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Quiz 5
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Q.1
Force of attraction between the plates of a parallel plate capacitor is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
let q be the charge on the plates We may consider secend plate is in the electric field of first. Thus second plate experience forec F=EqElectric field due to first plate E=σ/2εoK but σ=q/A Thus E=q/2εoAK F=q2/2εoAKAnswer: (a)
Q.2
six equal capacitors each of capacitance C are connected as shown in figure. Then equivalent capacitance between A and B is
0%
a) 6C
0%
b) C
0%
c)2C
0%
d)C/2
Explanation
Capacitor 1 to 5 forms a balanced Whetastone bridge. Current will not flow through capactor 3. Capacitor 1 and 4 are in series=C/2Capacitor 2 and 5 are in series=C/2Above combination is parallel to each other thus capacitance=C This resultant is parallel to capacitor 6 thus final resulatnet capacitance=C + C=2CAnswer: (c)
Q.3
A capacitor is charged by using a battery, which is then disconnected. A dielectric slab is then slipped between the plates which results in [ MPPMT 1995]
0%
a) reduction of charge on the plates and increase of potential difference across the plates
0%
b) increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates
0%
c)decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
0%
d)none of the above
Explanation
Answer:(c)
Q.4
the combined capacit of the parallel combination of two capacitor is four times their combined capacity when connected in series. This means that [ EAMCET 1994]
0%
a)their capacites are equal
0%
b) their capacities are 1µF and 2µF
0%
c)their capacities are 0.5µF and 2µF
0%
d)their capacities are infinite
Explanation
Answer: (a)
Q.5
Figure a shows two capacitors connected in series and joined to a battery. The graph in figure (b) shows the variation in potential as one moves from left to right on the branch containing the capacitors if. [ MPPMT 1999]
0%
a) C1 > C2
0%
b) C1=C2
0%
c)C1 < C2
0%
d)the information is not sufficient to decide the relation between C1 and C2
Explanation
Answer: (c)
Q.6
The capacity of a parallel plate condenser depends on [ MPPMT 1994]
0%
a) the type of metal used
0%
b) the thickness of plates
0%
c)the potential applied across the plates
0%
d)the separation between the plates
Explanation
Answer:(d)
Q.7
The force between the plates of parallel plate capacitor of capacitance C and distance of separation of plate d with potential difference V between the plates, is [ MPPMT 1999]
0%
a) CV2 / 2d
0%
b) C2V2 / 2d2
0%
c) C2V2 / d2
0%
d)V2d / C
Explanation
The force between the plates of a parallel plate capacitor is given by Answer: (a)
Q.8
The numerical value of the charge on either plate of the capacitor C shown in the figure is
0%
a)CE
0%
b) CER1 / (R2 + r )
0%
c)CER2 / (R2 + r )
0%
d) CER1 / (R1 + r )
Explanation
there will be no current in the branch containing capacitor , as the resistance offered by the capacitor in D.C. circuit is infinite. Therefore current drawn from the battery.I=E / (R2 + r )Potential difference across the terminal of the battery V=E -Ir=E - [E / (R2 + r )]V=ER2 / (R2 + r )Charge on capacitor q=CV q=C×ER2 / (R2 + r )Answer: (c)
Q.9
Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant K and thickness t is introduced between the plates, its capacitance becomes [ MPPMT 1989]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Electric field in air between the plates=E1=q / εoA Electric field in dielectric slab E2=q / KεoA The elctric field E1 between the plate exists in the distance (d-t) and E2 in the distance tHence , if the potential difference between the plates be V, thenV=E1(d - t) + E2t Answer: (c)
Q.10
Two metal spheres of capacitance C1 and C2 carry some charges. They are put in contact and then separated. the final charges Q1 and Q2 on then will satisfy [ MPPMT 1999]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
