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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.
| 1. |
What is the net flux of the uniform electric field of the above question through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? |
| Answer» All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero. | |
| 2. |
An electric dipole consists of charges `pm 2.0 xx 10^(8) C` separated by a distance of `2.0 xx 10^(-3)` m. It is placed near a long line charge of linear charge density `4.0 xx 10^(-4) C m^(-1)` as shown in figure (30-W4), Such that the negative charge is at a distance of `2.0 cm from the line charge. Find the force acting on the dipole. A. 7.2 N towards the line chargeB. 6.6 N away from the line chargeC. 0.6 N away from the line chargeD. 0.6 N towards the line charge. |
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Answer» Correct Answer - D The electric field at a distance r from the line charge of linear density `lamda` is given by `E=lamda/(2piepsilon_(0)r)` Hence, the field at the negative charge, `E_(1)=((4.0xx10^(-4))(2xx9xx10^(9)))/0.02=3.6xx10^(8)NC^(-1)` The force on the negative charge, `F_(1)=(3.6xx10^(8))(2.0xx10^(-8))=7.2N` towards the line charge Similarly, the field at the positive charge, i.e., at r = 0.022 m is `E_(2)=3.3xx10^(8)NC^(-1)`. The force on the positive charge, `F_(2)=(3.3xx10^(8))xx(2.0xx10^(-8))=6.6N` away from the line charge. Hence, the net force on the dipole = 7.2N - 6.6N =0.6N towards the line charge |
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| 3. |
A particle of mass `10^(-3)`kg and charge `5muC` is thrown at a speed of `20ms^(-1)` against a uniform electric field of strength `2xx10^(5)NC^(-1)`. The distance travelled by particle before coming to rest isA. 0.1 mB. 0.2 mC. 0.3 mD. 0.4 m |
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Answer» Correct Answer - B `F=qE=5xx10^(-6)xx2xx10^(5)=1N` Since, the particle is thrown against the field `therefore" "a=-F//m=-1/10^(-3)=-10^(3)ms^(-2)` `As" "v^(2)-u^(2)=2as` `therefore" "0^(2)-(20)^(2)=2xx(-10^(3))xxsors=0.2m` |
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| 4. |
A uniform electric field `E=2xx10^(3)NC^(-1)` is acting along the positive x-axis. The flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane isA. `20NC^(-1)m^(2)`B. `30NC^(-1)m^(2)`C. `10NC^(-1)m^(2)`D. `40NC^(-1)m^(2)` |
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Answer» Correct Answer - A Here, `E=2xx10^(3)NC^(-1)` is along + x-axis Surface area, `s=(10cm)^(2)=10^(2)xx10^(-4)m^(2)=10^(-2)m^(2)` when plane is parallel to yz plane, `theta=0^(@)` So `phi=Escostheta=2xx10^(3)xx10^(-2)cos0^(@)=20NC^(-1)m^(2)` |
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| 5. |
The electrostatic attracting froce on a small sphere of charge `0.2muC` due to another small sphere of charge `-0.4muC` in air is 0.4N. The distance between the two spheres isA. `43.2xx10^(-6)m`B. `42.4xx10^(-3)m`C. `18.1xx10^(-3)m`D. `19.2xx10^(-6)m` |
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Answer» Correct Answer - B Here, `q_(1)=0.2muC=0.2xx10^(-6)C` `q_(2)=-0.4muC=-0.4xx10^(-6)C,F=-0.4N` `As" "F=(q_(1)q_(2))/(4piepsilon_(0)r^(2)` `therefore" "r^(2)=(q_(1)q_(2))/(4piepsilon_(0)F)=(0.2xx10^(-6)xx0.4xx10^(-6)xx9xx10^(9))/0.4` `=1.8xx10^(-3)` `therefore" "r=(1.8xx10^(-3))^(1//2)=0.0424m=42.4xx10^(-3)m` |
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| 6. |
A point charge `4muC` is at the centre of a cubic Gaussian surface 10 cm on edge. Net electric flux through the surfaceisA. `2.5xx10^(5)Nm^(2)C^(-1)`B. `4.5xx10^(5)Nm^(2)C^(-1)`C. `4.5xx10^(6)Nm^(2)C^(-1)`D. `2.5xx10^(6)Nm^(2)C^(-1)` |
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Answer» Correct Answer - B Here, `q=4muC=4xx10^(-6)C,l=10cm` `phi=q/epsilon_(0)=(4xx10^(-6))/(8.85xx10^(-12))=4.5xx10^(5)Nm^(2)C^(-1)` |
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| 7. |
A point positive charge is brought near an isolated conducting sphere as shown in figure the electric field is best given by A. Figure (i)B. Figure (ii)C. Figure (iii)D. Figure (iv) |
