cm`, where `k = (1)/(4pi in_(0))` Again, `q_(2) = + 2 muC, r_(2) = 4.0 cm` `V_(2) = (q_(2))/(4pi in_(0) r^(2)) = (k(+2.0))/(4.0)= (k)/(2) muC//cm`, For the dipole, `theta = 90^(@)` Potential due to dipole at the center `V_(3) = (p cos theta)/(4pi in_(0) r^(2)) = (p cos 90^(@))/(4pi in_(0) r^(2)) = 0` `:.` Net electric potential at the center `V = V_(1) + V_(2) + V_(3) = (k)/(2) (k)/(0) + 0 = ` zero. |
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| 98. |
An electric dipole consists of two unlike charges of magnitude 2 × 10-6 C separated by 4 cm. The dipole is placed in an external field of 105 N/C. Find the work done by an external agent to turn the dipole through 180°. |
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Answer» Data: q = 2 × 10-6 , 2l = 4cm = 4 × 10-2 m. E = 105 N/C, θ = 180° + θ0 Let us assume the dipole is initially aligned parallel to the field, i.e., θ0 = 0. Then, θ = 180°. The work done by an external agent, W = pE(1 – cos θ) = q (2l) E(1 – cos 180°) = (2 × 10-6 C)(4 × 10-2m)(105 N/C) [1 – (-1)] = 16 × 10-3 J |
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| 99. |
`A,B,C` and `D` are identical, parallel , conducting plates arranged as shown, with equal separations between consecuting plates. `A` and `D` are connected to a cel. If `B` is now connected to `C`, which of the following will occur?A. Only that some charge will flow through the cell.B. Only that some charge will flow from `B` to `C`C. Only that there will be no electric field between `B` and `C`D. More than one of the above |
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Answer» Correct Answer - D |
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| 100. |
The enrgy of a charged capacitor is U. Another identical capacitor is connected parallel to the first capacitor, after disconnecting the battery. The total energy of the system of these capacitors will beA. `(U)/(4)`B. `(U)/(2)`C. `(3U)/(2)`D. `(2U)/(4)` |
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Answer» Correct Answer - B Common potential `(C_(1)V_(0)+C_(2)xx0)/(C_(1)+C_(2))=(C_(2)V_(0))/(C_(1)+C_(2))` `U_("before")=(1)/(2)C_(1)V_(0)^(2)` `U_("after")=(1)/(2)C_(1)[(C_(1)V_(0))/(C_(1)+C_(2))]^(2)+(1)/(2)C_(2)[(C_(1)V_(0))/(C_(1)+C_(2))]^(2)` `=(1)/(2)[(C_(1)V_(0))/(C_(1)+C_(2))]^(2)(C_(1)+C_(2))` `rArr" "(U_("before"))/(U_("after"))=(C_(1)+C_(2))/(C_(1))` `"Here, "C_(1)=C_(2)=C` `therefore" "(U_("before"))/(U_("after"))=(2C)/(C)" "rArr=(U)/(2)` |
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