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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.
| 501. |
As is known, all mater is made up of atoms/molecules. Every atom consits of a central of a central core, called the atomic nucleus, around which negatvely charged electrons revole in ciruclar orbits. Every atom is electrically neutral. Containing as many electron as the number of protons in the nucleas. Thus, even though normally, the materails are electrically neutral, they do contains charges, but thier charges are exactly balanced. The vast amount of charge in an object is usually hidden as the object is usually hidden as the object is said to be electracally neutral charge. With such an equality or balance of charges the object is said to be electrically neutral or uncharged. To electrify or charge a neutral body, actully transfer to the other body. The body which gains electrons become negatively charged and the body which loses electrons becomes positivelyh charged. Further, like charges repel adn unlike charges attract. Read the above passage and answer the following questions : (i) Every body, whether a conductor or an insulator is electrically neutral. Is it true ? (ii) Charging lies in charge imbalance, i.e, excess charge, comment. (iii) How do you visualize this principle being applied in our daily life ? |
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Answer» (i) Yes, it is true. Every conductor/inssularor is elecrtrically neutral, as it contains equal amounts of positive charge and negative charge. (ii) This statement is true. Charging lies really in charge imbalance. When a body loses some electrons, it becomes positively charged because it has excess of protons over electrons. The reverse is also true. (iii) Nature/God has created the universse. In original,all bodies are neutral with no forces fo attraction/repuslsion. When intersts of any two persons clash (i.e., two bodies are rubbed against eachother), they become charged. From the charging, arise the forces of attractions/repulsion, i.e., pulls and pressures of life. Nature/God wants us to live in peace without stress and tensions in life. We get charged over petty things in life and invite all sorts of pulls, pressures and tensions. |
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| 502. |
What is potential gradient at a distance of `10^(-12) m` from the centre of the platinum nucleas ? What is the potential gradient at the surface of the nucleas ? Atomic number fo platinum is 78 and radius of platinum nucleas is `5xx10^(-15) m`. |
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Answer» Correct Answer - `1.23xx10^(17) Vm^(-1) ; 4.5xx10^(21) Vm^(-1)` Here, `q = Ze = 78xx1.6xx10^(-19)C` Potential gradient at a point is numberically equal to electric field at that point, i.e., `(dV)/(dr) = E = (q)/(4pi in_(0) r^(2))` At `r = 10^(-12)m`, `(dV)/(dr) = E = (9xx10^(9)xx78xx1.6xx10^(-19))/((10^(-12))^(2))` `= 1.123xx10^(17) Vm^(-1)` At `r = 5xx10^(-15) m`, `(dV)/(dr) = E = (q)/(4pi in_(0) r^(2)) = (9xx10^(9)xx78xx1.6xx10^(-19))/((5xx10^(-15))^(2))` `= 4.5xx10^(21) Vm^(-1)` |
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| 503. |
On which factors does the capacitance of a capacitor depend / |
| Answer» It depends on geometry of the plates, distance between them and nature of dielectric medium spearting the plates. | |
| 504. |
If the plates of a charged capacitor be suddenly connected to each other by a wire, what will happen ? |
| Answer» The capacitor will be discharged immediately. | |
| 505. |
The distance between the plates fo a parallel plate capacitor is d. A metal plate of thickness `d//2` is placed between the plates, what will be the new capacity ? |
| Answer» As electrons field inside the metal plate is zero, d becomes `d//2`. Hencr C becomes twice (from `C = in_(0) A//d`). | |
| 506. |
Is the electrostatic potential necessarily zero at a point where the electric field strength is zero? Give an example to illustrate your answer,. |
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Answer» No, As we know that electric field is equal to negative of potential gradient: `E = - (dV)/(dr)` so, even if electric field at a point is zero, the potential may have some non zero constant value at that point. Example. Electric fied inside a charged conduting sphere is zero but potential at any point inside the sphere is same as that on the surface of sphere. |
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| 507. |
A point charge Q is placed at the point O as shown in Fig. Is the potential difference `(V)_(A) - V_(B))` positive, negative or zero if Q is (i) possible (ii) negative ? |
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Answer» From fig. `V_(A) - V_(B) = (kQ)/(OA) - (KQ)/(OB)` As `OA lt OB`, `:.` When Q is possitive `(V_(A) - V_(B))` is positive and when Q is negative , `(V_(A) - V_(B))` is negative. |
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| 508. |
