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Q.1
A hollow metal sphere of radius 10cm is charged such that the potential on its surface is 80V. The potential at the centre of the sphere is [ CBSE-PMT 1994]
0%
a) zero
0%
b) 80V
0%
c) 800V
0%
d) 8V
Explanation
Potential at centre of the sphere=potential on the surface=80V Answer: (b)
Q.2
Two parallel metal plates having charges +Q and -Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
0%
a)remain same
0%
b) become zero
0%
c)increases
0%
d)decrease
Explanation
Electric field E=σ / kε0 Since for kerosene oil k >1 electric field will be decreasedAnswer: (d)
Q.3
A capacitor C1 is charged to a potential difference V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor CThe potential difference across the combination is [ CBSE-PMT 2002]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Charge on capacitor Q=C1V After connecting capacitor total charge will remain same Capacitors are connected in parallel so C=C1 + C2 Thus potential difference=Q/C=C1V / (C1+ C2) Answer: (a)
Q.4
A parallel plate condenser has a uniform electric field E(V/m) in the space between the plates. if the distance between the plates is d(m) and area of each plate is A(m2) the energy (joules) stored in the condenser is. [ CBSE-PMT 2011]
0%
a)E2Ad / ε0
0%
b) ½ ( ε0 E2)
0%
c)ε0EAd
0%
d)½ ( ε0E2Ad)
Explanation
U=½ CV2C=(Aε0 /d )V=EdSubstituting values of C and V in equation of potential we get Answer:(d)
Q.5
Three capacitors each of capacity 4µF are to be connected in such a way that the effective capacitance is 6µF. This can be done by [ CBSE-PMT 2003]
0%
a) Connecting two in parallel and one in series
0%
b) connecting all of them in series
0%
c) connecting all of them in parallel
0%
d) connecting two of them in series and one in parallel
Explanation
Connect two in series thus C'=2µF Now connect this C' in parallel with third So C=2+4=6µF Answer: (d)
Q.6
A charge q is placed at the centre of the line joining two equal charges Q each. The system of three charges will be in equilibrium if q is equal to [ MPPMT 1999]
0%
a)-Q/2
0%
b) +Q/2
0%
c)-Q/4
0%
d)-4Q
Explanation
The system is in equilibrium if force on each charge is zeroIt is clear fro figure, the force on central charge is always zero, what so ever be the magnitude of this charge.Thus, for equilibrium FQQ + FQq=0Answer: (c)
Q.7
The electric potential at the surface of an atomic nucleus ( Z=50) of radius 9.0×10-15m..[ JIPMER 1998]
0%
a) 80 V
0%
b) 8×106 V
0%
c)9 V
0%
d)9×105 V
Explanation
Answer: (b)
Q.8
Two metal plates having a potential difference of 800V are 2cm apart. It is found that a particle of mass 1.96×10-15 kg remain suspended in the region between the plates. The charge on the particle must be ( e=elementary charge) : [ PET 1999]
0%
a) 3e
0%
b) 4e
0%
c)6e
0%
d)8e
Explanation
V=800, d=2×10-2 mE=V/d=800 / (2×10-2 )=4×104 V/mFor equilibrium Eq=mg q=mg / E Answer:(a)
Q.9
Total electric flux coming out of a unit positive charge put in air is [ MPPET 1995]
0%
a) εo
0%
b) εo-1
0%
c) (4πεo)-1
0%
d) 4πεo
Explanation
Answer: (b)
Q.10
Electric potential is given by V=6x - 8xy2-8y+6yz-4z2 then electric force acting on 2C point charge placed on the origin will be [ Raj.PET 1997]
0%
a)2N
0%
b) 6N
0%
c)8N
0%
d)20N
Explanation
We know that partial derivative of V gives electric field Eat origin x=y=z=0 Thus component of electric filed along X axis=6icomponent of electric filed along Yaxis=-8jcomponent of electric filed along Z axis=0E=6i -8j |E|=10N/CF=Eq=10×2=20 NAnswer: (d)
