WorksheetsUntitled form GGPS
Total questions: 103
Worksheet time: 2hrs 4mins
Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is
2σε0k
-2σε0k
4σε0k
-4σε0k
An electric dipole consists of two opposite charges, each of magnitude 1.0μC separated by a distance of 2.0 cm. The dipole is placed in an external electric field of 105NC-1. The maximum torque on the dipole is
0.2×10-3 N m
1×10-3 N m
2×10-3 N m
4×10-3 N m
A ring of radius R carries a charge Q, uniformly distributed along its circumference. What is the ratio of the electric field strength at a distance R to that at a distance R2 along the axis?
38
338
3342
2233
An infinitely long thin straight wire has uniform linear charge density of 13Cm-1. Then the magnitude of the electric intensity at a point 18 cm away is
0.33×1011 NC-1
0.66×1011 NC-1
3×1011NC-1
1.32×1011 NC-1
Two point charges are 3 m apart and their combined charge is 8μC. The force of repulsion between them is 0.012 N. Charges are
4μC,4μC
6μC,2μC
5μC,3μC
7μC,1μC
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will
increase four times
be reduced to half
remain the same
be doubled
The direction of electric field intensity (E) at a point on the equatorial line of an electric dipole of dipole moment (p) is
along the equatorial line towards the dipole
along the equatorial line away from the dipole
perpendicular to equatorial line,opposite to (p)
perpendicular to the equatorial line and parallel to (p)
What is the electric flux linked with closed surface?
1011 N m2C-1
1012 N m2C-1
1010 N m2C-1
8.86×1013 Nm2C-1
Which of the following figures cannot possibly represent electrostatic field lines?
Given below are two statements: Statement I: A point charge is brought in an electric field. At a point near to the charge may increase if the charge is positive. Statement II: An electric dipole is placed in a nonuniform electric field. The net electric force on the dipole will not be zero. Considering above statements choose the most appropriate answer from the options given below.
Both Statement I and Statement II are correct.
Both Statement I and Statement II are incorrect.
Statement I is correct, Statement II is incorrect.
Statement I is incorrect, Statement II is correct.
A charge Q is placed at the centre of the line joining two point charges +q and +q, as shown in the figure. What is the ratio Qq for the system to be in equilibrium?
4
14
-4
-14
In a region, the intensity of an electric field is given by E=2i+3j+k in NC-1. The electric flux through a surface S=10i m2 in the region is
5Nm2C-1
10Nm2C-1
15Nm2C-1
20Nm2C-1
Let ρ(r)=QrπR4 be the charge density distribution for a solid sphere of radius R, total charge Q. For a point P inside the sphere at a distance r1 from the centre of the sphere, the magnitude of electric field is
Q4πε0r12
Qr124πε0R4
Qr123πε0R4
zero
A charged ball B hangs from a silk thread S, which makes an angle θ with a large charged conducting sheet P as shown in figure. The surface charge density of the sheet is proportional to
cosθ
cotθ
sinθ
tanθ
A charge 10μC is placed at the centre of a hemisphere of radius R=10 cm as shown. The electric flux through the hemisphere (in MKS units) is
20×105
10×105
6×105
2×105
Two-point charges +10-7C and -10-7C are placed at A and B,20 cm apart as shown in the figure. Calculate the electric field at C,20 cm apart from both A and B.
1.5×10-5 N/C
2.2×104 N/C
3.5×106 N/C
3.0×105 N/C
The electric field due to an infinitely long straight uniformly charged wire at a distance r is directly proportional to
r
r2
1r
1r2
There is a point charge q located at the centre of a cube. What is the electric flux of this point charge, through a face of the cube?
qε0
q6ε0
q3ε0
It will depend upon the size of the cube.
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): A negative charge in an electric field moves along the direction of the electric field. Reason (R): On a negative charge a force acts in the direction of the electric field.
Both (A) and (R) are true and (R) is the correct explanation of (A).
Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(A) is true but (R) is false.
