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Quiz-1 (Electrostatics)
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1. Change in P.E. per unit charge in the electric field is:
Solution: AΔV = W/q = ΔU/q
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2. What happens when fast-moving electrons are stopped and fall on the metallic target in an evacuated glass bulb
Solution: DElectromagnetic waves are produced when fast-moving electrons are suddenly stopped by a metal target of a high atomic number.
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3. The electric force of repulsion between two electrons at a distance of 1m isA. 1.8NB. 1.5×1019C. 2.30×10-28 ND. 2.30×10-18 N
Solution: C
F = K (q2/r2)F = Ke2 = 9×109 × (1.6×10-19)2= 23.04×10-29F = 2.304×10-28 N
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4. The electric field in a region of space is given by E= 5 N/C. The electric flux due to this field through an area 2 m2 isA. 100 N/C m2B. 20 N/C m2C. 10 NC-1 m2D. 1000 NC-1 m2
Solution: CWe know thatφe = E.A = 5×2 = 10Nm2C-1
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5. The value of relative permittivity for all the dielectrics other than air or vacuum is always:
Solution: CThe value of relative permittivity for all the dielectrics other than air or vacuum is always greater than unity.
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6. An electric field lines provide the information about
Solution: DAn electric line of force is an imaginary continuous line or curve drawn in an electric field such that tangent to it at any point gives the direction of the electric force at that point and depends on the medium.
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7. The electric force between two charges placed in air is 2N. When placed in a medium of Er = 80 the force reduces to
Solution: BCoulomb force with permittivity is given by
Fr = F/ϵr = 2/80 = 0.025N
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8. The force between two point charges separated by air is 4N. When separated by a medium of relative permittivity 2, the force between them becomes:
Solution: CFmed = Fvac/εr = 4/2 = 2N
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9. Two capacitors are identical one is filled with air and the other with oil. Both capacitors carry the same charge. The ratio of the electric fields Eair/Eoil is:
Solution: DAs the charge is same it means that capacitor is charged and battery is disconnected. So, by inserting dielectric (oil) electric field decreases.
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10. A positive charge is moved against an electric field. Its P.E:
Solution: ADirection of electric field is from positive to negative, so when the ‘+’ charge is moved against it, its P.E willIncrease according to E = qΔv
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11. One volt is:
Solution: AΔV = W/q1 volt = 1J/1C =1 JC-1
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12. S.I unit of permittivity of free space is:A. faradB. C2/Nm2C. WeberD. C2/Nm
Solution: BIn units, the value of permittivity of free space is
ε0 = 8.85×10-12 C2 N-1m-2
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13. Potential gradient is defined as:
Solution: DThe quantity which gives the maximum value of rate of change of potential with distance is known as potential gradient:Potential gradient = ΔV/Δr
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14. Two parallel plates are placed 2cm apart the potential difference between them is 1000V. A proton is placed in between them the force acting on the proton is:A. 8.0×10-15NB. 4×10-15NC. 3.2×10-24ND. 6×10-18N
Solution: A
Plates are connected with a battery of 1000 V.
So, the potential difference between the two plates is 1000 Volt.
Given that the distance between the plates d is 2 cm= .02 m
Electric field E= V/d
E= 1000/0.02 =50,000 N/C
and Force F= E×q
Here q is charge on a proton= 1.6×10-19 C
So, force on proton is = 50,000×1.6×10-19
= 8×10-15 N
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15. When two charges are equal “q” each, force they exert on each other is “F”. When one of charge is double, the 2q charge exerts a force 2F on charge q. the force exerted by q on 2q is?
Solution: DCoulomb force is a mutual force.
F12 = -F21
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16. The minimum charge on any object cannot be less than:
A. 1.6×10-19 CB. 9.1×109 CC. 3.2×10-19 CD. no definite value exists
Solution: AThe charge on a body is integral multiple of a minimum value of charge equal to 1.6×10-19 C.
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17. The ratio between the charge stored and potential difference across theplates of a capacitor is known as:
Solution: DC = Q/V
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18. Which of the following laws define the quantitative relationship of the force between two-point charges and the distance separating them?
Solution: BThis law states that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.F = k (q1q2/r2)
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19. Two point charges 4 μ C and 10 μ C repel each other with a force of 40N. When a third charge of 2 μ C is entered mid way between them, the charges now repel each other with a force of:
Solution: CSince the charge is placed midway between them, where the field is opposing, the magnitude of force will remain same.
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20. When two electric field lines cross each other then we have
Solution: CIf two electric field lines intersect each other then there will be two directions of the electric field at that point which is not possible. Due to this contradiction two electric field lines never intersect each other.
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21. The force between two charges situated in the air is F. The force between the same charges if the distance between them is reduced to half and they are situated in a medium having dielectric constant 4 is:
Solution: AWhen charges are placed in a medium with a dielectric constant, the force between them changes. The force between two charges.
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22. The constant “k” in Coulomb’s Law depends upon:
Solution: D
F = k (q1q2)/r2
"k" is the constant of proportionality. Its value depends upon the nature of the medium between the two charges and the system of units.