On contact both the spheres will have same potential V=Q1 / C1=Q2 / C2Thus Q1 / Q2=C1 / C2 Answer:(b)
Q.11
Each capacitor in figure has capcity 5µF. The voltmeter, connected in parallel reads 100V. The charge on each plate of cpacirtor is
0%
a) 0.05 mC
0%
b) 0.5 mC
0%
c) 5 mC
0%
d) 1 mC
Explanation
As volymeter is paraellel to capacitor voltage across capacitor is 100 q=CV=5×10-6×100=0.5 mC Answer: (b)
Q.12
The capacity of paralel plate ccondenser is 10µF without dielectric. Dielectric of constant 2 is used to fill half the thickness between the plates the capacitance in µF is .. [ EAMCET 1995]
0%
a)10
0%
b) 20
0%
c)15
0%
d)13.33
Explanation
From the formulaAnswer: (d)
Q.13
An electron moving along the positive x-axis with speed of 3×106 m/s enters the region of a uniform electric field. The electron stops after travelling a distance of 90 mm in the field. The electric field strength is [ mass of electron 9.1×10-31 kg]
0%
a) 2.84 kV/m, along -ve x-axis
0%
b) 0.284 kV/m, along -ve x-axis
0%
c)0.284 kV/m, along +ve x-axis
0%
d)28.4 kV/m, along +ve x-axis
Explanation
Resistive force=Eq retardation a=Eq/m from equation v2=u2 - 2as 0=u2 - 2as a=u2 / 2asEq/m=u2 / 2as E=mu2 / 2qas on solving E=0.284 kV/m force on e should be along -ve x ais so E should be along +ve x ais Answer:(c)
Q.14
An electric dipole of moment p is placed normal to the line of force of electric field E, then work done in defleting it through an angle of 180 degree is
0%
a) pE
0%
b) +2pE
0%
c) -2PE
0%
d) zero
Explanation
Answer: (d)
Q.15
SI unit of electric permitivity is
0%
a) N m2C-2
0%
b) Am-1
0%
c) N/C
0%
d) C2/ N m2
Explanation
Answer: (d)
Q.16
A finite ladder is constructed by connecting several sections of 2µF, 4µF capacitor combinations as shown in figure. It is terminated by a capacitor C. What value should be chosen for C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between? [ MPPMT 1999]
0%
a) 4 µF
0%
b) 2 µF
0%
c)18µF
0%
d)6 µF
Explanation
Effective capacitance between A and B is C . for the infinite long ladder. Thus by adding one extra unit will not affect the capacitance as shown in figureCapacitance Across A' and B' will be C Thus C across A and B is is in series with 4µF capacitot effective capacitance C'=4C/ 4+C C' is in parallel with 2µF and resultant of this combionation is C Answer: (a)
Q.17
The value of one Farad in e.s.u. will be [ PET 1998]
0%
a) 3×1010
0%
b)9×1010
0%
c)(1/9)×10-11
0%
d)(1/3)×1010
Explanation
Answer:(b)
Q.18
Capacitance of a capacitor made by a thin metal foil is 2µF. If the foil is folded with paper of thickness 0.15 mm, and dielectric constant of paper is 2.5, width of paper is 40 mm, then length o foil will be [ Raj. PET 1997]
0%
a) 0.34 m
0%
b) 1.33 m
0%
c) 13.4 m
0%
d) 339 m
Explanation
Answer: (d)
Q.19
Condenser A has a capacity of 15µF when it is filled with a medium of dielectric constantAnother condener B has a capacity 1µF with air between the plates. Both are charged separately by battery of 100V. After charging, both are connected in parallel with the and dielectric material being removed. The common potential now is [ MNR 1994]
0%
a)400 V
0%
b) 800 V
0%
c)1200 V
0%
d)1600 V
Explanation
Charge on the first capacitor filled with dielectric medium=15×10-6 ×100=15×10-4Charge on second capacitor=1×10-6 ×100=1×10-4 Total charge on both the capacitor=16×10-4 Capacity of first capacitor before filling dielectric=C/k=15µF/15=1µFCharge on second capacitor after connecting in parallel=Now V Q2 / C2V=8×10-4 / (1×10-6=800VAnswer: (b)
Q.20
Between the plates of parallel plate condenser there is 1mm thick paper of delectric constant . It it is charged at 100 V. the eletric field in Volt/metre between the plates of the capacitor is... [ MPPMT 1994]
0%
a) 100
0%
b) 100000
0%
c)25000
0%
d)400000
Explanation
E=V/d=100 / ( 10-3 )=100,000Answer: (b)
Q.21