Both (A) and (R) are false.
A conducting sphere of radius 10 cm has an unknown charge. If the electric field at a distance 20 cm from the centre of the sphere is 1.2×103 NC-1 & points radially inwards. The net charge on the sphere is
-4.5×10-9C
4.5×109C
-5.3×10-9C
5.3×10-9C
An oil drop of 10 excess electrons is held stationary under a constant electric field of 3.65×104 NC-1 in Millikan's oil drop experiment. The density of oil is 1.26 g cm-3. Radius of the oil drop is (Take, g=9.8 m s-2,e=1.6×10-19C )
1.0×10-6 m
4.8×10-18 m
4.8×10-5 m
1.13×10-18 m
A positive charge Q is uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring as shown in figure. Then
if q>0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre.
if q<0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring.
if q<0 it will perform SHM for small displacement along the axis.
all of the above.
Under the action of a given coulombic force, the acceleration of an electron is 2.5×1022 m s-2. Then the magnitude of the acceleration of a proton under the action of same force is nearly
1.6×10-19 m s-2
9.1×1031 m s-2
1.4×1019 m s-2
1.6×1027 m s-2
A rod of length 2.4 m and radius 4.6 mm carries a negative charge of 4.2×10-7C spread uniformly over its surface. The electric field near the mid-point of the rod, on its surface is
-8.6×105 NC-1
8.6×104NC-1
-6.8×105 NC-1
6.8×104 NC-1
Assertion: Acceleration of charged particle in nonuniform electric field does not depend on velocity of charged particle. Reason: Charge is an invariant quantity. That is amount of charge on particle does not depend on frame of reference.
If both assertion and reason are true.
If assertion is true but reason is false.
If assertion is false but reason is true.
If both assertion and reason are false.
tric field does not depend on velocity of charged particle. Reason: Charge is an invariant quantity. That is amount of charge on particle does not depend on frame of reference.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
The electrostatic potential inside a charged spherical ball is given by ϕ=ar2+b where r is the distance from the centre; a,b are constants. Then the charge density inside the ball is
-24πaε0r
-6aε0r
-24πaε0
-6aε0
Electrical as well as gravitational effects can be thought to be caused by fields. Which of the following is true of an electrical or gravitational field?
The field concept is often used to describe contact forces.
Gravitational or electric field does not always exist in the space around an object.
Fields are useful for understanding forces acting through a distance.
There is no way to verify the existence of a force field since it is just a concept.
If an object of mass 1 kg contains 4×1020 atoms. If one electron is removed from every atom of the solid, the charge gained by the solid in 1 g is
2.8C
6.4×10-2C
3.6×10-3C
9.2×10-4C
Two point charges of 1μC and -1μC are separated by a distance of 100Å. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be
9NC-1
0.9 NC-1
90 NC-1
0.09 NC-1
Two charges ±20μC are placed 10 mm apart. The electric field at point P, on the axis of the dipole 10 cm away from its centre O on the side of the positive charge is
8.6×109 NC-1
4.1×106 NC-1
3.6×106 NC-1
4.6×105 NC-1
In a field free region, two electrons are released to move on a line towards each other with velocities 106 m s-1. The distance of their closest approach will be nearer to
1.28×10-10 m
1.92×10-10 m
2.56×10-10 m
3.84×10-10 m
A uniform electric field E=2×103 NC-1 is acting along the positive x-axis. The flux of this field through a square of 10 cm side whose plane is parallel to the yz plane is
20 NC-1 m2
30 NC-1 m2
10 NC-1 m2
40 NC-1 m2
Four point charges are placed at the corners of a square ABCD of side 10 cm, as shown in figure. The force on a charge of 1μC placed at the centre of square is
7 N
8 N
2 N
zero
If there were only one type of charge in the universe, then
∮ sE⋅ds≠0 on any surface
∮ sE⋅ds=0 if the charge is outside the surface
∮ sE⋅ds=qε0 if charges of magnitude q were inside the surface
both (b) and (c) are correct
A uniformly charged conducting sphere of 4.4 m diameter has a surface charge density of 60μCm-2. The charge on the sphere is
