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Quiz 1
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Q.1
At a certain distance from a point charge the electric field is 500 V/m and the potential is 3000 V. What is this distance
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a) 6 m
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b) 12 m
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c)36 m
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d)144 m
Explanation
Answer:(a)
Q.2
A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x=0 and positive plate is x=3d. The slab is equidistant from the plates. The capacitor is given some charge. As x goes from 0 to 3d
0%
a) the magnitude of the electric field remains the same
0%
b) the direction of the electric field remains the same
0%
c)the electric potential remains constant
0%
d)the electrical potential increases at first, then decreases and again increases
Explanation
Answer: (b)
Q.3
The diameter of the plate of parallel plate condenser is 6cm. If its capacity is equal to a sphere of diameter 200cm, the separation between the plates of the condenser is
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a) 4.5 ×10-4 m
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b) 2.25 ×10-4 m
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c) 6.75 ×10-4 m
0%
d) 9 × 10-4 m
Explanation
Answer: (b)
Q.4
The study of the effect associated with electric field at rest is known as [AFMC 1999]
0%
a) electrostatics
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b) electromagnetism
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c) magneto static
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d) none of these
Explanation
Since electric field is at rest therefore .The study of the effect associated with electric field at rest is known as electrostaticsAnswer (a)
Q.5
When we touch the terminal of a high voltage capacitor, even after a high voltage has been cut off. Then the capacitor has the tendency to [AFMC 2000]
0%
a) affect dangerously
0%
b) discharge energy
0%
c) restore energy
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d) both (a) and (b)
Explanation
Function of capacitor is to store energy in the form of electric field and charge in accumulated in it when connected to high voltage source.When such a charged capacitor is conned to resistors or any conductor charge flows through it.Our body is a conductor of electric charge. When we touch capacitor, it tries to send out charge through our body to ground. Which can give fatal shock to the body if discharge current is high.Answer: (d)
Q.6
In a region of space having a uniform electric field E, a hemi-spherical bowl of radius r is placed. The electric flux φ through the bowl is
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a) πr2E
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b) 2πr2E
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c)2πrE
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d)4πr2E
Explanation
φ=E.A=Eπr2Answer: (a)
Q.7
A short electric dipole has a dipole moment of 16 × 10-9 C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is : [NEET 2020]
0%
a) 400 V
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b) zero
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c) 50 V
0%
d) 200 V
Explanation
Answer is (d)
Q.8
Four capacitors each of 25 μF are connected as shown in diagram. The DC voltmeter reads 200 volt. The charge on each plate of capacitor will be [AFMC1997]
0%
a) 5×10-2C
0%
b)2× 10-2C
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c)5× 10-3 C
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d) 2× 10-3
Explanation
From the figure two capacitors are connected in parallel and such two combinations are connected in series.Now Voltmeter is connected parallel to one of the parallel combination of capacitor. Hence Potential difference across each capacitor in a parallel combination will be same i.e 200 V.Now C=Q/V Therefore Q=CV Q=25×10-6 x 200 here Q=25×10-6 and V=200 V givenQ=5 × 10-3 C. Since Another combination is similar to voltmeter combination and all the capacitors equal. Charge across each capacitor will be same. Answer is (c)
Q.9
Three capacitors of 2.0, 3.and 6.0 μF connected in series to a 10 V source. The charge on 3.0 μ F capacitor is [AFMC1998]
0%
a) 15μC
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b) 12μC
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c) 5 μ C