Find the resultant capacitance between A and B [ Raj. PET 1997]
0%
a)(2/3)µF
0%
b) (8/3)µF
0%
c)(6/5)µF
0%
d)(7/3)µF
Explanation
Answer:(b)
Q.22
The area of plates of parallel plate condenser is A and the distance between the plate is 10 mm. There are two dielectric sheets in it, one of dielectricconstant 10 and thick ness 6 mm and other of dielectric constant 5 and thickness 4mm. The capacity of the condenser is [ MPPMT 1997]
0%
a) (12/35) εoA
0%
b) (2/3) εoA
0%
c) (5000/7) εoA
0%
d) (12/35) εoA
Explanation
Arrangement is equivalent to two capacitors in series having capacity Answer: (c)
Q.23
A condenser of capacity C1 is charged to a potential Vo. The eletrcostatic energy stored in it is Uo. It is connected to another uncharged condenser of capacity C2 in parallel. The energy dissipated in the process is [ MPPMT 1994]
0%
a)C2Uo) / (C1 + C2)
0%
b) C1Uo) / (C1 + C2)
0%
c)
0%
d)
Explanation
Loss of energyAnswer: (a)
Q.24
An air capacitor of capacity C=10µF is connected to a constant voltage battery of 12V. Now, the space is filled with a liquid of dielectric constantThe charge that flow now from battery to the capacitor is [ MPPMT 1997]
0%
a) 120 µC
0%
b) 600 µC
0%
c)480 µC
0%
d)24 µC
Explanation
Charge before filling of liquid q1=CvV=10µF×12=120µC Charge after filling dilelectric=KCV=5×10µF×12=600µC∴ Charge flowing from battery=600-120=480µCAnswer: (c)
Q.25
An automobile spring extends 0.2m from 5000N load. The ratio of potential energy stored in this spring when it has been compressed by 0.2m to the potential energy stored in 10µF capacitor at a potential difference of 10,000V is [ B>HU 1996]
0%
a)2
0%
b)1/2
0%
c)1
0%
d)1/4
Explanation
Energy Storred in sprin=½ K x=½ 5000×0.2=500 J Energy stored in capacitor=½ CV2=½ ×10×-6×108=500 JThus ratio is one Answer:(c)
Q.26
A parallel plate capacitor C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is [ IIT 1995]
0%
a) zero
0%
b) 25CV2 / 6
0%
c) 3CV2 / 2
0%
d) 9CV2 / 2
Explanation
Initial charges on capacitors are q1=CV q2=(2C)(2V)=4CV After connecting these condensers togather such that the positive plate of one is connected to negative pg other, the net charge Q=4CV-CV=3V ∴ Energy stores it is U=½ [q2 / (2C+C)]=(3/2) CV2 Answer: (c)
Q.27
Two point charges placed at a cetrain distance r in air exerts a force of F on each other. Then the distance r' at which these charges will experience the same force in a medium of dlelectric constant K is
0%
a)r
0%
b) r/K
0%
c)r/√K
0%
d)r√K
Explanation
Answer: (c)
Q.28
An oil drop carrying a charge of 4 electrons has mass of 3.2×10-17 kg. It is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is ( g=10 ms-2)
0%
a) 2×103 V/m
0%
b) 1×103 V/m
0%
c)3×103 V/m
0%
d)8×103 V/m
Explanation
In first case mg=kv In second case, motion up with velocity v so resulatant force is kv upkv=Eq -kv Eq=2kvEq=2mgE=2mg /q On subtituting values we get E=103 V/m Answer: (b)
Q.29
Charges +q, -4q and +2q are arranged at the corners of an equilaterial triangle of side 0.15m. If q=1µC, their mutual potential energy is
0%
a) 0.4 J
0%
b) 0.5 J
0%
c)0.6 J
0%
d)0.8 J
Explanation
Potential energy of system Answer:(c)
Q.30
Due to a charge inside a cube the electric field is Ex=600x1/2 , Ey=Ez=The charge inside the cube is
0%
a) 600µC
0%
b) 60µC
0%
c) 7µµC
0%
d) 6µµC
Explanation
Since Ey=Ez=0, therefore flux is linked only with face 1 and face 2 According to Gauss's theorem E.A=q/εo q=εo E.A Angle between area vector of face 1 and electric field=0 and angle between area vector of face 2 and electric field is 180° x for face 1=0.2m and distance for face 2=0.1m q=ε[ ExA cos0 + ExA cos180] q=ε[ Ex cos0 + Ex cos180]A q=8.86×10-12 [ 600(0.2)1/2 - 600(0.1)1/2] (0.1)2 q=µµCAnswer: (c)
Q.31
Abullet of mass 2g is moving with a speed of 10 m/s. If the bullet has a charge of 2µC, through what potential must it be accelerated, starting from rest to acquire the same speed?