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23. The potential difference between the pair of similar conducting plates is known. What additional information is needed in order to find the electric field intensity?
Solution: BV = Ed
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24. If the potential difference on a surface is equal to zero b/w any two points, then surface is said to be:
Solution: CIf the potential difference on a surface is equal to zero b/w any two points, then surface is said to be an equi-potential surface.
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25. Which among the following statements is true with regard to electric field lines?
Solution: DElectric field lines can never intersect because tangent at any point on electric field lines represent the direction of electric field and if they intersect at a point it means that at that point there are two different directions for electric field which is not possible.
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26. If electric and gravitational forces on an electron placed in a uniform electric field balance each other, then the electric intensity will be?
Solution: AqE = mg ⇒ E = (mg)/q
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27. What is a possible unit for electric field strength?A. Cm-1B. Nc-1C. Nm-1D. NV-1
Solution: BThe SI unit of electric field strength is newton per coulomb (N/C) or volts per meter (V/m). The force experienced by a very small test charge q placed in a field E in a vacuum is given by E = F/q, where F is the force experienced.
THINGS TO REMEMBER:
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28. A particle carrying a charge of 2e falls through potential difference of 3.0V. Calculate the energy acquired by it:A. 9.6×10-16 JB. 9.6×10-15 JC. 9.6×10-20 JD. 9.6×10-19 J
Solution: DEnergy = qΔV =(2e)(3V) = 6eV= 6×1.6×10-19 J = 9.6×10-19 J
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29. The conventional direction of electric field is __________.
Solution: AThe conventional direction of field lines is from positive to negative. The field lines originate at the positive charge and terminate at the negative charge.
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30. The charge on an electron is known to be 1.6 x 10-19 Coulomb. In a circuit the current flowing is 1A. How many electrons will be flowing through the circuit in a second?A. 1.6×1019B. 1.6×10-19C. 6.25×1018D. 0.625×1012
Solution: CWe know that the number of electrons in one coulomb is the ratio of one coulomb to the charge on one electron.
When 1 Coulomb charge flow through a wire in 1 second then the current through the wire is 1 AMPERE.
I=Q/t
1 Ampere = 1 Coulomb /1 SecondCharge on 1 electron = e = 1.6 x 10-19 CCurrent I =1 A
time =1 s
I= Q/t = (ne)/t
n= (I t)/ei.e, 6.25 x 1018 electrons.
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31. If an electric field in a region can be regarded as zero then the possibility is:
Solution: BIf the potential is constant then the electric field is zero.
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32. The value of Coulomb’s constant ‘K’ in terms of permittivity of free space is given as:
Solution: AThe value of constant K in Coulombs law is 9×109 Nm2C-2And it depends on nature of medium and system of unit.
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33. There are two charges 1μC and 6μC, the ratio of forces acting on them will be:
Solution: BCoulomb force is a mutual force.F12 = - F21
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34. A line whose tangent at each point is in the direction of electric intensity at that point is called a line of:
Solution: CA line whose tangent at each point is in the direction of electric intensity at that point is called a line of electric force.
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35. The force of proton in electric field of magnitude 106 N/C is:A. 1.6×10-15 NB. 1.6×1013 NC. 1.6×109 ND. 1.6×10-13 N
Solution: DF = qE = (1.6×10-19)(106) = 1.6×10-13 N
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Quiz-2 (Electrostatics)
1. Electric field intensity and electric flux density have:
Solution: CElectric field = Electric flux density = φe/ABoth have same unit “N/C”.
2. Under static conditions, the electric field at all points on the surface of a conductor is
Solution: BThe electric field is defined as the gradient of potential and the surface of a conductor has a constant potential. Therefore, there is no field along the surface of the conductor and hence the electrostatic field at the surface of a charged conductor should be Normal to the surface at every point.
3. An electron of mass “m” and charged “e” is accelerated from rest through a potential different “V” in vacuum. Its final speed will be
4. Electric intensity due to a point charge varies as:A. rB. 1/r2C. r2D. 1/r
Solution: BE= k (q/r2) ⇒ E ∝ 1/r2
5. Two particles, X and Y, are 4m apart. X has a charge of 2Q and Y has a charge of Q. The force of X on Y:
Solution: CCoulomb force is a mutual force.FXY = -FYX
6. Electric field terminates at _________.
Solution: BElectric field originates at the positive charge and terminates at the negative charge. The conventional direction of the field is from positive to negative.
7. sec/ohm is equal to:
Solution: ARC = t ⇒ C = t/R
8. The electric intensity at infinite distance from the point charge is:
Solution: CE = k (q/r2) = k (q/∞2) = 0Note: Anything divided by infinity is equal to zero.
9. Charge “Q” is spread uniformly along the circumference of a circle of radius “R”. A point particle with charge "q” is placed at the center of this circle. The total force exerted on the particle can be calculated by Coulomb’s law:
Solution: DThe vector sum of all the forces acting on a charge placed at the center of the circle is zero.