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is ismilarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the poistive terminal of one is connected to the negative terminal of the other. The final energy of the configuration isA. (a) zeroB. (b) `3/2CV^2`C. (c) `25/6CV^2`D. (d) `9/2CV^2` |
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Answer» Correct Answer - B C and 2C are in parallel to each other. `:.` Resultant capacity=(2C+C)` `C_R=3C` Net potential `=2V-V` `V_R=V` `:.` Final energy `=1/2C_R(V_R)^2=1/2(3C)(V)^2=3/2CV^2` |
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| 509. |
A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase ro remain constant. |
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Answer» When the capacitance is connected to dc source, it gets charged. The battery is now disconnected So no more charge can flow in, i.e., charged stored will stay constant. On removing dielectric, capacitance decreases . Energy stored `= (Q^(2))/(2C)` will increase, Pot, `V = (Q)/(C)` would increase `E = (V)/(d)` will increase. |
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| 510. |
Prove that a closed equipotenitial surface with no charge within itself must enclose an equipotential volume. |
| Answer» Suppose a closed equipotential surface with no charge within itself does not enclose an equipotential volume. Therefore, potential just inside the surface would be different from potential at the surface, resulting in some potential gradient. Therefore, there would be field lines pointing inwards or outwards from the surface. This is possible only if other end of the lines are at a charges inside. As there is no charge inside, therefore, the entire volume inside the equipotential surface must be the same potential. | |
| 511. |
A test charge q is made to move in the electric field of a point charge Q along two different closed paths. Fig. First path has sections along and perpendicular loop of the same area as the first loop. How does the work done compare in the two cases? |
| Answer» From the knowledge of theory, we understand that electric field is conservative. Therefore, work done in moving a test charge over any closed path is zero. | |
| 512. |
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. `(16C_(1))/(n_(1) n_(2))` |
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Answer» Correct Answer - D `C_(s) = (C_(1))/(n_(1)) , C_(p) = n_(2) C_(2)` `E_(s) = (1)/(2) C_(s) (4V)^(2) = (1)/(2) (C_(1))/(n_(1)). 16 V^(2) = (8 C_(1) V^(2))/(n_(1))` `E_(p) = (1)/(2) C_(p) V^(2) = (1)/(2) n_(2) C_(2) V^(2)` As `E_(p) = E_(s) :. (1)/(2) n_(2) C_(2) V^(2) = (8 C_(1) V^(2))/(n_(1))` `C_(2) = (16C_(1))/(n_(1) n_(2))` |
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| 513. |
Two blocks A and B are connected by a spring made of a non conducting material. The blocks are placed on a non conducting smooth horizontal surface (see figure). The wall touching A is also non conducting. Block A carries a charge – q. There exists a uniform electric field of intensity `E_0` in horizontal direction, in the entire region. Find the value of minimum positive charge Q that we must place on block B and release the system so that block A subsequently leaves contact with the wall. Force constant of the spring is k. Neglect interaction between charges on the blocks. |
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Answer» Correct Answer - `Q=(q)/(2)` |
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| 514. |
Two point charges A and B of value of `+15 mu C and +9 mu C` are kept 18 cm apart in air. Calculate the work done when charge B is moved by 3cm towards A. |
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Answer» Correct Answer - `1.35 J` Here, `q_(1) = 15xx10^(-6)C, q_(2) = 9xx10^(-6)C` `r_(1) = 18xx10^(-2) m , r_(2) = (18 - 3) 10^(-2)m` work done = final P.E. - initial P.E. `w = (q_(1) q_(2))/(4pi in_(0) r_(2)) -(q_(1) q_(2))/(4pi in_(0) r_(1)) = (q_(1) q_(2))/(4pi in_(0)) [(1)/(r_(2)) - (1)/(r_(1))]` `= 9xx10^(9)xx15xx10^(-6)xx9xx10^(-6)xx9xx10^(-6) [(10^(2))/(15) - (10^(2))/(18)]` W = 1.35 joule |
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| 515. |
How is force between two charges affected when each charge is doubled and distance between them is also doubled? |
| Answer» As `F prop (|q_(1)|| q_(2)|)/(r^(2)) :.` F becomes `((2) (2))/(2^(2))` time =1 time, i.e., force remains the same, | |
| 516. |
Charge Q is given by the displacement `r=a hati+bhat j` in an electrif field `E=E_(1)hati+E_(2)hatj`. The work done isA. `Q((E_(1)a)+(E_(2)b))`B. `Qsqrt((E_(1)a)^(2)+(E_(2)b)^(2))`C. `Q(E_(1)+E_(2))sqrt(a^(2)+b^(2))`D. `Q(sqrt(E_(1)^(2)+E_(2)^(2)))sqrt(a^(2)+b^(2))` |
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Answer» Correct Answer - A Work done , `W=F.R = QE.r = Q(E_(1)hati+E_(2)hatj).(ahati+bhatj)` `=QE_(1)q+QE_(2)b=Q(E_(1)a+E_(2)b)` |
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| 517. |