0%
a)5 kV
0%
b) 5 V
0%
c)50 V
0%
d)50 kV
Explanation
(1/2) mV2=qV (1/2) ×0.2×10-3×100=2×10-6×V V=50kVAnswer: (d)
Q.32
A hollow sphere of copper is having a uniform charge density of 0.5µC/mIts radius is 0.1 m. ThE potential at the centre of the sphere is
0%
a) zero
0%
b) 1800π volts
0%
c)180π volts
0%
d)4.5 kV
Explanation
Potential at the cetre is equal to that on the surfaceV=σr / εoV=0.5××10-6×0.1×4π×9××109 [Here 1 /ε=4π×9××109] V=1800πAnswer: (b)
Q.33
The force between two short electric dipoles separated by a distance r varies as
0%
a) r2
0%
b) r4
0%
c)r-2
0%
d)r-4
Explanation
Answer: (d)
Q.34
Two thin concentric hollow conducting spheres of radii R1 and R2 bears changes Q1 and Q2 respectively. If R1 > R2 then the potentil at point distance r such that : ( R1 > r > R2)
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
From figure point P is exterior for inner sphere of radisu R2 While for outer sphere pint P is interior thus potential at point P dure to outer sphere is equal to potential at its surface since potential is a scalar quantity we will add potentilas due to both the spheres at point PAnswer: (d)
Q.35
Two small spheres carry charge of +3 nC and -12 nC respectively. The charges are distance d apart. The force they exert on one another is FThe spheres are made to touch one another and then separated to distance d apart. The force they exerts on one another is now F2 . Then F1 /F2
0%
a) 1
0%
b) 2
0%
c)1/2
0%
d)16:9
Explanation
Force F1 ∝ (3)(-12) On touching and separating total change on both spheres=-12+3=-9 nCon sepataring charge on one sphere is=9/2 Force F2 ∝ (4)(4) Thus F1 /F2=(3)(-12) / [(9/2)(9/2)]=16:9Answer: (d)
Q.36
A cylinder of radius R and length L, is placed in the region of uniform electric field E as shown in figure. The electric flux linked with cylinder is
0%
a) R/L
0%
b) R2/L
0%
c)zero
0%
d)infinity
Explanation
Curved surface area vector is perpendicular to electric file thus flux through curvered surface is zeroFlux entaring cylinder through circular end is equal to flux leaving cylinder through circular end thus flux throgh circular end is zero. Total flux=0 Answer:(c)
Q.37
Equal and opposite sharges q are placed at point A and B as shown in figure P1 and P2 are equidistant from O. The ratio of P1 . P2 is
0%
a) 1
0%
b) b/a
0%
c) zero
0%
d) infinity
Explanation
potential at equatorial point P1 is zero thus ratio is zero Answer: (c)
Q.38
A is a spherical conductor placed concentrically inside a hollow spherical conductor B, +Q charge is given to A and B is earthed. The electric field intensity is not zero
0%
a)inside A
0%
b) otside B
0%
c)on the surcae of B
0%
d)between A and B
Explanation
Answer: (c)
Q.39
An insulated sphere of radius R has a uniform charge density λ. The electric field at point P inside the sphere and at distance r from the centre is
0%
a) zero
0%
b) Rλ / 3εo
0%
c)rλ / 3εo
0%
d)(2/3) (rλ / εo)
Explanation
Charge enclosed in the shaded sphere=λ(4πr3 /3) By using Gauss's theorem Answer: (c)
Q.40
A ball of mass 600mg and having charge 5 µC is suspended by a thread in between two parallel plates as shown in figure. The distance between the two plates is 10cm and the potential difference between them is 30V. The tention in the string is
0%
a)0.667 m
0%
b) 0.5 m
0%
c)0.4 m
0%
d)0.6 m
Explanation
Tension is either (Eq + mg ) or (mg - Eq) Answer:(d)
Q.41
The work done in carrying a charge of 5µC from point A and B is 8mJ. The difference of potential between A and B is
0%
a) 160 V
0%
b) 16 V
0%
c) 1.6kV
0%
d) 16 kV
Explanation
Answer: (c)
Q.42
the uncharged matallic sphere A suspended as shown in figure is given a push so that it moves towards +ve plate. Which one of the following statement is correct?
