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Quiz 11
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Q.1
Two condensers of capacitance 2C and C are joined in parallel and charged up to potential V. The battery is removed and the condenser of capacity C is filled with medium of dielectric constant K. The p.d. across the capacitor will now be
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a)6CV/(K+2)
0%
b) 3CV / K
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c)VC /(K+2)
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d)VC/K
Explanation
Total capacitance in combination=C+2C=3CTotal charge of combination=3C×VWhen dielectric medium of constant k is placed in capacitor with capacity C potential across both the capacitor will charge and charge will flow till potential across both capacitor becomes same let potential be V'and charge on capacitor with kC capacitance be q1 and on other be q2Answer: (a)
Q.2
Two condensers of capacity 0.3µF and 0.6µF respectively are connected in series. The combination is connected across a potential of 6volts. The ratio of energies stored by the condenser will be
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a) 1/2
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b) 2
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c)1/4
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d)4
Explanation
USe formula energy=(1/2)Q2/ C , q will be same for as series combinationAnswer: (b)
Q.3
In the following circuit the resultant capacitance between A and B is 1µF. Then value of C is
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a)(32/11) µF
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b) (11/32)µF
0%
c)(23/32)µF
0%
d)(32/23)µF
Explanation
Answer:(d)
Q.4
0.2F capacitor is charged to 600V by a battery. On removing the battery, it is connected with another parallel plates condenser ( 1.0F). The potential decreases to
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a) 100 V
0%
b) 120 V
0%
c) 300 Volts
0%
d) 600 Volts
Explanation
USe formula and V2=0 Answer: (a)
Q.5
Separation between the plates of parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and a slab of thickness t ( t < d) is introduced between the plates, its capacitance becomes
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a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.6
Four capacitors of each capacity 3µF are connected as shown in figure . The ratio of equivalent capacitance between A and B , and between A and C will be
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a) 4 : 3
0%
b) 3 : 4
0%
c)2 : 3
0%
d)3 : 2
Explanation
Answer: (a)
Q.7
The capacity of the capacitors are shown in figure. The equivalent capacitance between the point A and B and the charge on the 6µF capacitor will be
0%
a) 27µF, 540µC
0%
b) 15µF, 270µC
0%
c)6µF, 180µC
0%
d)15µF, 90µC
Explanation
Answer:(c)
Q.8
In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state
0%
a) VAB=100 V, VBC=100 V
0%
b) VAB=75 V, VBC=25 V
0%
c) VAB=40 V, VBC=60 V
0%
d) VAB=50 V, VBC=50 V
Explanation
Reduced circuit is Now potential cross AB=VAB=Potential across 6µF capacitor=Q/ 6¯Fpotential across BC=VBC=potential across 4µF capacitor=Q/ 4µFVAB + VBC=100 VAnswer: (c)
Q.9
Two capacitors 3pF and 6pF are connected in series and a potential difference of 5000V is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is
0%
a)2250 V
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b) 2222 V
0%
c)2.25V×106 V
0%
d)1.1 ×106V
Explanation
Capacitance of combination=2pF charge through the combination=C×V=2pF×5000=10-8 Potential across capacitor 6pF V1=Q/C=10-8 / 6×10-12 V1=104/6 Potential across capacitor 3pF V1=Q/C=10-8 / 3×10-12=104/3 Now from the formula when connected parallel Answer: (b)
Q.10
The capacitance of two conductors are C1 and C2 and their respective potential V1 and V2IF they are connected by a thin wire, then loss of energy will be given by
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.11
The plates of a parallel plate capacitor are charged with a battery so that the plates of the capacitor have acquired the P.D. equal to e.m.f of the battery and The ratio of the energy stored in the capacitor and the work done by the battery will be
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a) 2 : 1
0%
b) 1 : 1
0%
c)1 : 2
0%
d)1 : 4
Explanation
Answer:(a)
Q.12
A capacitor C1=4µF is connected in series with another capacitor C2=1µF. The combination is connected across a d.c. source voltage 200V. The ratio of potential cross C1 and C2 is
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a) 1 : 4
0%
b) 4 : 1
0%
c) 2 : 1
0%
d) 1 : 2
Explanation
Answer: (a)
Q.13