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Answer» Correct Answer - A When a point positive charge is brought near an isolated conductiong sphere, then there develops some negative charge on left side of the sphere. Electric lines of forces emmanting from the point positive charge end normally on the left side of the sphere. And due to positive charge on the right side of the sphere, the electric field is best given by figure (a). |
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| 8. |
A coin is made up of Al and weighs 0.75 g. It has a square shape and its diagonal measures 17 mm. It is electrically neutral and contains equal amounts of positive and negative charges. The magnitude of these charges is (Atomic mass of Al = 26.98 g)A. `3.47xx10^(4)C`B. `3.47xx10^(2)C`C. `1.67xx10^(20)C`D. `1.67xx10^(22)C` |
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Answer» Correct Answer - A Mass of the coin = 0.75 g, Atomic mass of aluminium = 26.98 g Number of Al atoms in the coin, `N=(6.02xx10^(23))/(26.98)xx0.75=1.67xx10^(22)` As charge number (z) Al is 13, each atom of Al contain 13 protons and 13 electrons. Magnitude of positive and negative charges is one paisa coin `=NZe=1.67xx10^(22)xx13xx1.6xx10^(-19)C=3.47xx10^(4)C` |
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| 9. |
In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is , however, not constant but increases uniformly along the positive z-direction. At the rate of `10^(5)NC^(-1)m^(-1)` . What are the force and torque experienced by system having a total dipole moment equal to `10^(-7)`Cm in the negative z-direction? |
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Answer» Dipole moment of the system `p=qxxdl=-10^(-7)Cm` Rate of increase of electric field per unit length, `(dE)/(dl)=10^(+5)NC^(-1)` Force (F) experienced by the system is given by the relation F=Qe `F=a(dE)/(dl)xxdl` `=pxx(dE)/(dl)` `-10^(-7)xx10^(-5)` `-10^(-2)N` The force is `-10^(-2)N` in the negative z-direction i.e., opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is `180^(@)`. Torque (τ) is given by the relation, `tau = pE sin180^(@)`. =0 Therefore, the torque experienced by the system is zero. |
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| 10. |
A point charge causes an electric flux of `-1.0xx10^(3) N m^(2)//C` to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface ? (b) What is the is the value of the point charge ? |
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Answer» (a) Electric flux, `phi=-1.0xx10^(3)Nm^(2)//C` Radius of the Gaussian surface, r= 10.0 cm Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., `-10^(3)Nm^(2)//C`. (b) Electric flux is given by the relation, `phi=(q)/(in_(0))` Where, q = Net charge enclosed by the spherical surface `in_(0)` = Permittivity of free space = `8.854xx10^(-12)N^(-1)C^(2)m^(-2)` `thereforeq=phiin_(0)` `=-1.0xx10^(3)xx8.854xx10^(-12)` `=-8.854xx10^(9)C` `=-8.854nC` ltBrgt Therefore, the value of the point charge is −8.854 nC. |
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| 11. |
A conducting sphere of radius 10 cm has unknown charge. If the electric field at a distance 20 cm from the centre of the sphere is `1.2xx10^(3)NC^(-1)` and points radially inwards. The net charge on the sphere isA. `-4.5xx10^(-9)C`B. `4.5xx10^(9)C`C. `-5.3xx10^(-9)C`D. `5.3xx10^(9)C` |
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Answer» Correct Answer - C Here, distance of point from the centre of the sphere, r = 20 cm = 0.2 m Electric field, E = `-1.2xx10^(3)NC^(-1)` As `E=q/(4piepsilon_(0)r^(2))` `therefore" "q=(4piepsilon_(0)r^(2))E=((0.2)^(2)xx(-1.2xx10^(3)))/(9xx10^(9))=-5.3xx10^(-9)C` |
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| 12. |
A spherical insulator of radius R is charged uniformly with a charge Q throughout its volume and contains a point charge `Q/16` located at its centre. Which of the following graphs best represents qualitatively, the variation of electric field intensity E with distance r from the centre?A. B. C. D. |
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Answer» Correct Answer - A For `(x lt R),=(Qx)/((4)/(3)piR^(3)3epsi_(0))+(Q//16)/(4piepsi_(0)x^(2))` `(dE)/(dx)=0implies(Q)/(4piepsi_(0))=((1)/(R^(3))-(2)/(16x^(3)))=0` `(d^(2)E)/(dx^(2))=(6Q)/(4piepsi_(0)(16x^(4)))gt0` `thereforeAt x=(R)/(2),E` is minimum. For `(x gt R), E=((Q+Q//16))/(4piepsi_(0)x^(2))` |