7.3×10-3C
3.7×10-6C
7.3×10-6C
3.7×10-3C
Assertion: If an electron and proton possessing same kinetic energy enter an electric field in a perpendicular direction, the path of the electron is more curved than that of the proton. Reason: Electron forms a larger curve due to its small mass.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
A dipole of electric dipole moment p is placed in a uniform electric field of strength E. If θ is the angle between positive directions of p, E, then potential energy of the electric dipole is largest when θ is
π4
π2
π
zero
A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net electric field E at the centre O is
q2π2ε0r2j
q4π2ε0r2j
-q4π2ε0r2j
-q2π2ε0r2j
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 )
3.47×104C
3.47×102C
1.67×1020C
1.67×1022C
An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge upto a radius R. The atom as a whole is neutral. The electric field at a distance r from the nucleus is (r>R)
Ze4πε01r2-rR3
Ze4πε01r3-rR2
Ze4πε0rR3-1r2
Ze4πε0rR3+1r2
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 105 NC-1 m-1. The force experienced by the system having a total dipole moment equal to 10-7Cm in the negative z-direction is
-10-2 N
10-2 N
10-4 N
-10-4 N
The tracks of three charged particles in a uniform electrostatic field is shown in the figure. Which particle has the highest charge to mass ratio?
A
B
C
A and B
Given below are two is statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: If an electric dipole of dipole moment 30×10-5Cm is enclosed by a closed surface, the net flux coming out of the surface will be zero. Reason R: Electric dipole consists of two equal and opposite charges.
A is false but R is true.
A is true but R is false.
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is NOT the correct explanation of A.
Three charges of equal magnitude q is placed at the vertices of an equilateral triangle of side l. The force on a charge Q placed at the centroid of the triangle is
3Qq4πε0l2
Qq4πε0l2
Qq2πε0l2
zero
A cup contains 250 g of water. Find the number of positive charges present in the cup of water.
1.34×1019C
1.34×107C
2.43×1019C
2.43×107C
Flux of the electric field E=24i+30j+28kNC-1 through an area of 20 m2 on the yz plane is
480 N m2C-1
600 N m2C-1
560 N m2C-1
1640 N m2C-1
An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2×104 NC-1. The time taken by the electron to fall this distance is
1.3×102 s
2.1×10-12 s
1.6×10-10 s
2.9×10-9 s
e of 1.5 cm in a uniform electric field of magnitude 2×104 NC-1. The time taken by the electron to fall this distance is
1.3×102 s
2.1×10-12 s
1.6×10-10 s
2.9×10-9 s
The ratio of magnitude of electrostatic force and gravitational force between an electron & a proton is
6.6×1039
2.3×1039
6.6×1029
2.3×1029
The nucleus of helium atom contains two protons that are separated by distance 3.0×10-15 m. The magnitude of the electrostatic force that each proton exerts on the other is
20.6 N
25.6 N
15.6 N
12.6 N
A particle of mass m, charge -q enters the region between the 2 charged plates initially moving along x-axis with speed vx as shown in figure. The length of plate is L, a uniform electric field E is maintained between the plates. The vertical deflection of the particle at the far edge of the plate is
qEL22mvx2
qEL22mvx
2mvx2qEL2
2mvxqE2L
The electric field at C due to charge +10-7C at A is E1=14πε010-7(0.2)2 along AC. The electric field at C due to charge -10-7C at B is E2=14πε010-7(0.2)2 along CB. As E1=E2. By symmetry the vertical components will cancel out and horizontal components will add. The resultant electric field at C is
oton due to same force F is ap=Fmp#(ii) where mp is the mass of the proton. Divide (ii) by (i), we get apae=memp∴ap =aememp=2.5×1022 m s-29.1×10-31 kg1.67×10-27 kg =13.6×1018 m s-2≈1.4×1019 m s-2
(c) : Here, l=2.4 m,r=4.6 mm=4.6×10-3 m q=-4.2×10-7C Linear charge density, λ=ql=-4.2×10-72.4=-1.75×10-7Cm-1 Electric field E=λ2πε0r =-1.75×10-72×3.14×8.854×10-12×4.6×10-3 =-6.8×105 NC-1
(b) : a=qEm as E varies so does a, it does not depend on its velocity.