Q.11
The ratio of momentum of an electron and an alpha particle which are accelerated from rest by a potential difference of 100V is [ MNR 1994]
0%
a) 1
0%
b)
0%
c)
0%
d)
Explanation
energy of charge=Vqand momentum in terms of energy E is P=√(2Em) For electron P=√( 2×e×100×me For alpha particle P'=√( 2×2e×100×mα Thus P/P'=√(me /2mα Answer: (d)
Q.12
A region surrounding a stationary electric dipole has [ MPPMT 1994]
0%
a) Magnetic field only
0%
b) Electric field only
0%
c)Both electric and magnetic fields
0%
d)No electric and magnetic fields
Explanation
Answer:(b)
Q.13
An electron is moving around infinite linear positive charge in the orbit of 0.1m. If the linear change density is 1µC/m, the velocity of electron will be [ RajPET 1996]
0%
a) 0.562×107 m/sec
0%
b) 5.62×107 m/sec
0%
c) 5.2×105 m/sec
0%
d) 0.0562×107 m/sec
Explanation
Centripetal force=Electrical force Answer: (b)
Q.14
The dielectric constant K of an insulator can not be [ JIPMER 1998]
0%
a)4
0%
b) 6
0%
c)∞
0%
d)9
Explanation
Answer: (c)
Q.15
A solid metallic sphere has a charge +3Q concentric with this sphere is a conducting spherical shell having charge -Q. The radius of a sphere is 'a' and that of the spherical shell is 'b'. (b > a). What is the electric field at a distance R ( a < R< b) from the centre? [ PMT 1995]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Given point is inside the hollow shell and out side the solid sphere thus electric field will be only due to sphereAnswer: (c)
Q.16
A flat circular disc has a charge +Q uniformly distributed on the disc. A charge +Q is thrown with kinetic energy E, towards the disc along its normal axis. The charge Q will [ PMT 1995]
0%
a) hit the disc at the centre
0%
b) return back along its path after touching the disc
0%
c)return back along its path, without touching the disc
0%
d)any of the above three situations is possible, depending on the magnitude of E
Explanation
Answer:(d)
Q.17
An electric dipole is placed along x-axis at the origin O. A point P is situated at a distance 20 cm from this origin such that OP makes an angle π/3 with the X-axis. If the electric filed at P makes an angle θ with the X-axis the value of θ would be [ MPPMT 1997]
0%
a) π/3
0%
b) π/3 + tan-1(√3 / 2)
0%
c) 2π/3
0%
d) tan-1(√3 / 2)
Explanation
θ=π/3 +α Now Answer: (c)
Q.18
A charge q is placed at the centre of a cube of side a. The electric flux passing through any one face of the cube is [ CPMT 1993]
0%
a)q/ εo
0%
b) q/ (3εo)
0%
c)q/ (6εo)
0%
d)6q/εo
Explanation
Answer: (c)
Q.19
Three charges +Q each are placed at the corners A,B and C of an equilateral triangle. At the circum centre O, the electric fieldstrenght is ; ( r be the distance between corners and circumcentre) [ CPMT 1998] k=1 /4πεo
0%
a) zero
0%
b) 3kQ/ r2
0%
c)kQ/ r2
0%
d)kQ2/ r2
Explanation
Due to symmetric distribution of charge electric siled at circum centre O is zeroAnswer: (a)
Q.20
Two metallic spheres of radii 1cm and 2cm are given charges 10-2C and 5×10-2C respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is : [ CBSE 1995]
0%
a) 3×10-2C
0%
b)1×10-2C
0%
c)4×10-2C
0%
d)2×10-2C
Explanation
Electrons will flow from higher potential to lower potential spheres till potential of both the spheres becomes equal Answer:(d)
Q.21
There are 27 drops of a conducting fluid. Each has a radius a and they are charged to potential Vo. These are combined to form a bigger drop. Its potential will be [ MPPMT 1994]
0%
