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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.
| 1001. |
The insulation property of air breaks down at `E=3xx10^(6) "volt"//meter`. The maximum charge that can be given to a sphere of diameter `5m` is approximately (in coulombs)A. 3 nCB. 20 nCC. 1.5 nCD. 2 nC |
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Answer» Correct Answer - D Here, `R = 3 mm = 3xx10^(-3)m` `"As, "E=(Q)/(4pi epsilon_(0)R^(2)) or Q=4pi epsilon_(0)R^(2)E` For the maximum value of `E=10%` dielectric strength `therefore" E"=10%" of " 2xx10^(7)=2xx10^(6)NC^(-1)` `Q=(1)/(9xx10^(9))(3xx10^(-3))^(2)xx2xx10^(6)` `=2xx10^(-9)C=2nC` |
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| 1002. |
The unit of physical quantity obtained by the line integral of electric field isA. `NC^(-1)`B. `Vm^(-1)`C. `JC^(-1)`D. `C^(2)N^(-1)m^(-2)` |
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Answer» Correct Answer - C The unit of physical quantity obtained by the line integral of electric field is `JC^(-1)`. |
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| 1003. |
What is relative permittivity? |
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Answer» i. Relative permittivity or dielectric constant is the ratio of absolute permittivity of a medium to the permittivity of free space. It is denoted as K or εr. i.e., K or εr = ε/ε0 ii. It is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium. i.e., K or εr = \(\frac{F_{vacuum}}{F_{medium}}\) iii. It is also called as specific inductive capacity or dielectric constant. |
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| 1004. |
The unit of physical quantity obtained by the line integral of electric field isA. `NC^(-1)`B. `Vm^(-1)`C. `JC^(-1)`D. `C^(2)N^(-1) m^(-2)` |
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Answer» Correct Answer - C |
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| 1005. |
The electrostatic force between two charges is a central force. Why? |
| Answer» The electrostanic force acts along the line joining the two charges. So, it it a central force. | |
| 1006. |
Two point charges +4 µC and +2 µC repel each other with a force of 8 N. If a charge of -4 µC is added to each of these charges, the force would be(A) zero (B) 8 N (C) 4 N (D) 12 N |
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Answer» Correct option is: (A) zero |
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| 1007. |
Two point charge `+2 C` and `+6 C` repel each other with a force of `12 N` . If a charge of `-2 C` is given to each other of these charges , the force will now beA. 4 N (repulsive)B. 4 N (attractive)C. 12 N (attractive)D. 8 N (repulsive) |
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Answer» Correct Answer - B |
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| 1008. |
If two objects repel one another, you know both carry either positive charge or negative charge. How would you determine whether these charges are positive or negative ? |
| Answer» To determine the polority of charge on the two objects, we bring one of the objects near a positively charged glass rod. If the object is repelled away from the rod, it must be positibely charged (as like charges repel). However, if the object is attracted towards the glass rod, it must be negatively charged (as unlike charges attract). | |
| 1009. |
Two equal like charges in air repel eachother with a force F. By what percentage should each charge be reduced so that the force between them in medium of dielectric constant 2 reduces by `28`%`? |
| Answer» This display of light is often called sparking. It occurs due to electric discharge of the electrostatic changes produced in chewing the winter green life saver. | |
| 1010. |
Two charges are at a distance `d` apart. If a copper plate (conducting medium) of thickness `d//2` is placed between them , the effictive force will beA. `2 F`B. `F//2`C. `0`D. `sqrt(2) F` |
| Answer» Correct Answer - 3 | |
| 1011. |
Two point charge `+2 C` and `+6 C` repel each other with a force of `12 N` . If a charge of `-2 C` is given to each other of these charges , the force will now beA. zeroB. `8 N` (attractive)C. `8 N` (repulsive)D. none |
| Answer» Correct Answer - 1 | |
| 1012. |
Two protons are a distance of `1 xx 10^(-10) cm` from each other. The force acting on them areA. nuclear force and gravitational forceB. nuclear force and coulomb forceC. coulomb force and gravitational forceD. nuclear, coulomb and gravitational force |