A small charged ball is in state of equilibrium at a height h above a large horizontal uniformly charged dielectric plate having surface charge density of `sigma C//m^2`. (a) Find the acceleration of the ball if a disc of radius r (ltlt h) is removed from the plate directly underneath the ball. (b) Find the terminal speed `(V_0)` acquired by the falling ball. Assume that mass of the ball is m, its radius is x and coefficient of viscosity of air is `eta`. Neglect buoyancy and assume that the ball acquires terminal speed within a short distance of its fall. |
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Answer» Correct Answer - (a). `(gr^(2))/(2h^(2))` (b). `V_(0)=(mgr^(2))/(12 pi eta xh^(2))` |
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| 518. |
A small part of `dl` length is removed from a ring having charge per unit length `lamda`. Find electric field at centre due to remaining ring.A. zeroB. `(-lambda l)/(4 pi epsi_(0)a^(2))`C. infinityD. `(lambda)/(4 pi epsi_(0)l)` |
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Answer» Correct Answer - B |
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| 519. |
A sermicircular ring of radius R carries a uniform linear charge of `lamda`. P is a point in the plane of the ring at a distance R from centre O. OP is perpendicular to AB. Find electric field intensity at point P. |
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Answer» Correct Answer - `(1)/(4pi in_(0))(lamda)/(R)ln(sqrt(2)+1)` |
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| 520. |
Two equal insulating threads are placed parallel to each other. Separation between the threads (= d) is much smaller than their length. Both the threads have equal and opposite linear charge density on them. The electric field at a point P, equidistant from the threads (in the plane of the threads) and located well within (see figure) is `E_0`. Calculate the field at mid point (M) of line AB. |
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Answer» Correct Answer - `(E_(0))/(2)` |
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| 521. |
Two point charges 'q1' and 'q2' are placed at a distance 'd' apart as shown in the figure. The electric field intensity is zero at a point 'P on the line joining them as shown. Write two conclusions that you can draw from this. |
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Answer» (i) The two point charges (q1 and q2) should be of opposite nature. (ii) Magnitude of charge g, must be greater than that of charge q2. |
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| 522. |
In the following circuit, the resultant capacitance between `A` and `B` is `1muF`. Then value of `C` is A. `32/11muF`B. `11/32muF`C. `23/32muF`D. `32/23muF` |
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Answer» Correct Answer - D From figure, equivalent capacitance of `6muF, 12muF and 4muF` condensers is, `C_(1)=8muF`. Similarly equivalent capacitance of `8muF, 2muF and 2muF` condensers is, `C_(2)=8//3muF`. The equivalent capacitance of `8muF and 1muF` condensers is, `C_(3)=8//9muF`. The equivalent of `C_(2) and C_(3)` is, `32//9muF`. Now, equivalent of C and `32//9muF` condenser is `1muF`. Thus, `C=32//23muF`. |
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| 523. |
The resultant capacitance between the points A and B in the figure below is(A) 1 μF (B) 1.5 μF (C) 2 μF (D) 3 μF. |
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Answer» Correct option is (A) 1 μF |
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| 524. |
A parallel-plate air capacitor of a plate area A and plate separation d has capacitance C0 . A dielectric of thickness d, width A/2 and relative permittivity k, is inserted between the plates as shown. The capacitance of the capacitor is(A) 2(k + 1) C0 (B) (k + 1 )C0 (C) kC0(D) \(\cfrac{k+1}2\)C0 |
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Answer» Correct option is (D) \(\cfrac{k+1}2\)C0 |
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| 525. |
If the electric field given by `(5hat(i)+4hat(j)+9hat(k))`, then the electric flux through a surface of area 20 unit lying in the `yz`- plane will beA. 100 unitB. 80 unitC. 180 unitD. 20 unit |
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Answer» Correct Answer - A Electric flux is equal to the product of an area element and the perpendicular component of E. As the surface is lying in Y-Z plane `therefore" E.dA"=phi=(5)(20)="100 unit"` |
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| 526. |
The electric field in a region of space is given by, `E=5veci+2vecj N//C` The electric flux through an area `2m^(2)` lying in the YZ plane in SI unit isA. 10B. 20C. 5D. 15 |
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Answer» Correct Answer - A Along X axis E is `5N//C`, YZ plane is perpendicular to X axis `therefore" "phi=Eds=5xx2=10` |
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| 527. |
Two small spheres, each carrying a charge q are placed r m apart and they interact with force F. If one of the sphere is taken around the other once in a circular path, the work done will be equal to |
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Answer» Correct Answer - A Work done in equipotential surface is zero. |
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| 528. |