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| 13. |
A conducting sphere fo radius 10 cm has an unknown charge. If the electric field 20 cm from the center of the sphere is `1.5xx10^(3) N//C` and points radially inwards, what is the net charge on the sphere ? |
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Answer» Electric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation, `E=(q)/(4piin_(0)d^(2))` where, ltBrgt q = Net charge = `1.5xx10^(3)N//C` d = Distance from the centre = 20 cm = 0.2 m `in_(0)` = Permittivity of free space And, `(1)/(4piin_(0))=9xx10^(9)Nm^(2)C^(-2)` `thereforeq=E(4piin_(0))d^(2)` `=(1.5xx10^(3)xx(0.2)^(2))/(9xx10^(9))` `=6.67xx10^(9) C` `=6.67nC` ltBrgt Therefore, the net charge on the sphere is 6.67 nC. |
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| 14. |
A system consits of a uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density `rho=alpha/r`, where `alpha` is a positive constant and r is the distance from the centre of the sphere. Find the charge of the sphere for which the electric field intensity E outside the sphere is independent of R.A. `alpha/(2epsilon_(0))`B. `2/(alphaepsilon_(0))`C. `2pialphaR^(2)`D. None of these |
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Answer» Correct Answer - C Using Gauss theorem for spherical surface of radius r outside the sphere with a uniform charge density `rho` and a charge q `underset(s)intE.ds=(Q_(enc))/(epsi_(0))` `E4pir^(2)=(1)/(epsi_(0))(q+underset(R)overset(r)int(alpha)/(r)(4pir^(2))dr),` `E4pir^(2)=((q-2pialphaR^(2)))/(epsi_(0))+(4pialphar^(2))/(2 epsi_(0))` The intensity E does not depend on R if `(q-2pialphaR^(2))/(epsi_(0))=0or q=2pialphaR^(2)` |
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| 15. |
A uniformly charged conducting sphere of 2.4m diameter has a surface density of `80.0 mu C//m^(2)`. (a) Find the charge on the sphere (b) What is the total electric flux leaving the surface of the sphere ? |
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Answer» (a) Diameter of the sphere, d = 2.4 m ltBrgt Radius of the sphere, r = 1.2 m Surface charge density, `sigma=80.0muC//m^(2)=80xx10^(-6)C//m^(2)` Total charge on the surface of the sphere, Q = Charge density `xx` Surface area `=sigmaxx4pir^(2)` `=80xx10^(-6)xx4xx3.14xx(1.2)^(2)` `=1.447xx10^(-3)C` Therefore, the charge on the sphere is `1.447xx10^(-3)C` (b) Total electric flux `(phi_(Total))` leaving out the surface of a sphere containing net charge Q is given by the relation, `phi_(Total)=(Q)/(in_(0))` Where, `in_(0)` = Permittivity of free space ltBrgt `=8.854xx10^(-12)N^(-1)C^(2)m^(-2)` `Q=1.447xx10^(-3)C` `phi_(Total)=(1.44xx10^(-3))/(8.854xx10^(-12))` `=1.63xx10^(8)NC^(-1)m^(2)` Therefore, the total electric flux leaving the surface of the sphere is `1.63xx10^(8)NC^(-1)m^(2)`. |
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| 16. |
Repulsion is the sure test of charging than attraction. Why? |
| Answer» A charged body may attract a neutral body and also an oposiet charged body. But it always repels like a charged body. Hence repulsion is the sure test of electrification. | |
| 17. |
Assertion: If there exists coulombic attracation between two bodies both of them may not be charged. Reason: In coulombic attraction two bodies are oppositely charged.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B Coulomb attraction exists even when one body is charged, and the other is uncharged. |
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| 18. |
(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e, where `vec(E) = 0`) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart. |
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Answer» (a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable. (b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions. |
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| 19. |
Two large, thin plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude `16xx10^(-22)C m^(-2)`. The electric field between the plates isA. `1.8xx10^(-10)NC^(-1)`B. `1.9xx10^(-10)NC^(-1)`C. `1.6xx10^(-10)NC^(-1)`D. `1.5xx10^(-10)NC^(-1)` |