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Answer» Correct Answer - C |
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| 1013. |
Find the ratio of the magnitude of the electric force to the grativational force acting between two protons.A. `10^(40)`B. `10^(38)`C. `10^(36)`D. `10^(34)` |
| Answer» Correct Answer - 3 | |
| 1014. |
Which of the following is correct regarding electric charge ? (i) If a body is having positivr charge i.e. shortage of electrons (ii) If a body is having negative charge i.e. excess of electrons (iii) Minimum possible charge `= +- 1.6 xx 10^(-19) C` (iv) Charge is quantised i.e. `Q = +- n e`, where `n = 1 , 2, 3, 4 ....`A. (i) and (ii)B. (ii) and (iii)C. (i) ,(ii) and (iii)D. All |
| Answer» Correct Answer - 4 | |
| 1015. |
Which of the following is correct regarding electric charge ? (i) If a body is having positivr charge i.e. shortage of electrons (ii) If a body is having negative charge i.e. excess of electrons (iii) Minimum possible charge `= +- 1.6 xx 10^(-19) C` (iv) Charge is quantised i.e. `Q = +- n e`, where `n = 1 , 2, 3, 4 ....`A. both (i) and (ii)B. Both (ii) and (iii)C. (i), (ii), (iii)D. All of these |
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Answer» Correct Answer - D |
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| 1016. |
One metallic sphere `A` is given positive charge whereas another identical metallic sphere `B` of exactly same mass as of `A` is given equal amount of negative charge. ThenA. mass of A and mass of B still remain equalB. mass of A increaseC. mass of B decreaseD. mass of B increase |
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Answer» Correct Answer - D |
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| 1017. |
Is there any kind of material which when placed between the plates of capacitor reduces its capacitance ? |
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Answer» No, dielectric constant of a material is always greater than I. As `K = (C_(m))/(C_(0)) and K gt 1 :. C_(m) gt C_(0)` i.e., the capacitance with dielectric slab between the plates is greater than with vacuum between the plates. So there is no such material which when placed between the plates of capacitor, redicues its capacitance. |
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| 1018. |
A copper sphere of mass `2g` contains nearly `2xx10^(22)` atoms. The charge on the nucleus of each atom is `29 e`. What fraction of the electrons must be removed from the sphere to give it a charge fo `+ 2muC` ? |
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Answer» Total number of electrons in the sphere `= 29xx2xx10^(22)` No. of electrons removed `= (q)/(e) = (2xx10^(-6))/(1.6xx10^(-19)) = 1.25xx10^(13)` Fraction of electrons removed `= (1.25xx10^(13))/(29xx2xx10^(22))= 2.16xx10^(-11)` |
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| 1019. |
What is the dielectric constant of metal ? |
| Answer» The dielectric constant of a metal si infinity. The electric field inside a conductor is zero so the dielectric costant, which is the ratio of applied electric field to the reduced electric filed, will be infinite for a metallic conductor. | |
| 1020. |
A capacitor is charged through a potential difference of 200V, when 0.1 C charge is stored in it. How much energy will it release, when it is discharge ? |
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Answer» Here, V = 200 vol t` q = 0.1 C` Energy relased on discharging = energy stored on charging `= (1)/(2) qV = (1)/(2) xx 0.1xx200 = 10J` |
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| 1021. |
Two chareged sperical conductors of radill `R_(1) and R_(2)` when connected by a connecting wire acquire charges `q_(1) and q_(2)` respectively. Find the ratio of their charge densities in terms of their radil ? |
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Answer» Two spherical conductors, when connected by a wire will share charges till, they common potential. Thus, `(q_(1))/(4pi in_(0) R_(1)) = (q_(2))/(4pi in_(0) R_(2)) or (q_(1))/(q_(2)) = (R_(1))/(R_(2))` Ratio of surface density of charges is `(sigma_(1))/(sigma_(2)) = (q_(1)//4pi in_(0) R_(1)^(2))/(q_(2)//4pi in_(0) R_(2)^(2)) = (q_(1))/(q_(2)) xx (R_(2)^(2))/(R_(1)^(2)) = (R_(1))/(R_(2)) xx (R_(2)^(2))/(R_(1)^(2)) = (R_(2))/(R_(1))` |
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| 1022. |
Give four properties of electric charges. |
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Answer» (i) Like charges repel and unlike charges attract each other. (ii) Charge is conserved. (iii) Charge on a body is not afffected by its motion. |