(d) : ϕ=ar2+b Electric field, E=-dϕdr=-2ar According to Gauss's theorem, ∮ E⋅dS=qinside ε0qinside =-8ε0aπr3 Charge density inside the ball is ρinside =qinside 43πr3∴ρinside =-8ε0aπr343πr3 ρinside =-6aε0
(b) : Here, number of electrons removed = number of atoms in 1 g or n=4×1020103=4×1017 ∴ charge, q=ne=4×1017×1.6×10-19C =6.4×10-2C
(d) : The point lies on equatorial line of a short dipole. ∴ E=ql4πε0r3=9×109×10-6×10-810-13=9×10-2 NC-1
(c) : Here, q=±20μC=±20×10-6C 2a=10 mm=10×10-3 m r=OP=10 cm=10×10-2 m |p|=q×2a=20×10-6×10×10-3=2×10-7Cm The electric field along BP,E=2pr4πε0r2-a22 As a
(c) : At the distance of closest approach (r) K.E. = P.E. ∴12mv2+12mv2=14πε0e2r mv2=14πε0e2rr=14πε0e2mv2=9×109×1.6×10-1929×10-31×1062 =2.56×10-10 m
(a) : Here, E=2×103 NC-1 along +x-axis Surface area, S=(10 cm)2=102×10-4 m2=10-2 m2. When plane is parallel to yz plane, θ=0∘ So ϕ=EScosθ=2×103×10-2cos0∘ =20 NC-1 m2
(d) : From figure, length of diagonal of the square =AC=BD=102+102=102 cm OA=OB=OC=OD=1022=102 cm Forces of repulsion on 1μC charge at O due to 3μC charges, at A and C are equal and opposite. So they cancel each other. Similarly, forces of attraction on 1μC charge at O due to -4μC charges at B and D are also equal and opposite. So they also cancel each other. Hence the net force on the charge of 1μC at O is zero.
(d) : According to Gauss's theorem in electrostatics ∮ E⋅ds=qε0 Here q is charge enclosed by the surface. If the charge is outside the surface, then qinside =0 Also, ∮ E⋅ds=0. So, both (b) and (c) are correct.
(d) : Here, D=2r=4.4 m, or r=2.2 m σ=60μCm-2 Charge on the sphere, q=σ×4πr2 =60×10-6×4×227×(2.2)2=3.7×10-3C
(d) : As qE=mv2r,r=mv2qE=2K.E.qEK.E.=12mv2 i.e., r∝1q. Since electron and proton have equal charge, r is the same for both of them. Here, r depends on K.E. (which is given to be the same for both the particles) and as such not on their masses.