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Answer» Correct Answer - A Here, `E=sigma/epsilon_(0)=(16xx10^(-22))/(8.854xx10^(-12))=1.8xx10^(-10)NC^(-1)` |
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| 20. |
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed `v_x` (like particle 1 in Figure). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is `(qEL^2)/(2m v_(x)^2)`. Compare this motion with motion of a projectile in gravitational field. |
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Answer» Charge on a particle of mass m = − q Velocity of the particle = `V_(X)` Length of the plates = L Magnitude of the uniform electric field between the plates = E Mechanical force, F = Mass (m) `xx` Acceleration (a) `a=(F)/(m)` However, electric force, `F=qE` Therefore, acceleration, `a=(qE)/(m)" "...(1)` Time taken by the particle to cross the fields of length L is given by, `t=("Length of the plate")/("Velocity of the particle")=(L)/(v_(x))" "...(2)` In the vertical direction, initial velocity, u = 0 According to the third equation of motion, vertical deflection s of the particle can be obtained as, `s=ut+(1)/(2)at^(2)` `s=0+(1)/(2)((qE)/(m))((L)/(v_(x)))^(2)` `s=(qEL^(2))/(2mv_(x)^(2))" "...(3)` Hence, vertical deflection of the particle at the far edge of the plate is `qEL^(2)//(2mv_(x)^(2))`. This is similar to the motion of horizontal projectiles under gravity. |
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| 21. |
Figure shows tracks of three charged particles crossing a uniform electrostatic field with same velocities along horizontal. Give the sign of the three charges. Which particle has the highest charge to mass ratio? A. AB. BC. CD. A and B |
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Answer» Correct Answer - C Particles A and B have negative charges because they are being deflected towards the positive plate of the electrostatic field. Particle C has positive charge because it is being deflected towards the negative plate. `therefore" "` Deflection of charged particle in time t in y-direction `h=0xxt+1/2at^(2)=1/2(qE)/mt^(2)" "i.e.,hpropq//m` As the particle C suffers maximum deflection in y-direction, so it has highest charge to mass `q//m` ratio. |
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| 22. |
(a) Explain the meaning of the statement ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges? |
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Answer» (a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge. (b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous. |
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| 23. |
A polythene piece rubbed will wool is found to have a negative charge of `3.0xx10^(-7)C`. (a) Estimate the number of electrons transferred (from which to which )? (b) Is there a transfer of mass from wool to polythene? |
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Answer» (a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged. Amount of charge on the polythene piece, `q = −3 × 10^(−7) C` Amount of charge on an electron, `e = −1.6 × 10^(−19) C` Number of electrons transferred from wool to polythene = n can be calculated using the relation, q = ne `n=(q)/(e)` ` =(-3 xx 10^(-7))/(-1.6 xx 10 ^(-19))` `= 1.87 xx 10^(12)` Therefore, the number of electrons transferred from wool to polythene is `1.87 × 10^(12)`. (b) Yes There is a transfer of mass taking place. This is because an electron has mass, `me = 9.1 × 10^(−3) kg` Total mass transferred to polythene from wool, `m = m_(e) xx n` `= 9.1 xx 10 ^(-31) xx 1.85 xx 10^(12)` `= 1.706 xx 10^(-31) xx 10^(-18) kg` Hence, a negligible amount of mass is transferred from wool to polythene. |
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| 24. |
An electric dipole is placed at an angle of `30^(@)` with an electric field intensity `2xx10^(5)N//C`. It experiences a torque equal to `4Nm`. The charge on the dipole, if the dipole is length is `2 cm`, isA. 8 mCB. 4 mCC. 6 mCD. 2 mC |
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Answer» Correct Answer - D Here, `E=2xx10^(5)NC^(-1),l=2cm,tau=4Nm` Torque, `vectau=vecpxxvecE,tau=pEsintheta` `therefore" "4=pxx2xx10^(5)xxsin30^(@)orp=4xx10^(-5)Cm` `therefore" "` Charge, `q=p/l=(4xx10^(-5)Cm)/(0.02m)=2xx10^(-3)C=2 mC` |