10. An alpha particle has twice the charge of a proton. Two protons separated by a distance “d” exert a force “F”. on each other. What must be the separation between the alpha particles so that they also exert a force “F” on each other?
11. If 500 J of work are required to carry a charged particle between two points with a potential difference of 20V, the magnitude of the charge on the particle is:
Solution: DΔV = W/q ⇒ q=W/ΔV = 500/20 = 25C
Solution: CThe energy stored in a capacitor in term of electric field between the plates can be found by Energy = ½ ε0εrE2 (Ad)
12. In capacitor, energy is stored in:
13. Choose the correct statement concerning electric field lines:
Solution: BThe line are closer where the field is strong, the lines are farther apart where the field is weak.
14. Which, among the following, is the field where electric charge experiences a force?
Solution: ACharges experience a force in an electric field because charges come under the influence of a field that already has charges- an electric field.
15. Electric field originates at __________.
Solution: AElectric field originates at the positive charge and terminates at the negative charge. The conventional direction of the field is from positive to negative.
16. An isolated charged particle produces an electric field with magnitude “E” at a point 2m away from the charge. A point at which the field magnitude is E/4 is:
Solution: AE = k (q/r2) ⇒ E ∝ 1/r2
17. The electric field at a distance of 10cm from an isolated point particle with a charge of 2×10-9 C is:
18. The no. of electrons in one coulomb charge is equal to:A. 6.02×1018B. 1.6×10-19C. 6.2×1018D. 6.2×1021
Solution: CQ = nen= Q/e = 1/(1.6×10-19) = 6.25×1018
19. A uniform electric field and a uniform magnetic field are produced, pointed in the same direction. An electron is projected with its velocity pointed in the same direction:
Solution: DElectric and magnetic filed are in same direction and electron projected in same direction. The angle between B and Ve is o. and Sin 0 = 0.So the magnetic field does not work on the electron. Then Electron move due to the electric field. Which is from positive to negative charge, So electron and electron repel and magnitude of velocity decreases.
20. If the distance between the two charged bodies is halved, the force between them becomes:
Solution: B
F = k (q1q2)/r2 ⇒ F ∝ 1/r2
21. A method for charging a conductor without bringing a charge body in contact with it is called:
Solution: BA method for charging a conductor without bringing a charge body in contact with it is called electrostatic induction.
22. The electric intensity is expressed in unit of N/C or:
23. An electron traveling north enters a region where the electric field is uniform and points north. The electron:
Solution: CWhen electron moves in the direction of electric field, its speed decreases.
24. A field that spreads outwards in all directions is __________.
Solution: BA radial field is one which spreads in all directions. This field is known as the radial field because it spreads out radially from a source.
25. Charges of +2μc and -2μc are placed at points (P) and (Q) respectively. Tell the location at which electric potential is zero:
Solution: DLet 2d is separation between +2μC and -2μC. Net potential due to both charges at midway between “P” and “Q” is
26. Electric and gravitational forces:
Solution: BBoth gravitational and electric forces are conservative.
27. The potential at a point, where a charge of 1×10-3 C is placed at a distance of 10 m is:
28. The expression of energy stored in a capacitor is given by:A. E = CV2B. E = ½ C2 VC. E = ½ CV2D. E = ½ (CV)2
Solution: CElectrical potential energy = (Average p.d)(Charge)Energy = (½ V)(Q) = ½ VQ = ½ V (CV) = ½ CV2
29. A 5MΩ resistor is connected with a 2µf capacitor. The time constant of the circuit is:
Solution: Dt = RC = (5×106 )(2×10-6) = 10s
30. Electric lines of forces are parallel and equally spaced, then electric field is:
Solution: DWhen the electric lines are equally spaced and parallel, then the field is uniform. In this case the same number of lines pass through per unit area.
31. In the time constant “RC” circuit, how much charge is stored, out of maximum charge q0:A. 0.37q0B. 0.63qoC. 0.51q0D. 0.90q0
Solution: BCapacitive time constant is the time required by the capacitor to deposit 0.63 times the equilibrium charge “qo” (63% of q∘).
32. The ratio of the force between two small spheres with constant charges A in air, B in a medium of dielectric constant K is:A. K2 : 1B. K : 1C. 1 : KD. 1 : K2
33. A hollow metal sphere. is charged to a potential "V". The potential at its center is:
Solution: AInside a hollow charge conducting sphere potential is constant and it has the same value at that of surface.Vsurface = Vcenter
34. The unit of εre
A. Nm-1 A-1B. Nm2 C2C. N-1 m1C2D. No unit
Solution: Dεr = Fvac/Fmed = Evac/Emed = Vvac/Vmed = Cmed/Cvac
(unit less and dimension less)
35. Two large parallel conducting plates are separated by a distance d, placed in a vacuum, and connected to a source of potential difference “V”. An oxygen ion, with charge “2e”, starts from rest on the surface of one plate and accelerates to the other. If “e” denotes the magnitude of the electron charge, the final kinetic energy of this ion is:
Solution: DK.E = qΔV = 2eV