The `500 muF` capacitor is charged at a steady rate of `100 mu C//s`. The potential difference across the capacitor will be 10 V after an interval ofA. 5 sB. 0.5 sC. 0.05 sD. 50 s |
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Answer» Correct Answer - D `V=q/C=((dq//dt)xxt)/C" "thereforet=(V*C)/((dq//dt))=50s` |
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| 529. |
A small electric dipole is placed at origin with its dipole moment directed along positive x-axis .The direction of electric field at point `(2,2sqrt(2),0)`A. along z-aixsB. along y-axisC. along negative y-axisD. along negative z-axis |
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Answer» Correct Answer - B |
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| 530. |
Two conducting concentric, hollow spheres A and B have radii a and b respectively, with A inside B. Their common potentials is V. A is now given some charge such that its potential becomes zero. The potential of B will now be |
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Answer» Correct Answer - B |
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| 531. |
A positive charge is fixed at the origin of coordinates. An electri dipole which is free to move and rotate is placed on the positive `x`-axis. Its moment is directed away from the origin. The dipole willA. move towards the originB. move away from the originC. rotate by `pi//2`D. rotate by `pi` |
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Answer» Correct Answer - A |
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| 532. |
There are two infinite slabs of charge, both of thickness d with the junction lying on the plane x = 0. The slab lying in the range 0 lt x lt d has a uniform charge density `+rho` and the slab lying in the region – d lt x lt 0 has uniform charge density `-rho`. Find the Electric field everywhere and plot its variation along the x axis. Note: This can be used to model the variation of electric field in the depletion layer of a p – n junction. |
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Answer» Correct Answer - `E=-(rho)/(epsilon_(0))(d+x)" "-d lt x le0` `=-(rho)/(epsilon_(0))(d-x)" "0lexltd` `=0" "|x|gt d` |
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| 533. |
A nonconducting sheet of large surface area and thickness d contains uniform charge distribution of density `rho`. Find the electric field at a point P inside the plate, at a distance x from the central plane. Draw a qualitative graph of E against x for `0ltxltd`. |
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Answer» Correct Answer - `2:1` |
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| 534. |
A solid conducting sphere of radius R is placed in a uniform electric field E as showo in figure. Due to electric field non uniform surface charges are induced on the surface of the sphere. Consider a point A on the surface of sphere at a polar angle `theta` from the direction of electric field as shown in figure. Find the surface density of induced charges at point A in terms of electric field and polar angle `theta` : lt brgt |
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Answer» Correct Answer - `[3 in_(0)Ecostheta]` |
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| 535. |
A particle (A) having charge Q and mass m is at rest and is free to move. Another particle (B) having charge q and mass m is projected from a large distance towards the first particle with speed u. (a) Calculate the least kinetic energy of the system during the subsequent motion. (b) Find the final velocity of both the particles. Consider coulomb force only. |
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Answer» Correct Answer - (a). `(1)/(4)m u^(2)` (b). A moves to right with velocity u. B is at rest . |
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| 536. |
A circular metal plate of radius 10 cm is given a charge of `20muC` on its surface. The charge density of the plate isA. `3.185xx10^(-6)C//m^(2)`B. `2xx10^(-6)C//m^(2)`C. `3xx10^(-9)C//m^(2)`D. `3.184xx10^(-4)C//m^(2)` |
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Answer» Correct Answer - D `sigma=q/(2piR^(2))=(20xx10^(-6))/(2xx3.14xx10^(-2))` `= 3.184xx10^(-4)C//m^(2)` |
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| 537. |
Below the fixed end O of the insulating horizontal thread OB, there is a fixed charge A of `Q = 20muc`. At the end B of the thread there is a small mass m carrying charge `Q = 20moc`. The mass is released from the position shown and it is found to come to rest when the thread becomes vertical. Assume that the thread does not hit the fixed charge at A. `[g = 10 m//s^2]` (a) Find mass m. (b) Find tension in the thread in the equilibrium position when the thread is vertical. (c) Is the equilibrium mentioned in (b) stable or unstable? |
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Answer» Correct Answer - (a). 72g (b). 4.32N (c) unstable |
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| 538. |