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| 1023. |
Dielectric constant of a medium is unity. What will be its permittively ? |
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Answer» We know that dielectric constant of a medium is `K = in_(r) = (in)/(in_(0))` `:. In = K in_(0) = 1xx8.854xx10^(-12)` `= 8.854xx10^(-12) C^(2) N^(-1) m^(-2)` |
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| 1024. |
A parallel palte capacitor is filled by a dielectric whose relative permittively varies with the applied voltage (U) as `epsilon = alpha U` where alpha `= 2V^(-1)`. A similar capacitor with no dielectric is charged to `U_(0) = 78V`. It is then is connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors. |
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Answer» On connecting the two given capacitors,let the final voltage be U. If capacity of capacitor of capacitor without the dielectric is C, then the charge on this capacitor is `Q_(1) = CU` The other capacitor with dielectric has capacity `in C` . Therefore, charge on it is `Q_(2) = in CU`. As `in = alpha U`, therefore, `Q_(2) = alpha C U^(2)` The initial charege on the capacitor (without the dielectric is C, then the charge on this capacitor is `Q_(1) = CU` The other capacitor with dielectric has capacity `in C`. Therefore, charge on it is `Q_(2) = in CU`. AS `in = alpha U`, therefore, `Q_(2) = alpha C U^(2)` The initial charge on the capacitor (without dielectric) that was charged si `Q_(0) = CU_(0)` From the conservation of charge, `Q_(0) = Q_(1) + Q_(2)` `CU_(0) = CU + alpha CU^(2) or alpha U^(2) + U - U_(0) = 0 :. U = (-1 +- sqrt(1+4 alpha U_(0)))/(2 alpha)` Using `alpha = 2 V^(-1) and U_(0) = 78 vol t`, we got `U = (-1 +- sqrt(1+4xx2xx78))/(2xx2) = (-1 +- sqrt(625))/(4)` As U is positive, therefore , `U = (sqrt(625) -1)/(4) = (24)/(4) = 6vol t` |
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| 1025. |
Write down the expression for the capacitance of a spherical capacitor. |
| Answer» `C = 4pi in_(0) (ab)/(b - a)` , where a and b are the radii of the inner and outer spheres respectively. | |
| 1026. |
An air -filled parallel-plate capacitor is to be constructed which can store`12 mu C` of charge when operated at `1200V.`What can be the minimum plate area of the capacitor?The dielectric strength air is `3xx10^6Vm^(-1).` |
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Answer» Here `Q = 12 muC, V = 1200 V,A = ?` Dielectric Strength `= 3xx10^(7) V//m` The electric field between the planes should not exceed `10%` of the dielectric strength i.e., `E = (sigma)/(in_(0)) = (Q)/(A in_(0)) = (10)/(100) xx 3xx10^(7) = 3xx10^(6) V//m` `:. A = (Q)/(in_(0)xx3xx10^(6)) = (12xx10^(-6))/(8.85xx10^(-12)xx3xx10^(6))` `= 0.45 m^(2)` |
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| 1027. |
For a given potential difference, does a capacitor store more or less charge with a dielectric than it does without a dielectric. |
| Answer» A capacitor with a dielectric would store more charge, as its capacity increases. | |
| 1028. |
Is there any condutor which can be given almost unlimited charge ? |
| Answer» Yes, earth can be given almost unlimited charge, because its capacity is very large. | |
| 1029. |
A large plane sheet of charge having surface charge density `5xx10^(-16) cm^(-2)` lies in XY plane. Find electric flux through a circular area of radious 1cm Given normal to the circular area makes an angle of `60^(@)` with Z-axis. |
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Answer» Here, `sigma = 5xx10^(-16) cm^(-2), phi = ?` `r = 1 cm = 10^(-2) m, theta = 60^(@)` `theta = E (Delta S) cos theta = ((1)/(2 in_(0))) pi r^(2) cos 60^(@)` `= (5xx10^(-16)xx3.14(10^(-2))^(2)xx1//2)/(2xx8.85xx10^(-12))` `4.44xx10^(-9) Nm^(2) C^(-1)` |
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| 1030. |
A plastic rod of length 2.2m and radius 3.6 mm carries a negative charge of `3.8xx10^(-7) C` spread uniformly over its surface. What is the electric field near the mid-point of the rod, at a point on its surface? |
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Answer» Here, l = 2.2m, `r = 3.6 mm = 3.6xx10^(-3)m`. `q = -3.8xx10^(-7)C.E = ?` Linear charge density, `lambda = (q)/(1) = (-3.8xx10^(-7))/(2.2)` `=-1.73xx10^(-7) Cm^(-1)` As `E = (lambda)/(2pi in_(0) r) = (2 lambda)/(4pi in_(0) r)` `:. E = (2(-1.73xx10^(-7)) xx9xx10^(9))/(3.6xx10^(-3))` `= -8.6xx10^(5) N//C` |
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| 1031. |