a) Vo
0%
b) 3Vo
0%
c) 9Vo
0%
d) 27Vo
Explanation
Volume of big drop=27×( volume of small drop) Thus radius of big drop R=3r potential of small drop Vo=kq/r Potential of big drop V'=k(27q) / R V'=k(27q) / 3r V'=9 (kq/r)=9o Answer: (c)
Q.22
Two identical thin rings, each of radius R m are co-axially placed at distance R metre apart. If Q1 and Q2 C are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is [ MPMT 1999]
0%
a)Zero
0%
b)
0%
c)
0%
d)
Explanation
Potential at O1=V1 Potential at O2=V2 Work W=( V1 - V2)q Answer: (b)
Q.23
A sphere of radius R has a uniform distribution of electric charge in its volume. At a distance x from its centre, for x
0%
a) 1/x2
0%
b) 1/x
0%
c)x
0%
d)x2
Explanation
Answer: (c)
Q.24
Charges of +(10/3)×10-9 C are placed at each of the four corners of a square of side 8cm. The potential at the intersection of diagonals is [ BIT 1993]
0%
a) 150√2 V
0%
b) 1500√2 V
0%
c)900√2 V
0%
d)900 V
Explanation
Point of intersection of diagonal is at 4√2 All the charges are same thus total potential=4× Potential due to single charge Answer:(b)
Q.25
Equal charges q are placed at the four corners A,B,C and D of a square of length a. The magnitude of the force on the charge at D will be [ MPPMT 1994]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Force on D due to resultant of force F1, F2 F3 Force F2 and F3 are equal in magnitude but perpendicular to each other Thus Direction of F' is along F1 thus there magnitude will ge added then Answer: (c)
Q.26
Electric charges q, q and -2q are placed at the corners of an equilateral triangle ABD of side l. The magnitude of electric dipole moment of the system is : [ MPPMT 1994]
0%
a)ql
0%
b) 2ql
0%
c)√3 (ql)
0%
d)4ql
Explanation
from figure |P1|=|P2|=q×l Thus Answer: (c)
Q.27
An electric dipole consists of two opposite charges each of magnitude 1.0µC separated by distance of 2.0cm. The dipole is placed in an external field of 1.0×105N/C. The maximum torque on the dipole is [ CPMT 1990]
0%
a) 0.2×10-3 N-m
0%
b) 1.0×10-3 N-m
0%
c)2.0×10-3 N-m
0%
d)4.0×10-3 N-m
Explanation
τmax=pE=10-6 ×2×10-2×1.0×105=2×10-3 N-mAnswer: (c)
Q.28
A charged conductor B is kept inside another hollow conductor A. B is in contact with A. What will be effect of charge on B
0%
a) Total charge on B is transferred to A and comes out on the outer surface of A
0%
b) Total charge on B is transferred to A and remains on the 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 t all
Explanation
Answer:(b)
Q.29
The following figure shows two parallel equipotential surfaces A and B kept at a small distance r from each other. A point charge of -q coul is taken from the surface A to B. The amount of net work W done will be given by : [ CPMT 1988] k=1/ 4πεo
0%
a) -kq/r
0%
b) kq/r2
0%
c) -kq/r2
0%
d) zero
Explanation
W=ΔV since both the plate s are at same potential ΔV=0, W=0 Answer: (d)
Q.30
When a charge of 3 C is placed in a uniform electric field, it experiences a force of 3000N, with this field, potential difference between two pints separated by distance of 1 cm is [ MP 1986]
0%
a)10 V
0%
b) 90 V
0%
c)1000 V
0%
d)3000 V
Explanation
E=F/q=3000/3=1000 V/mV=E.d V=1000×0.01=10VAnswer: (a)
Q.31
Three charges 2Q, -q- and -q are located at the vertices of an equilateral triangle. At the circumcentre of the triangle [ MPPMT 1986]
0%
a) the field is zero but potential is not zero
0%
b) the field is non zero but potential is zero
0%
c)both field and potential are zero
0%
d)both, field and potential are non zero