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d) 10 μ C
Explanation
All the capacitors are connected in series hence charge on each capacitor is same Equivalent capacitor On substituting the value and taking reciprocal we get C=1 μFNow Q=CVTherefore Q=1 μF × 10VQ=10 μ C Answer: (d)
Q.10
There is an electric field E in X-direction. If the work done in moving a charge 0.2 C through a distance of 2m along a line making an angle 60o with x-axis is 4J, the value of E is [AFMC1998]
0%
a) 2√3 N/C
0%
b) 5 N/C
0%
c) 4 N/C
0%
d) None of these
Explanation
Electric field is along X-axis and charge is moved along the direction makings angle of 60o with positive X-axis there fore Electric filed along given direction will be Ecos60Work done in the given direction W=Force × displacement Work done in the given direction W=(qEcos60 )× d 4=[(0.2)E (1/2)]×(2)On simplification we get E=20 N/CAnswer : 20N/C (d)
Q.11
The equivalent capacitance of the combination of three capacitors, each of capacitance C between A and B as shown in the given figure, is
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a) C/2
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b) 3C/2
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c) C/3
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d) 2C
Explanation
Capacitor C3 is connected across, hence shorted, no current will flow through it. So it will not work.Now remaining capacitors are parallel to each other Resultant capacitance CR=C1 + C2 CR=C + C CR=2CAnswer : (d)
Q.12
The electric field intensity on the surface of a charged conductor is
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a) zero
0%
b) directed normally to the surface
0%
c)directed tangetially to the surface
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d)directed along 45° to the surface
Explanation
Answer:(b)
Q.13
The capacitors of capacitance 4 μF and 6 μF are connected in series. A potential difference of500 volt is applied to the outer plates of two capacitor system. Then the charge on each plate of each capacitor is numerically [AFMC 2000]
0%
a) 6000 μC
0%
b) 1200 μC
0%
c) l2000 μC
0%
d) 6000 μC
Explanation
Both the capacitors are in series and connected across the source of 500V. Hence charge on the each capacitor will be same.Effective capacitance On substituting the values of both the capacitors , after solving for C, we get c=2.4μFNow Q=CV Q=2.4μF ×500V Q=1200μC Answer:(b)
Q.14
The concentric spheres of radii R and r have similar charges with equal surface densities (σ). What is the electric potential at their common center? [AFMC 2001]
0%
a)
0%
b)
0%
c) Rσ /ε0
0%
d) σ /ε0
Explanation
Formula for electric potential at the center of Hollow charged sphere is V=kQ/R : Here k=1/4πε0 Q :charge and R radiusSpheres are concentric hence Potential will get added algebraically.Therefore potential at centre of spheres is V=V1 +V2 --- (1)Here V1 Potential at center due to :RV2 Potential at center due to :r here Q=Area of surface × σ Surface charge density Q=4πR2σ V1=σR/ε0 similarly V2=σr/ε0Substituting values of V1 and V2 in equation 1 we get HenceAnswer : (b)
Q.15
A charge Q µC is placed at the center of a cube the flux coming out from any surface will be [AFMC 2001]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Flux passing through the closed surface according to Gauss law Surface area of Cube=6× ( Area of one surface) Now flux are spread symmetrically , flux passing through any one face is one sixth of the total flux through closed surfaceTherefore flux passing through any one surfaceAnswer : (c)
Q.16
A capacitor C1 is charged by a potential difference Vo as shown in the figure. This charging battery is then removed and the capacitor is connected to an uncharged capacitor CWhat is the final potential difference V across the combination? [AFMC 2001]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
When C1 is completely charged Charge on C1 is Q=C1V0 By connecting charge capacitor C1 with C2 charge will get distributed according to value of capacitor let it be q1and q2Potential difference across each capacitor will be same let it be VNow V=q1/C1=q2/C2 Therefore q1/q2=C1/C2 OR Total charge of the combination Q=C1V0 Equivalent capacitance C=C1+C2 as they are parallel Common voltage V=Q/C Answer : (c)
Q.17
The electric field required to keep a water drop of mass m just to remain suspended, when charged with one electron is [AFMC 2001]
0%
a) em/g
0%
b) mg/e
0%
c) emg
0%
d) mg
Explanation