0%
a)A touches +ve plate and remain in contact with it
0%
b) A touches +ve plate and then moves towards -ve plate and remains in contact with it
0%
c)A moves to and fro between the two plates with a constant time period
0%
d)A moves to and fro between the two plates with an increasing time-period
Explanation
Answer: (c)
Q.43
the number of electric lines of force crossing the surface of an arbitrary curve of surface area S and enclosing a charge q is
0%
a) εo
0%
b)qεoS
0%
c)q/εo
0%
d)qV/εo
Explanation
Answer: (c)
Q.44
A particle of mass 2gm and charge 1µC is held at rest on a frictionless horizontal surface at a distance of 1 m from the fixed charge of 1 mC. If the particle is released it will be repelled. the speed of the particle when it is at a distance of 10 m from the fixed charge is
0%
a) 100 m/s
0%
b) 90 m/s
0%
c)60 m/s
0%
d)45 m/s
Explanation
Initail potential energy Final potential energy By conservation of energy ΔK=ΔK (1/2)mv2=(9-0.9) (1/2)×2×10-3v2=8.1 v2=8100 v=90 m/s Answer:(b)
Q.45
Charges of 2µC and -3µC are placed at two point A and B distance 1 m as shown in figure. The distance of the point from A, where net potential is zero, is
0%
a) 0.667 m
0%
b) 0.5 m
0%
c) 0.4 m
0%
d) 0.6 m
Explanation
Let the distance of the point of zero potential from A be x. Then Answer: (c)
Q.46
In a certain region of space there exists a uniform electric field of 2×103 k V/m. A rectangular coil of dimensions 10cm ×20cm is placed in XY plane. The electric flux through the coil is
0%
a)zero
0%
b) 4×105
0%
c)40
0%
d)4
Explanation
Pcoil is in XY plane therefore area vector is along z axis Φ=EAcosθΦ=(2×103×200×10-4Φ=40Answer: (c)
Q.47
Electric potential V at some point in space is zero. This means
0%
a) electric field at that point is necessarily zero
0%
b) electric field at that point is necessarily non-zero
0%
c)electric field at that point may or may not be non-zero
0%
d)none of above
Explanation
Answer: (c)
Q.48
A charged conductor B is kept inside another hollow uncharged conductor A. B is in contact with A. What will be the effect of charge on B
0%
a) Total charge on B is transferred to A and comes on the outer surface of A
0%
b) Total charge on B is transferred to A and remains on inner surface of A
0%
c)Opposite charge is induced on the outer surface of A
0%
d)Charge on B is not transferred to A at all.
Explanation
Answer:(b)
Q.49
Charges are placed at the corners of a square of side a as shown in figure. The charge A is in equilibrium. The ratio q1 / q2
0%
a) 1
0%
b) √2
0%
c) 1/√2
0%
d) 2√2
Explanation
Force between charge at A and C is repulsive while force between A and B is attractive also force between A and D is attractive since charge is in equilibrium resultant force on A is zero FAB=FAC + FAD Answer: (d)
Q.50
Mechanical force per unit area of a charged conductor having surface charge density σ is proportional to
0%
a)1 / σ2
0%
b) σ2
0%
c)σ
0%
d)√σ
Explanation
Answer: (b)
0 h : 0 m : 1 s
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