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude `19xx10^(-22) Cm^(-2)`. What is E (a) to the left of the plates (b) to the right of the plates (c) between the plates ? Here, `sigma = 19xx10^(-22) Cm^(-2)` |
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Answer» (a) On the left, the fields of the two plates are equal and opposite, So E = Zero. (b) On the right, the fields of the two plates are equal and opposite, So the resultant field is `E = (sigma)/(2 in_(0)) + (sigma)/(2 in_(0)) = (sigma)/(in_(0)) = (19xx10^(-22))/(8.85xx10^(-12))` `= 2.14xx10^(-10) NC^(-1)` |
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| 1032. |
Two large this metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and magnitude `17.0xx10^(-22) C//m^(2)`. What is `vec(E)` : in the outer region of the first plate. (b) in the outer region of the secound plate, and (c) between the plates ? See Fig. |
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Answer» Here, `sigma = 17.0xx10^(-22) C m^(-2)`, As discussed in theory , in region 1, to the left of the plates , `E = 0`. And in region III to the right of the plates , again, `E = 0`. However,in region II inbetween the plates, `E = (sigma)/(in_(0)) = (17.0xx10^(-22))/(8.85xx10^(-12)) = 1.92xx10^(-10) NC^(-1)` |
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| 1033. |
Which of the following figures cannot possibly represent electrostatic field lines A. I, ii, iii, ivB. I, ii, iii onlyC. I, iii, iv onlyD. ii, iii, iv only |
| Answer» Correct Answer - C | |
| 1034. |
The number of electric lines of force passing normally through unit area is calledA. electric fluxB. magnetic fluxC. flux densityD. none of these |
| Answer» Correct Answer - A | |
| 1035. |
The charge deposited per unit area of the surface is calledA. linera charge densityB. surface charge densityC. charge densityD. all of these |
| Answer» Correct Answer - B | |
| 1036. |
Consider a region inside which there are various types of charges but the total charge is zero ,.At points outside the regionA. the electric field is necessarily zeroB. the electric field is due to the dipole moment of the charge distribution onlyC. the dominant electric field is `prop (1)/(r^(3))`, for large `r`, where `r` is the distance from a origin in this regionD. the work done to move a charged particle along a closed path, away from the region, will be zero. |
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Answer» Correct Answer - C::D When there are various types of charges in a region, but the total charge is zero, the region can be supposed to contain a number of electric dipoles. Therefore ,at a points outside the region (which may be anywhere `w.t.t` electric dipoles), the dominant electric field `prop (1)/(r^(3))` for large `r`. Choice (c) is correct. Further, as electric field is conservative, work done to move a charged particle along a closed pth, away from the region will be zero, choice (d) is also correct. |
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| 1037. |
The charge deposited per unit length of cylinder is called asA. linear charge densityB. surface charge densityC. charge densityD. all of these |
| Answer» Correct Answer - A | |
| 1038. |
A metal surface of area `1 m^(2)` is charged with `sqrt(8.85)muC` in air. The mechanical force acting on it isA. 1 NB. 0.5 NC. 10 ND. 50 N |
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Answer» Correct Answer - B `F=(sigma^(2)ds)/(2in_(0)k)=q^(2)/(2in_(0)kds)` `=((sqrt(8.85)xx10^(-6))^(2))/(2xx8.85xx10^(-12)xx1xx1)=0.5N` |
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| 1039. |
The mechanical force acting on a unit area of a charged conductor isA. `f=(sigma^(2))/(2in_(0)k)`B. `f=(sigma)/(2in_(0)k)`C. `f=(sigma^(2))/(in_(0)k)`D. `f=(sigma)/(in_(0)k)` |
| Answer» Correct Answer - A | |
| 1040. |
The Electric field at a point isA. always continousB. continous if there is no charge at that pointC. discontinous only if there is a negative charge at that pointD. discontinous if there is a charge at that point |
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Answer» Correct Answer - B::D From the genral knowledge of theory, electric field at a point is continous If there is no charge at that point. And the field is dicontinous if there is charge at that point. Choices (b) and (d) are correct. |
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| 1041. |