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| 25. |
Which of the following statement is not a similarity between electrostaic and gravitational forces?A. Both forces obey inverse square law.B. Both forces operate over very large distances.C. Both forces are conservative in nature.D. Both forces are attractive in nature always. |
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Answer» Correct Answer - D Electrostatic forces are both attractive and repulsive depending upon the type of charge, but gravitational force is always attractive. |
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| 26. |
The nucleus of helium atom contains two protons that are separated by distance `3.0xx10^(-15)m`. The magnitude of the electrostatic force that each proton exerts on the other isA. 20.6 NB. 25.6 NC. 15.6 ND. 12.6 N |
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Answer» Correct Answer - B Charge of proton is `q_(p)=1.6xx10^(-19)C` Distance between the protons is, `r=3xx10^(-15)m` The magnitude of electrostatic force between protons is `F_(e)=(q_(p)q_(p))/(4piepsilon_(0)r^(2))=(9xx10^(9)xx1.6xx10^(-19)xx1.6xx10^(-19))/((3xx10^(-15))^(2))=25.6N` |
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| 27. |
There are two types of electric charges positive charges and negative charges. The property which differentiates the two types of charges isA. field of chargeB. amount of chargeC. strength of chargeD. polarity of charge |
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Answer» Correct Answer - D The property which differentiates the two types of charge is called the polarity of charge. |
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| 28. |
The electric field components in Fig. are `E_(x)=ax^(1//2), E_(y)=E_(z)=0`, in which `alpha=800N//C m^(1//2)`. Calculate the flux through the cube |
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Answer» Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and `Delta S` is `pm pi//2`. Therefore, the flux `phi=E. Delta S` is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is `E_(L)=x^(1//2)=alpha a^(1//2)` (x = a at the left face). The magnitude of electric field at the right face is `E_(R)=alpha x^(1//2)=alpha (2a)^(1//2)` (x = 2a at the right face). The corresponding fluxes are `phi_(L)=E_(L).Delta S=Delta SE_(L). hat(n)_(L)=E_(L) Delta S cos theta=-E_(L) Delta S`, since `theta=180^(@)` `=-E_(L) a^(2)` `phi_(R)=E_(R).Delta S=E_(R) Delta S cos theta =E_(R)Delta S`, since `theta=0^(@)` `=-E_(R)a^(2)` Net flux through the cube. `=phi_(R)+phi_(L)=E_(L)a^(2)-E_(L)a^(2)=a^(2) (E_(R)-E_(L))=alpha a^(2)[(2a)^(1//2)-a^(1//2)]` `=alpha a^(5//2) (sqrt(2)-1)` `=800(0.1)^(5//2) (sqrt(2)-1)` ,brgt `=1.05N m^(2) C^(-1)` |
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| 29. |
A point charge `+10 mu C` is at distance of 5cm directly above the center of a square of side 10 cm as shown in Fig. What is the magnitude of the electric flux through the square ? (Hint. Think of the square of the square as one face of a cube with edge 10 cm) |
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Answer» The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces. `phi_(Total)=(q)/(in_(0))` Hence, electric flux through one face of the cube i.e., through the square, `phi=(phi_(Total))/(6)` `=(1)/(6)(q)/(in_(0))` Where, `in_(0)` = Permittivity of free space `=8.854xx10^(-12)N^(-1)C^(2)m^(-2)` `q=10muC=10xx10^(-6)C` `thereforephi=(1)/(6)xx(10xx10^(-6))/(8.854xx10^(-12))` `=1.88xx10^(5)Nm^(2)C^(-1)` Therefore, electric flux through the square is `1.88xx10^(5)Nm^(2)C^(-1)`. |
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| 30. |
Assertion : Total flux through a closed surface is zero if no charge is enclosed by the surface. Reason : Gauss law is true for any closed surface, no matter what its shape or size is.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B Gauss law implies that the total electric flux through a closed surface is zero is no charge is enclosed by the surface and it is true for any closed surface, independent of its shape and size. |
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| 31. |