A radio capacitor of variable capacitance, is made of n plates, each of area(A) and separated from each other by a distance(d). The alternate plate is fixed while the other is movable. The maximum capacitance of the system is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.14
A condenser of capacitance 10µF has been charged to 100V. It is now connected to another uncharged condenser. The common potential become 40V. The capacitance of another condenser is
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a) 5 µF
0%
b) 10 µF
0%
c)15 µF
0%
d)20 µF
Explanation
Use formula V=(C1V1 + C1V1) / (C1 + C2)Answer: (c)
Q.15
Calculate the reading of voltmeter between X and Y then ( Vx - Vy) is equal to
0%
a)10 V
0%
b) 13.33 V
0%
c)3.33 V
0%
d)10.33 V
Explanation
Charge flowing through branch AXB=resultant capacitance × Voltage=1×20=20 C Potential across 2F=Q/ C=20/2=10 V Thus VA - Vx=10 V Charge flowing through branch AYB=resultant capacitance × Voltage=2×20=40 C Potential across 3F=Q/ C=40/3 V Thus VA - VY=40/3 V Vx - Vy=[VA - VY] - [VA - Vx] Vx - Vy=40/3-10=10/3=3.33 V Answer:(c)
Q.16
A1 is a spherical conductor of radius (r) placed concentrically inside a thin spherical hollow conductor A2 of radius (R). A1 is earthed and A2 is given a charge +Q then the charge induced on A1 is
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a) -Q
0%
b) Qr/R
0%
c) -QR/r
0%
d) -Q(R-r) / R
Explanation
Answer: (c)
Q.17
For circuit, the equivalent capacitance between P and Q is
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a) 6C
0%
b) 4C
0%
c)3C/2
0%
d)3C/4
Explanation
Answer: (d)
Q.18
For equal capacitor, each with a capacitance (C) are connected to a battery of e.mf 10Volts as shown in the adjoining figure. The mid point of the capacitor system is connected to earth. Then the potentials of B and D are respectively
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a)+10 volts, zero volts
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b) +5 volts, -5volts
0%
c)-5 volts, +5 volts
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d)zero volts, 10 volts
Explanation
Answer:(b)
Q.19
Two spherical conductors A1 and A2 of radii r1 and r2 ( r2 > r1) are placed concentrically in air. A1 is given a charge +Q while A2 in earthed. Then the capacitance of the system is
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a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (a)
Q.20
Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become … [ NEET 2013]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let height be 2h , mass of ball be m , let charge q on each, given separation is r Now balls are in equilibrium Thus electrostatic force = component of tension mg=Tcosθ Now height is half let separation be r’ Take ratio of (i) and (ii) Answer:(b)
Q.21
A, B and C are three points in a uniform electric field. The electric potential is[ NEET 2013]
0%
a) Maximum at A
0%
b) Maximum at B
0%
c) Maximum at C
0%
d) Same at all the three points A, B and C
Explanation
Electric field is directed along decreasing potential VB > VC > VA. Answer:(b)
Q.22
In a region, the potential is represented by V(x, y, z) = 6x – 8xy– 8y+6yz,whereVisinvolts and x,y, z are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is [AIPMT 2014]
0%
a) 24 N
0%
b) 4√35 N
0%
c) 6√5 N
0%
d) 30 N
Explanation
Force = Eq X-component of electric filed and by substituting (1,1,1) Thus Force = 2×2√35=4√35 Answer:(b)
Q.23
Two thin dielectric slabs of dielectric constants K1 and K2 (K1< K2) are inserted between plates of a parallel plate capacitor, as shown in figure. The variation of electric field ‘E’ between the plates with distance ‘d’ as measured from P is correctly shown by …[NEET 2014]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Electric field in the dielectric E1= E/K1 and E2= E/K2 As K1 < K2 , E1 > E2 E > E1 > E2 Answer:(a)
Q.24
A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are [NEET 2014]
0%
a)
0%
b) both are zero
0%
c)
0%
d)
Explanation
Electric potential is uniform inside the sphere and electric filed is zero Answer:(d)
Q.25
he electric field in a certain region is acting radially outward and is given by E = Ar . A charge contained in a sphere of radius ‘a‘ centred at the origin of the field, will be given by : … [ AIPMT 2015]
0%
a) a) ε0Aa3
0%
b) 4 πε0 Aa2
0%
c) Aε0 a2
0%
d) 4 πε0 Aa3
Explanation
Electric filed by point change which is in the sphere of radius a, thus electric field E= Aa Q=Eπε0 a2 Now E = Aa Q=Aa4πε0 a2 = A4πε0 a3 Answer:(d)
Q.26
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect? [AIPMT 2015]
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a) The charge on the capacitor is not conserved.
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b) The potential difference between the plates decreases K times.