Let electric filed be E Water drop is subjected to upward electric force FE=eE Down word gravitational force FG=mg To keep water drop suspended resultant force must be zero FE=FG On substituting values of FE and FG eE=mg E=mg/eAnswer : (b)
Q.18
The formation of dipole is due to two equal and dissimilar point charges placed at a [ CBSE-PMT 1996]
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a)short distance
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b) long distance
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c)above each other
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d)none of these
Explanation
Answer: (a)
Q.19
A charge Q is placed at the corner of a cube. The electric flux through all the six faces of the cube is .. [ CBSE-PMT 2000]
0%
a) Q/ 3 ε0
0%
b) Q/ 6 ε0
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c)Q/ 8 ε0
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d)Q/ ε0
Explanation
Many students calculate option "c", and calculation is on the basis of charge at corner is shared by 8 boxes Gauss Law states about charge enclosed, if charge is on the surface, can not be considered as enclosed. Now charge is not like atom to be shared, charge is a point . so either it will be enclosed or on the surface of cube Possible answers are Option "d" if enclosed or zero if it is on the surface According to Gauss's theorem, electric flux through a close surface=Q/ε0Where Q is charge enclosedAnswer: (d)
Q.20
An electric dipole of moment P is lying along a uniform electric field E. The work done in rotating the dipole by 90° is .. [ CBSE-PMT 2006]
0%
a) PE/2
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b) 2PE
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c)PE
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d)PE√2
Explanation
Work done in rotating a dipole=PE( 1-cosθ)If θ=90°, work done=PE(1-0)=PE Answer:(c)
Q.21
The mean free path of electrons in metal is 4×10-8 m. The electric field which can give on an average 2eV energy to an electron in the metal will be in units of V/m [ CBSE-PMT 2009]
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a) 5×10-11
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b) 8×10-11
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c) 5×107
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d) 8×107
Explanation
E=V/d E=2/ (4×10-8) E=5×107 Vm-1Answer: (c)
Q.22
From a point charge, there is a fixed point A. At A, there is an electric field of 500 V/m and potential difference of 3000V. Distance between point charge and A will be ... [ CBSE-PMT 1997]
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a)6m
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b) 12m
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c)16m
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d)24m
Explanation
from formula E=V/d d=V/Ed=3000/500=6mAnswer: (a)
Q.23
If a dipole moment P is placed in uniform electric field E, then torque acting on it is given by [ CBSE-PMT 2001]
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a) τ=P ⋅E
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b) τ=P ×E
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c)τ=P + E
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d)τ=P - E
Explanation
Answer: (b)
Q.24
A charge Q µC is placed at the centre of a cube, the flux coming out from any surface will be
0%
a)(Q/6ε0) ×10-6
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b) (Q/6ε0) ×10-3
0%
c)(Q/24ε0)
0%
d)(Q/24ε0)
Explanation
Total flux out of all six faces as per Gauss's theorem should be (Q ×10-6) / ε Therefore, flux coming out of each face=(1/6ε) (Q ×10-6) Answer:(a)
Q.25
The electric intensity due to a dipole of length 10 cm and having a charge of 500µC, at a point on the axis at a distance 20cm from one of the charge in air, is [ CBSE-PMT 2001]
0%
a) 6.25×107 N/C
0%
b) 9.28×107N/C
0%
c) 13.1×1011 N/C
0%
d) 20.5×107 N/C
Explanation
Asked to find electric field at a axial pointGiven length of dipole 2a=10cm ∴ a=5cm=5×10-2 Dipole moment P=q(2a)=500×10-6 ×0.1=5×10-5 distance of point from mid-point of dipole, r= 5cm +20 cm = 25cm Electric field intensity due to dipole at equatorial point is Answer: (a)
Q.26
Three point charges +q, -q and +q are placed at point (x=0,y=a,z=0), (x=0, y=0,z=0) and (x=a, y=0, z=0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are .. [ CBSE-PMT 2007]
0%
a)qa√2 along the line joining points (x=0, y=0,z=0) and (x=a, y=a, z=0)
0%
b) qa along the line joining points ( x=0, y=0,z=0) and (x=a, y=a, z=0)