A circular metal plate of radius 10 cm is given a charge of `20muC`. The outward pull on the plate in the vacuum isA. 560 NB. 370 NC. 360 ND. 630 N |
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Answer» Correct Answer - C `F=q^(2)/(2epsi_(0)kdS)=q/(2epsi_(0)k2piR^(2))=q^(2)/(4piepsi_(0)kR^(2))` `=(400xx10^(-12)xx9xx10^(9))/((10xx10^(-2))^(2))=360N.` |
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| 539. |
When a parallel plate capacitor is connected to a source of constant potential differenceA. all the charge drawn from source is stored in the capacitorB. the potential difference across capacitor grows very rapidly initially and this rate decreases to zero eventuallyC. only half of the energy drawn from the source is dissipated outside the capacitorD. all of these |
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Answer» Correct Answer - D Initially when potential difference is high, hence rate of flow of charge is high. But when potential difference across capacitor reaches the applied potential difference, this rate tends towards zero. Energy drawn from source = QV, but energy stored in capacitor = `QV//2` |
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| 540. |
Calculate the capacitance of a parallel plate capacitor having circular discs of radii `0.05m` each. The separation between the discs is `1 mm`. |
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Answer» Correct Answer - `0.69xx10^(-10) F` Here, `r = 0.05m, d = 1 mm = 10^(-3) m` Capacitor `C = (in_(0) A)/(d) = (in_(0) pi r^(2))/(d)` `= 8.85xx10^(-12) xx(22)/(7) xx ((2.05)^(2))/(10^(-3)) = 0.69xx10^(-10) F` |
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| 541. |
Two metallic conducors have net charge of `+70 pC and -70 pC`, which result in a potential difference of `20 V` between them. What is the capacitanace of the system ? |
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Answer» Correct Answer - `3.5 pF` Here, `q = 70 pC = 70xx^(-12)C, V = 20 V`, `C = (q)/(V) = (70xx10^(-12))/(20) = 3.5xx10^(-12) F = 3.5 pF` |
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| 542. |
(a) Calculate the potential at a point P due to a charge of `4xx10^(-7) C` located 9 cm away. (b) Hence obtain the work done in bringing a charge of `2xx10^(-9) C` from infinity to the point P. Does the answer depend on the path along which the charge is brought ? |
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Answer» (a) Here, `q = 4xx10^(-7) C`, `r = 9 cm = 9xx10^(-2)m , V = ?` `V = (q)/(4pi in_(0) r) = 9xx10^(9) ((4xx10^(-7)))/(9xx10^(-2))` `= 4xx10^(4)` volt (b) `W = q xx v = (2xx10^(-9))xx4xx10^(4)` `= 8xx10^(-5) J` The answer does not depend upon the path along which the charge is brought because electrostatic forces are conservative forces. |
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| 543. |
The capacity of a parallel plate condenser is inversely proportional to theA. dielectric constant of the mediumB. area of the plateC. length of the plateD. distance between the two plates |
| Answer» Correct Answer - D | |
| 544. |
The capacity of a parallel plate condenser is given byA. `C=Q/V`B. `C=(Akin_(0))/d`C. `C=d/(Akin_(0))`D. C = A.d. |
| Answer» Correct Answer - B | |
| 545. |
The relation between electric charge, electric potential and capacity isA. `C=Q/V`B. `C=V/Q`C. V=QCD. all of these |
| Answer» Correct Answer - A | |
| 546. |
One pico Farad is equal toA. `10^(-9) F`B. `10^(-19) F`C. `10^(-109)F`D. `10^(-12)F` |
| Answer» Correct Answer - D | |
| 547. |
Force between two charges separated by a certain distance in air is F. If each charge is doubled and the distance between them is also doubled, force would be(A) F (B) 2 F (C) 4 F (D) F/4 |
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Answer» Correct option is: (A) F |
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| 548. |
A hollow metal sphere of radius `5 cm` is charged so that the potential on its surface is `10 V`. The potential at the centre of the sphere isA. 0 VB. 10 VC. same as at a point 5 cm away from the surfaceD. same as at a point 25 cm away from the surface |
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Answer» Correct Answer - B Electric potential near the surface, on the surface and inside the surface is same. |
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| 549. |
The charge density on the surface of a conducting sphere is `64xx10^(-7)C//m^(2)` and the electric intensity at a distance of 2 m from the centre of the sphere is `4pixx10^(4)N//C`. The radius of the sphere isA. 0.83 mB. 0.4 mC. 0.6 mD. 0.38 m |
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Answer» Correct Answer - A `R^(2)=(Ein_(0)kr^(2))/sigma=(4pixx10^(4)xx8.85xx10^(-12)xx4)/(64xx10^(-7))` `therefore" "R=0.83m` |
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| 550. |
Define ‘capacitance’. Give its unit. |
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Answer» The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors. C = \(\frac{q}{V}\) or Q∝V The SI unit of capacitance is coulomb per volt or farad (F). |
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