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c) The energy stored in the capacitor decreases K times
0%
d) The change in energy stored is
Explanation
Answer:(a)
Q.27
A parallel plate air capacitor has capacity 'C' distance of separation between plates is 'd' and potential difference 'V' is applied between the plates force of attraction between the plates of the parallel plate air capacitor is : [Re AIPMT 2015]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Potential difference is V, thus average potential is V/2 Work done in moving charge w = VQ/2 Work = F×d Q = CV Answer:(c)
Q.28
If potential (in volts) in a region is expressed as V (x,y,z) = 6xy - y + 2yz, the electric field(in N/C) at point (1,1,0) is : [ReAIPMT 2015]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
On substituting values of x, y,z as 1, 1, 0 we get Answer:(c)
Q.29
acitor of 2μF is charged as shown in the diagram. When the switch S is turned to position2, the percentage of its stored energy dissipated is:
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a) a 0%
0%
b) 20%
0%
c) 75%
0%
d) 80%
Explanation
Energy store in 2μF = Energy stored in combination when switch is turned to position 2 Capacitor 2μF and 8μF gets connected in parallel thus effective capacitance = 10μF Now charge is conserved, so charge of combination is same as of 2μF Energy store in 10μF = Energy lost = E-0.2E = 0.8E or 80% Answer:(d)
Q.30
Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d <
0%
a) v ∝ x(1/2)
0%
b) v ∝ x
0%
c) v ∝ x(-1/2)
0%
d) v ∝ x(-1)
Explanation
As θ is very small tanθ = θ and x = lθ As l and mg are constant x ∝ Fe q2 ∝ x3 q ∝ x(3/2) Given that dq/dt is constant v ∝ x(-1/2) Answer:(c)
Q.31
A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k1,k2, k3 and k4 as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by :-…[NEET II 2016]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
We have to find equivalent capacitance From figure it is clear that three capacitance having dielectric medium k1, k2, k3 are parallel. This combination of three is in series with k4 CP and C4 are in series Answer:(a)
Q.32
An electric dipole is placed at an angle of 30° with an electric field intensity 2×105 N/C. It experiences a torque equal to 4 Nm. The charge on the dipole ,if the dipole length is 2 cm, is … [ NEET II : 2016]
0%
a) 5 mC
0%
b) 7 μC
0%
c) 8 mC
0%
d) 2 mC
Explanation
length of dipole be l = 2cm = 2×10-2 m τ = PE sinθ τ = ql E sin sinθ 4 = q × 2 × 10–2 × 2 × 105 sin 30° ∴ q = 2 mC Answer:(d)
Q.33
The diagrams below show regions of equi-potentials A positive charge is moved from A to B in each diagram.
0%
a) Maximum work is required to move q in figure (c).
0%
b) In all the four cases the work done is the same.
0%
c) Minimum work is required to move q in figure (a).
0%
d) Maximum work is required to move q in figure (b).
Explanation
Work done w = qΔV ΔV is same in all the cases so work is done will be same in all the cases. Answer:(b)
Q.34
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system … [NEET 2016]
0%
a) Increases by a factor of 4
0%
b) Decreases by a factor of 2
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c) Remains the same
0%
d) Increases by a factor of 2
Explanation
When Other capacitance is connected parallel , charge conserved Now two capacitance are in parallel thus new C’ = 2C Answer:(b)
Q.35
Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then …. [ AIIT Advance 2017]
0%
a) 5C1 = 3C2
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b) 3C1 = 5C2
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c) 3C1 + 5C2 = 0
0%
d) 9C1 = 4C2
Explanation
Potential = 0 on connecting them together i.e. Q = 0 i.e. C1V1 = C2V2 [capacitance is positive but they are connected with opposite polarity] 120 C1 = 200 C2 3C1 = 5C2 Answer:(b)
Q.36
Two charges, each equal to q, are kept at x = − a and x = a on the x−axis. A particle of mass m and charge q0/2 is placed at the origin. If charge q0 is given a small displacement (y << a)along the y−axis, the net force acting on the particle is proportional to … [ IIT Mains 2013]
0%
a) y
0%
b) -y
0%
c) 1/y
0%
d) -1/y
Explanation
F'=2Fcosθ R2 = a2 + y2 = a2 as y<
Q.37
A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is … [IIT Mains 2013]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Charge density λ = Q/L Consider very small element of dl at distance l; charge of small element =λdl Electric potential at O due to small element On integrating from L to 2L Answer:(d)
Q.38
than one correct answer Q539) In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to charge the capacitor CAfter some time, S2 is released and then S3 is pressed. After some time, ….[ IIT Advance 2013]
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a) the charge on the upper plate of C1 is 2CV0.
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b) the charge on the upper plate of C1 is CV0.
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c) the charge on the upper plate of C2 is 0.
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d) the charge on the upper plate of C2 is −CV0.