(c) : The potential energy of an electric dipole in a uniform electric field is U=-p⋅EU=-pEcosθ For U to be maximum cosθ=-1⇒θ=π
(d) : E=∫0π dEsinθ ⇒E=∫0π kdqr2sinθ =kr2∫0π qdθπsinθ as dq=qdθπ =kqπr2∫0π sinθdθ=q2π2ε0r2 as ∫0π sinθdθ=2 or E=-q2π2ε0r2j (as E is directed along -Y-axis)
(a) : Mass of the coin =0.75 g,ZAl=13 Atomic mass of aluminium =26.98 g Avogadro's number =6.02×1023 Number of Al atoms in the coin, N=6.02×102326.98×0.75=1.67×1022 As charge number of Al is 13, each atom of Al contain 13 protons and 13 electrons. Magnitude of positive and negative charges in one coin =NZAle =1.67×1022×13×1.6×10-19C =3.47×104C
(a) : Charge on nucleus =+Ze Total negative charge =-Ze ( ∵ atom is electrical neutral] Negative charge density, ρ= charge volume =-Ze43πR3 i.e. ρ=-34ZeπR3 Consider a Gaussian surface with radius r. By Gauss's theorem ϕ=E(r)×4πr2=qε0 Charge enclosed by Gaussian surface q=Ze+4πr33ρ=Ze-Zer3R3 (Using (i)) From (ii) E(r)=q4πε0r2=Ze-Zer3R34πε0r2=Ze4πε01r2-rR3
(a) : Consider an electric dipole with -q charge at A and +q charge at B, placed along z-axis, such that its dipole moment is in negative z direction. i.e., pz=-10-7Cm The electric field is along positive direction of z-axis, such that dEdz=105 NC-1 m-1 From F=qdE=(q×dz)×dEdz =pdEdz Force experienced by the system in the negative z-direction F =-p×-dEdz =10-7×-105=-10-2 N
(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. Deflection of charged particle in time t in y-direction As the particle C suffers maximum deflection in y-direction, so it has highest charge to mass q/m ratio.
(c) : Both A and R are true and R is the correct explanation of A. As electric dipole is a system of two equal and opposite charges separated by a length. Net flux is equal to 1ε0 time charge stored and the net charge is equal to zero in case of a dipole so flux is also zero.
(d) : From figure, AD=ABcos30∘=l32 Distance AO of the centroid O from A =23AD=2l332=l3 ∴ Force on Q at O due to charge q1=q at A F1=14πε0Qq(l/3)2=3Qq4πε0l2, along AO Similarly, force on Q due to charge q2=q at B F2=3Qq4πε0l2 along BO and force on Q due to charge q3=q at C F3=3Qq4πε0l2, along CO Angle between forces F2 and F3=120∘ By parallelogram law, resultant of F2 and F3=3Qq4πε0l2 along OA ∴ Total force on Q=3Qq4πε0l2-3Qq4πε0l2=0
(b) : Mass of water =250 g Molecular mass of water =18 g Number of molecules in 18 g of water (Avogadro's Number) =6.02×1023 ∴ Number of molecules in one cup of water =25018×6.02×1023 Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e. 10 electrons and 10 protons. ∴ Total positive and negative charge has the same magnitude and is =25018×6.02×1023×10×1.6×10-19C=1.34×107C
(a) : Here, E=24i+30j+28kNC-1 S=20i m2 Electric flux, ϕ=E⋅S =24i+30j+28kNC-1⋅20i m2 =480 N m2C-1
(d) : In figure the field is upward. So the negatively charged electron experiences a downward force. ∴ The acceleration of electron is ae=eEme#(i) The time required by the electron to fall through a distance h is te =2hae=2hmeeE#(i) =2×1.5×10-2×9.11×10-311.6×10-19×2×10412 =2.9×10-9 s#(i)
(b) : Here, between an electron and a proton Electrostatic force, Fe=14πε0e2r2 Gravitational force, Fg=Gmempr2 ∴ FeFg =14πε0e2Gmemp =9×109×1.6×10-1926.67×10-11×9×10-31×1.66×10-27 =2.3×1039
(b) : Charge of proton, qp=1.6×10-19C Distance between the protons, r=3×10-15 m The magnitude of electrostatic force between protons is Fe=qpqp4πε0r2=9×109×1.6×10-19×1.6×10-193×10-152=25.6 N
(a) : Here, in the vertical direction, initial velocity, v=0 acceleration, a=Fm=qEm ..(i) (∵F=qE) Time taken to cross the field, t= distance velocity =Lvx#(ii) ( ∵ velocity along the horizontal direction is constant) Now, s=vt+12at2 ∴ deflection, y=0+12qEmLvx2 [Using (i) and (ii)] ∴y=qEL22mvx2