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c)qa√2 along +ve x direction
0%
d)qa√2 along +ve y direction
Explanation
Now direction of both the dipoles are along the postive axis and are mutually perpendicularDipole moment of each dipole=qa ( a is distance between two charges)Net resultant dipole moment=[(qa)2 + (qa)2]1/2=qa√2And direction will be 45° along positive x-axis.. Now line joining points (x=0, y=0,z=0) and (x=a, y=a, z=0) will make an angle of 45° with x-axis Answer: (a)
Q.27
The mean free path of electrons in a metal is 4×10-8 m. The electric field which can give on an average 2eV energy to an electron in metal will be in units of V/m [ CBSE-PMT 2009]
0%
a) 5×10-11
0%
b) 8×10-11
0%
c)5×107
0%
d)8×107
Explanation
Energy =2eV Hnece potential fifference is 2V E=V/d E=2 / (4×10-8)W=5×107 Vm-1Answer: (c)
Q.28
A piece of cloud has area 25×106 m2 and electric potential of 105 vots. If the height of cloud is 0.75 km, then energy density of electric field between earth and cloud will be [ Ra.PET 1997]
0%
a) 250 J
0%
b) 750 J
0%
c)1225 J
0%
d)1475 J
Explanation
Formula for energy density Answer:(d)
Q.29
A hollow cylinder has a charge q coulomb within it. If Φ is the electric flux in units of voltmeter associated with the curved surface B, the flux linked with the plane surface A in units of voltmeter will be.. [ CBSE-PMT 2007]
0%
a) q /2ε0
0%
b) Φ / 3
0%
c)
0%
d)
Explanation
Let ΦA be the flux associated with surface ALet ΦB be the flux associated with surface BLet ΦC be the flux associated with surface CLet Let ΦA=Let ΦA=Φ'Now total flux Φt=ΦA + ΦB + ΦC Therefore q/ε0=2Φ' + ΦB Answer:(d)
Q.30
An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force F between the two is ... [ CBSE-PMT 2003]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Charge on electron is -e and charge on proton is +e Force vector form is Answer:(c)
Q.31
A hollow insulated conduction sphere is given a positive charge 10µC. What will be the electric field at the centre of the sphere if its radius is 2metres? [ CBSE-PMT 1998]
0%
a)zero
0%
b) 5µCm-2
0%
c)20µCm-2
0%
d)8µCm-2
Explanation
Answer: (a)
Q.32
A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is .. [ CBSE-PMT 2008]
0%
a) E along KO
0%
b) E along OK
0%
c)3E along KO
0%
d)3E along OK
Explanation
ABy symmetry of figure , the electric field at O due to AKB is towards KO while by ACDB is along OKAnswer: (b)
Q.33
There is an electric field E in x-direction. If the work done on moving a charge 0.2C through a distance of 2m along a line making an angle 60° with x-axis is 4J, then what is the value of E? [ CBSE-PMT 1995]
0%
a) 3 N/C
0%
b) 4 N/C
0%
c)5 N/C
0%
d)20 N/C
Explanation
Work done in moving the charge is W=Fdcosθ W=Eqdcosθ Now W=4Jq=0.2Cd=2mθ=60°On substituting the value in equation we getE=20 N/C Answer:(d)
Q.34
Intensity of an electric field E depends on distance r, due to a dipole, is related as .. [ CBSE-PMT 1996]
0%
a) E ∝ (1/r)
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b) E ∝ (1/r2)
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c) E ∝ (1/r3)
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d) E ∝ (1/r4)
Explanation
Answer: (c)
Q.35
An electric dipole , consisting of two opposite charges of 2×10-6C each separated by a distance 3cm is placed in an electric field of 2×105 N/C. Torque acting on the dipole is .. [ CBSE-PMT 1995]
0%
a)12×10-1N-m
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b) 12×10-2 N-m
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c)12×10-3 N-m
0%
d)12×10-4 N-m
Explanation
torque τ=qEdBy substituting values τ=2×10-6×2×105×3×10-2 τ=12×10-3 N-mAnswer: (c)
Q.36
A charge q is placed at the centre of the line joining two exactly equal positive charges Q. The system of three charges will be in equilibrium, if q is equal to .. [ CBSE PMT 1995]
0%
a) -Q/4
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b) +Q
0%
c)Q
0%
d)Q/2
Explanation
At equilibrium, net force is zeroAnswer: (a)
Q.37
A semi-circular arc of radius 'a' is charged uniformly and the charge per unit length is λ. The eletrcic field at the centre of the arc is [ CBSE-PMT 2000]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Electric field at O due to element dx is Horizontal electric field.ie. perpendicular to AO, will be cancelled.Hence electric field=addition of all electric field along the direction AO=∑dEsinθ From the formula for length of the arc x= a × θ Answer:(a)
Q.38
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will [ CBSE-PMT 2011]