If `oint_(s) E.ds = 0` Over a surface, thenA. the electric field inside the surface and on it is zeroB. the electric field inside the surface is necessarly uniformC. the number of flux lines entering the surface must be equal to the number of flux lines leaving itD. all charges must necessarily be outside the surface |
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Answer» Correct Answer - C::D `oint vec(E) .vec(ds)` represents electric flux over the closed surface, when `oint vec(E) .vec(ds) = 0`, it means the number of flux lines enetering the surface must be equal to number of flux lines leaving it. Further, as `oint vec(E) .vec(ds) = (q)/(in_(0))` where `q` is charge enclosed by the surface. When `oint vec(E) .vec(ds) = 0, q = 0, i.e.,` net charge enclosed by the surface must be zero. Therefore, all other charges must necessarily be outside the surface. This is because charges outside the surface do not contibute to the electric fllux, Choices (c) and (d) are correct. |
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| 1042. |
A charged conductor of area ds is placed in electric field of strength E. The force acting on unit area of the conductor isA. `(in_(0)kE^(2))/2`B. `(in_(0)k)/(2E)`C. `E^(2)/(2in_(0)k)`D. `in_(0)kE^(2)` |
| Answer» Correct Answer - A | |
| 1043. |
What do you understand by ECG and EEG ? What is their basis ? |
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Answer» ECG stands fpr electr- encephalo graph. Both are based on the fact that human body is a good conducter of electricity, but not a perfect conductor. Therefore, musle activity leads to small differences in electric potenitial form one point of the body to another. As human heart beats, potential differences in the range of 1 milli volt develop. These potential differences are detected and displayed with an insturment known as ECG. Further, nerve impulses in the brain lead to potential differences in the range of 0*1 milivolt. They are deteced and displayed with an insturment knwon as EEG. |
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| 1044. |
The solid angle substended at any point inside the surface due to small area ds is given byA. `(dscostheta)/(r^(2))`B. `(dsr^(2))/(costheta)`C. `dscosthetar^(2)`D. `(r^(2))/(dscostheta)` |
| Answer» Correct Answer - A | |
| 1045. |
Three charges `+q, 2 q and -4q` are placed on the three vertices of an equale-laterail triangle of each side`0*1m`. Calculate electrostatic potential energy of the system, take `q = 10^(-7) C` |
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Answer» Potential Energy of system of charges = `(1)/(4pi in_(0) r) [q(2q) + 2q + (-4q) + (-4q)q]` `= (10q^(2))/(4pi in_(0) r) = (9xx10^(9)xx10(10^(-7))^(2))/(0*1) = 9xx10^(-3) J`. |
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| 1046. |
A surface element `vec(ds) = 5 hat(i)` is placed in an electric field `vec(E) = 4 hat(i) + 4 hat(j) + 4 hat(k)`. What is the electric flux emanting from the surface ? |
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Answer» Here, `vec(ds) = 5 hat(i) , vec(E) = 4 hat(i) + 4 hat(j) + 4 hat(k)` Electric flux, `phi = vec(E). vec(ds) = (4 hat(i) + 4 hat(j) + 4 hat(k))* 5 hat(i)` `phi = 20 units` |
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| 1047. |
The dot product of E and normal area ds, calculated over the closed surface, isA. electric fieldB. electric fluxC. electric potentialD. all of these |
| Answer» Correct Answer - B | |
| 1048. |
Electric flux in an electric field `vecE` through area vector `vec(ds)` is given byA. `vecExxvec(ds)`B. `vecE*vec(ds)`C. `vecE//vec(ds)`D. `vec(ds)//vecE` |
| Answer» Correct Answer - B | |
| 1049. |
A point charge +Q is place just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct ? A. Total flux through the curved and the flat surface is `Q/epsi_(0)`B. The component of the electric field normal to the flat surface is constant over the surfaceC. The circumference of the flat surface is an equipotentialD. The electric flux passing through the curved surface of the hemisphere is `-Q/(2 epsi_(0)) (1-1/sqrt(2))` |
| Answer» Correct Answer - C::D | |
| 1050. |
Explain the disadvantage of static charge. |
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Answer» 1. When charge transferred from one body to other is very large, sparking can take place. For example, lightning in sky. 2. Sparking can be dangerous while refuelling your vehicle. 3. One can get static shock, if charge transferred is large. 4. Dust or dirt particles gathered on computer or TV screens can catch static charges and can be troublesome. |
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