Explanation
Potential is algebraic sum thus becomes zero, but position of charges are such that electric field will not be zeroAnswer: (b)
Q.32
There is a solid sphere of radius R of metal having uniform distributed charge. What is the relation between electric field 'E' and distance 'r' from the centre ( r < R) [ BHU 1998]
0%
a)E ∝ r-2
0%
b) E ∝ r-1
0%
c)E ∝ r
0%
d)E ∝ r2
Explanation
Answer:(c)
Q.33
A sphere of 4cm radius is suspended within a hollow sphere of 6 cm radius. the inner sphere is charged to a potential of 3 e.s.u. When the outer sphere is earthed, the charge on the inner sphere is [ JIPMER 1998]
0%
a) 54 esu
0%
b) 1/4 e.s.u
0%
c) 30 e.s.u
0%
d) 36 e.s.u
Explanation
Let the charge on the inner sphere br +q. Then the charge on the inner surface of outer sphere will be -q and on the outer surface of outer sphere there is no charge. Therefore the potential of the inner sphere 3=(q/4) - (q/6) esu q=36 e.s.u Answer: (d)
Q.34
There is an electric field E in x-direction. If the work done on moving charge 0.2C through a distance of 2 meters along a line making an angle 60° with x-axis is 4.0, what is the value of E? [ CBSE 1995]
0%
a)√3 N/C
0%
b) 4 N/C
0%
c)5 N/C
0%
d)none of these
Explanation
W=Fd cosΘ4 EqDcos604=E ×0.2×2×(1/2) E=4/02=20 N/CHence Answer is none of theseAnswer: (d)
Q.35
Two equal charges q are placed at a distance of 2a, and a third charge -2q is placed at the mid-point. The potential energy of the system is [ MPPMT 1997]
0%
a) q2 / 8πεoa
0%
b) 6q2 / 8πεoa
0%
c)- 7q2 / 8πεoa
0%
d)9q2 / 8πεoa
Explanation
Answer: (c)
Q.36
A soap bubble is given a negative charge then its radius{MNR 198]
0%
a) decreases
0%
b) increases
0%
c)remains unchanged
0%
d)nothing can be predicted as information is insufficient
Explanation
Answer:(b)
Q.37
A hollow charged metal sphere has radius r. If the potential difference between its surface and a point at distance 3r from the centre is V, then the electric field intensity at a distance 3r from the centre is [ MPPMT 1985]
0%
a) V/6r
0%
b) V/4r
0%
c) V/3r
0%
d) V/2r
Explanation
Answer: (a)
Q.38
Four charges are placed on corners of a square as shown in figure having sides 5cm. If q is one micro coulomb then the electric field intensity at centre will be [ Raj. PET 1997]
0%
a)1.02 ×107 N/C upward
0%
b) 2.04 ×107 N/C downwards
0%
c)2.04 ×107 N/C upward
0%
d)1.02 ×107 N/C downwards
Explanation
Distance of all charges from the centre is same for all charges let it be r Now Charge at Diagonal A is q and at C is 2q thus electric fields will oppose. resultant field at centre will be in the direction from C to A and magnitude will be Charges at diagonal B is -2q and at D is -q thus electric field at cenre will be from D to B Angle between E' and E" is 90° Thus resultant From the geometry of figure r=5/√2 cmDirection upwardsAnswer: (a)
Q.39
In an electric gun, the electrons are accelerated through a p.D. of V volts. Taking electronic charge and mass to be e and m respectively, the maximum velocity attained by them is [ BHU 1995]
0%
a) 2eV/m
0%
b) √(2eV/m)
0%
c)(2m/eV)
0%
d)(V2/2em)
Explanation
P.E=Kinetic energy Ve=½ mv2 v2=2Ve/m v=√(2Ve/m)Answer: (b)
Q.40
A small sphere of radius r is kept inside a hollow sphere of radius R ( R < r) concentrically. The large sphere are charged by charges Q and q respectively. Both the spheres are separated from each other. The potential difference depend on [ Raj. PET 1996]
0%
a) only on charge q
0%
b) only on charge Q
0%