0%
a) increase four times
0%
b) be reduced to half
0%
c) remain the same
0%
d) be doubled
Explanation
Since charge enclosed by the Surface is same electric flux will remain same Answer: (c)
Q.39
The electric potential V at any point (x,y,z) all in meters in space is given by V=4x2 volt. The eletric field at the point (1, 0, 2) in volt/meter is .. [ CBSE-PMT 2011]
0%
a)8 along positive x-axis
0%
b) 16 along negative x-axis
0%
c)16 along positive x-axix
0%
d)8 along negative x-axis
Explanation
∴ E(1,0,2)=-8i V/mAnswer: (d)
Q.40
The electric field at a distance 3R/2 from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is .. [ CBSE PMT 2010]
0%
a) E/2
0%
b) Zero
0%
c)E
0%
d)E/2
Explanation
Electric field inside the charged conducting spherical shell is zeroAnswer: (b)
Q.41
A square surface of side L meter in the plane of the paper is placed in a uniform electric field E volt/m acting along the same plane at an angle θ with the horizontal side of the square as shown in Figure. The electric flux linked to the surface in units of volt m is .. [ CBSE-PMT 2010]
0%
a) EL2
0%
b) EL2cosθ
0%
c)EL2sinθ
0%
d)zero
Explanation
Electric flux Φ=EA cosθ, where θ angle between E and normal to the surface.=π/2Φ=0 Answer:(d)
Q.42
Two positive ions,each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be ( e charge on electron) [CBSE-PMT 2010]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let n be the number of electrons missing Answer: (c)
Q.43
Point charges +4q, -q and +4q are kept on the x-axis at points x=0, x=a and x=2a respectively.. [ CBSE-PMT 1988]
0%
a)none of the charges is in equilibrium
0%
b) only -q is in stable equilibrium
0%
c)all the charges are in unstable equilibrium
0%
d)all the charges are in stable equilibrium
Explanation
Net force on each other of the charge due to the other charge is zero. However, disturbance in any direction other than along the line on which the charges lie, will not make the charges return Answer: (c)
Q.44
Point Q lies on the perpendicular bisector of an electric dipole of dipole moment P. if the distance of Q from the dipole is r( much larger than the size of the dipole), then the electric field at Q is proportional to .. [ CBSE-PMT 1998]
0%
a) P-1 and r-2
0%
b) P and r-2
0%
c)P2 and r-3
0%
d)P and r-3
Explanation
Answer: (d)
Q.45
When air is replaced by a dielectric medium of force constant K, the maximum force of attraction between two charges, separated by a distance.. [ CBSE-PMT 1999]
0%
a) decreases K times
0%
b) increases K times
0%
c)remains unchanged
0%
d)becomes 1/ K2 times
Explanation
Force when dielectric medium of constant K is between two changes is given by equation Answer:(a)
Q.46
Three charges , each +q, are placed t the corners of an isosceles triangle ABC of sides BC and AC of length 2a. . D nd E are the mid points of BC and CA. The work done in taking a charge Q from D to E is [ CBSE - PMT 2011]
0%
a)
0%
b)
0%
c) zero
0%
d)
Explanation
AC=BC VE=VD Now W=Q(VE - VD) ∴ W=0Answer: (c)
Q.47
If the potential of a capacitor having capacity 6µF is increased from 10V to 20V, then increase in its energy will be [ CBSE-PMT 1995]
0%
a)4×10-4 J
0%
b) 6×10-4 J
0%
c)9×10-4
0%
d)12×10-6 J
Explanation
The increase in energy ΔU =½ C ( V22 -V12)=½ ×( 6×10-6)×(202 - 102 ) =9×10-4 J Answer: (c)
Q.48
The capacitance of a parallel plate condenser is 10µF, when the distance between its plates is 8cm. If the distance between the plates is 8cm. If the distance between plates is reduced to 4cm then the capacity of this parallel plate condenser will be [ CBSE-PMT 2001]
0%
a) 5µF
0%
b) 10µF
0%
c)20µF
0%
d)40µF
Explanation
C ∝ 1/d Since distance reduced by half then capacitance will be doubledC'=20 µFAnswer: (c)
Q.49
The electric field at a point on equatorial line of dipole, and direction of the dipole moment [ MPPMT 1995]
0%
a) will be parallel
0%
b) will be in opposite direction
0%
c) will be perpendicular
0%
d) are not related
Explanation
Answer: (b)
Q.50
The electric potential at a point (x, y, z) is given by V=-x2y -xz3 +4 . The electric field E at that point is [ CBSE-PMT 2009]
0%
a) E=2xy + (x2+y2)j+(3xz - y2)k
0%
b)E=z3 i + xyxj +z2 k
0%
c)E=(2xy-z3)i + xy2j+3z2xk
0%
d)E=( 2xy +z3)i +x2j+3xz2
Explanation
The electric field at a point is equal to negative of potential gradient at that point y and z are constant when we take derivative with respect to xE=(2xy+z3)i + x2 j +3xz2k Answer:(d)
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