When s person combs his hair, static electricity is sometimes generated by what process?A. Contact between the comb and hair results in a charge.B. Friction between the comb and hair results in a charge.C. Deduction between the comb and hair.D. Induction between the comb and hair. |
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Answer» Correct Answer - B While combing hair, friction between the comb and the hair results is the transfer of electrons. |
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| 32. |
Assertion : Electrostatic field lines start at positive charges and end at negative charges. Reason : Field lines are continuous curves without any breaks and they form closed loop.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C Electrosatic field lines are continuous curves without any breaks. They start at positive charges and end at negative charges. They cannot form closed loops. |
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| 33. |
Match the following and find the correct option. A. A - q, B - r, C - pB. A - p, B - r, C - pC. A - r, B - p, C - qD. A - r, B - q, C - p |
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Answer» Correct Answer - A Linear charge density, `lamda`= (charge)/(length), `A to q` Surface charge density, `sigma`= (charge)/(area), `B to r` Volume charge density, `rho` = (charge)/(volume), `C to p` |
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| 34. |
In question 55, which region or regions of the figure could the electric field be zero?A. Near AB. Near BC. Near CD. Now here |
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Answer» Correct Answer - A From the figure, it is clear that, there cannot be a neutral point between a positive and a negative charge. Neutral point can be only there between two like charges i.e., A and C. As magnitude of charge A is smaller than the magnitude of charge C, therefore, neutral point would lie closer to the charge A. |
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| 35. |
A non-uniform electric field is represented by the diagram. At which of the following points the electric field is greatest in magnitude? A. AB. BC. CD. D |
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Answer» Correct Answer - D Electric field strength is maximum where the electric field lines are closer. |
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| 36. |
In the question number 74, torque experienced by the system isA. `10^(2)N`B. `10^(-2)N`C. zeroD. `10^(3)N` |
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Answer» Correct Answer - C The force on the dipole is along negative direction of z-axis, so `theta=180^(@)` `therefore" "` Torque on dipole, `tau=PEsin180^(@)=0` |
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| 37. |
The dimensional formula of electric intensity isA. `[M^(1)L^(1)T^(3)A^(-1)]`B. `[ML^(-1)T^(-3)A^(1)]`C. `[M^(1)L^(1)T^(-3)A^(-1)]`D. `[M^(1)L^(2)T^(1)A^(1)]` |
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Answer» Correct Answer - C Since `E=E/q,[E]=([M^(1)L^(1)T^(-2)])/([AT])=[M^(1)L^(1)T^(-3)A^(-1)]` |
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| 38. |
Which of the following figures represents the electric field lines due to a single positive charge ?A. B. C. D. |
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Answer» Correct Answer - A The field lines of a single positive charge are radially outward. |
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| 39. |
Which of the following figure represents the field lines due to a single negative charge ?A. B. C. D. |
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Answer» Correct Answer - B The field lines of single negative charge are radially inward. |
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| 40. |
The dimensional formula of electric flux isA. `[M^(1)L^(1)T^(-2)]`B. `[M^(1)L^(3)T^(-3)A^(-1)]`C. `[M^(2)L^(2)T^(-2)A^(-2)]`D. `[M^(1)L^(-3)T^(3)A^(1)]` |
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Answer» Correct Answer - B Electric flux, `phi=intvecE*vecs` The dimension of `phi` = Dimension of E `xx` dimension of s `[phi]=[M^(1)L^(1)T^(-2)][AT]^(-1)[L^(2)]=[M^(1)L^(3)T^(-3)A^(-1)]` |
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| 41. |
Sketch the electric field lines for a unifomly charged hollow cylinder shown in Fig. A. B. C. |