c)both on charge q and Q
0%
d)does not depends upon any of them
Explanation
Potential at smaller sphere will be always more than the larger sphere thus option 'a' is correct Answer:(a)
Q.41
Two point charges 100µC and 5µC are placed at point A and B respectively, with AB=40cm. The work done by the external force in displacing the charge 5µC from B to C where BC=30cm, angle ABC=π/2 and (1/4πεo )=9×109 [ MPPMT 1997]
0%
a) 9 J
0%
b) (81/20)J
0%
c) (9/25)J
0%
d) (-9/4)J
Explanation
Charge 5µC is moved in electric field of 100µC change Potential at B=VB Potential at C=VB WBC=(VC - VB)q WBC=(-2.5×10+ +1.8×106)(5×10-6=-9/4 JAnswer: (d)
Q.42
For a given surface, the Gauss's law is stated as From this, we can conclude that [ PMT 1995]
0%
a)E is necessarily zero on the surface
0%
b) E is perpendicular to the surface at every point
0%
c)the total flux through the surface is zero
0%
d)the flux is only going out of the surface
Explanation
Answer: (c)
Q.43
A charged water drop whose radius is 0.1µm is in equilibrium in an electric field. If charges on it is equal to charges of an electron, then intensity of electric field will be 9 g=10 m/s2)
0%
a) 1.61 N/C
0%
b) 26.2 N/C
0%
c)262 N/C
0%
d)1610 N/C
Explanation
Electric force=Gravitational force E×e=volume ×density ×gAnswer: (c)
Q.44
A charge of 5C is given a displacement of 0.5m and work done in the process is 10J. The difference of potential between the two points is [ MPPET 1990]
0%
a) 2 V
0%
b) 1 V
0%
c)0.25 V
0%
d)4 V
Explanation
W=Vq 10=V(5) V=2V Answer:(a)
Q.45
A particle A has charge +q and the particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through the same electrical potential difference, the ratio of their speed VA/ VB will become [ MNR 1991]
0%
a) 2:1
0%
b) 1:2
0%
c) 1:4
0%
d) 4:1
Explanation
Potential energy of both the charges is different given by qV and 4qV which is equal to kinetic energy Answer: (c)
Q.46
A hollow metal sphere of radius 5cm is charged such that potential on its surface is 10V. The potential at a distance of 2cm from the centre of the sphere is [ MPPMT 1991]
0%
a)zero
0%
b) 10V
0%
c)4V
0%
d)10/3 V
Explanation
Potential inside the surface=Potential on the surfaceAnswer: (b)
Q.47
Two point charges 20 mC and 80 mC are separated by a distance 10 cm. The point at which the intensity of electric filed will be zero along the line joining two charges will be [ Raj. PET 1996]
0%
a) -0.1 m
0%
b) -0.04 m
0%
c)0.003 m
0%
d)0.033 m
Explanation
Since the charges are of same sign, the point P where intensity is zero, lies on the line joining the two charges, in between the charges. Let x be the distance of point P from 20 mC charge Thus Answer: (d)
Q.48
With a rise in temperature, the dielectric constant K of a liquid.. [ MPPET 1999]
0%
a) increases
0%
b) decreases
0%
c)remains unchanged
0%
d)changes erratically
Explanation
Answer:(b)
Q.49
The electric potential V as a function of distance x ( in metre) is given by V=(5x2 + 10x - 9) volt. The value of electric field at x=1m would be [ MPPMT 1999]
0%
d) -20 volt/m
0%
a) 20 volt/m
0%
b) 6 volt/m
0%
c) 11 volt/m
Explanation
By taking partial derivative of potential with respect to x, we get E=-[ 10x+10] for x=1 E=- 20 volt/m Answer: (d)
Q.50
At a certain distance from a point charge, the electric field is 500V/m and the potential is 3000V. What is the distance? [ PMT 1995]
0%
a)6 m
0%
b) 12 m
0%
c)36 m
0%
d)144 m
Explanation
E=V/dAnswer: (a)
0 h : 0 m : 1 s
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