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Answer» Correct Answer - B For a uniformly charged hollow cylinder, the electric field lines are as shown in figure (b). |
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| 42. |
The SI unit of electric flux isA. `NC^(-1)m^(2)`B. `NCm^(-2)`C. `NC^(-2)m^(2)`D. `NC^(-1)m^(-2)` |
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Answer» Correct Answer - A As electric flux, `phi=E*Deltas` `therefore" "` unit of `phi` is `NC^(1)m^(2)` |
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| 43. |
Which of the following curves represent electrostatic field lines correctly?A. B. C. D. |
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Answer» Correct Answer - C Electrostatic field lines start or end only at `90^(@)` to the surface of the conductor. Closed electric field lines are not possible in the given situation. |
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| 44. |
In question 55, which charge has the largest magnitude?A. AB. BC. CD. B and C have equal magnitude |
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Answer» Correct Answer - C From the given figure it is clear that, maximum number of electric lines of force are associated with charge C, so it must have the largest magnitude. |
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| 45. |
Object may acquire an excess or deficiency of charge byA. electric forceB. heatingC. shakingD. by rubbing |
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Answer» Correct Answer - D Object can be charged by rubbing. During rubbing some of the free electrons may get transferred from one object to other. |
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| 46. |
Figure shows the electric field lines around three point charges `A, B` and `C`. Which of the following charges are positive? A. Only AB. Only CC. Both A and CD. Both B and C |
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Answer» Correct Answer - C In the given figure, the electric lines of force emanate from A and C. Therefore, charges A and C must be positive. |
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| 47. |
Which of the following statements is not true about electric field lines?A. Electric field lines start from positive charge and end at negative charge.B. Two electric field lines can never cross each other.C. Electrostatic field lines do not form any closed loops.D. Electric field lines cannot be taken as continuous curve. |
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Answer» Correct Answer - D In a charge-free region, electric field lines can be taken to be continuous curves without any breaks. |
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| 48. |
An object is charged when it has a charge imbalance, which means theA. object contains no protonsB. object contains no electronsC. object contains equal number of electrons and protonsD. object contains unequal number of electrons and protons |
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Answer» Correct Answer - D An object is said to be charged, if it has a deficiency of electron or excess of electrons. That means it contains an unequal number of protons and electrons. |
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| 49. |
The electric lines of force do not intersect. Why ? |
| Answer» They do not intersect because if they intersect, at the point of intersection, intensity of electric field must act in two different directions, which is impossible. | |
| 50. |
What will happen when we rub a glass rod with silk cloth?A. Some of the electrons from the glass rod are transferred to the silk cloth.B. The glass rod gets positive charge and silk cloth gets negative charge.C. New charge is created in the process of rubbing.D. both (a) and (b) are correct. |
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Answer» Correct Answer - D When a glass rod is rubbed with silk cloth then some of the electrons from the glass rod get transferred to silk cloth and thus the glass rod gets positive charge and the silk gets negative charge. No new charge is created in the process of rubbing. |
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