Explanation
S1 closed charge on the capacitor = 2V0C S1 is open and Now S2 closed charge on C2 = V0C and on C1 = V0C upper plates are positive Now S3 closed magnitude of charge on C2 will not change but , since upper plate is positive and now connected to negativeterminal of battery thus after some time C2 upper plate becomes negative –CV0 Answer:(b, d)
Q.39
than one correct answer Q540) Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities ρ1 and ρ2 respectively, touch each other. The net electric field at a distance 2Rfrom the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio ρ1/ρ2 can be ...
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a) -4
0%
b) -32/25
0%
c) 32/25
0%
d) 4
Explanation
When spheres are touched charges will not flow as both are non-conducting. If V is the volume of first sphere then charge on first sphere,= Q1 =V |ρ1| Then Charge on second sphere Q2 = 8V|ρ2| Electric filed at point is zero Thus E1 = E2 E1 = Electric field produced due to shall sphere E2 = Electric field produced by big sphere Case I if point is left of small sphere Changes on the spheres must be opposite On substituting values of Q1 and Q2 Case II Note point is inside the big sphere. Now charge enclosed by sphere of radius R of big sphere will produce electric filed, inside big sphere. Thus Q2 = ρ2V If both charge density are same then electric filed zero possible Answer:(b, d)
Q.40
than one correct answer Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densities +ρ and −ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region,[ IIT Advance 2013 ]
0%
a) the electrostatic field is zero.
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b) the electrostatic potential is constant.
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c) the electrostatic field is constant in magnitude.
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d) the electrostatic field has same direction
Explanation
Spheres are non-conducting therefore charges will not flow Option a is wrong as both the spheres have unequal charge Option b is wrong as charge in overlapping region is not uniform Option c is correct as, Electric field is calculated considering charge is located at the centre of sphere. If we consider point P in the overlap region, which is r1 from left sphere centre and r2 from right sphere centre, Electric filed is now depends of OO’ thus will be same for all the points in overlap region Option c is correct Option d: from explanation to option c direction of electric filed is same Answer:(c, d)
Q.41
A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is CWhen the capacitor is charged, the plate area covered by the dielectric gets charge Q1 and the rest of the area gets charge QThe electric field in the dielectric is E1 and that in the other portion is EChoose the correct option/options, ignoring edge effects.[ IIT Advance 2014]
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a) E1/E2 =1
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b) E1/E2 =1/k
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c) Q1/Q2 =3/k
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d) C/C2 =(2+k)/k
Explanation
Both capacitors are connected parallel C=C1 + C2 = (k+2)C’ Option d correct Capacitors are connected parallel potential across both capacitor is same thus E1 = E2 Option a is correct and b is wrong Q = CV Q1=C1V and Q2=C2V Option c wrong Answer:(a, d)
Q.42
1(r), E2(r) and E3(r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density λ, and an infinite plane with uniform surface charge density σ. If E1(r0) = E2(r0) = E3(r0) at a given distance r0, then … [ IIT Advance 2014]
0%
a) Q = 4σπr02
0%
b)
0%
c)
0%
d)
Explanation
Electric filed due to point charge Electric filed due to infinitely long wire with constant linear charge density λ Electric filed due to infinite plane with uniform surface charge density σ Option “a” Given E1(r0) = E2(r0) Q=2λr0 Option (a) wrong Option “b” Given E2(r0) = E3(r0) Option b wrong Option “c” But Q=2λr0 From (i) and (ii) Option c is Correct Option “d” From (a) λ=πσr0 From (iii) and (iv) option “d” is wrong Answer:(c)
Q.43
Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and3 of radii R/2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are E1, E2 and E3 respectively, then …[ IIT Advance 2014]
0%
a) E1 > E2 > E3
0%
b) E3 > E1 > E2
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c) E2 > E1 > E3
0%
d) E3 > E2 > E1
Explanation
Sphere I, electric filed at R Sphere II, electric filed at R Sphere III , electric filed at R Thus E2 > E1 > E3 Answer:(c)
Q.44
The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density λ are kept parallel to each other. In their resulting electric field, point charges q and -q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is (are)…[ IIT Advance 2015]
0%
a) Both charges execute simple harmonic motion.
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b) Both charges will continue moving in the direction of their displacement.
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c) Charge +q executes simple harmonic motion while charge -q continues moving in the direction of its displacement.
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d) Charge -q executes simple harmonic motion while charge +q continues moving in the direction of its displacement.
Explanation
The charge experiences resting force whereas -ve charge experiences force in the direction of displacement Answer:(c)
Q.45
A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When two dielectrics of different relative permittivity’s (ε1 = 2 and ε2 = 4) are introduced between the two plates as shown in the figure, the capacitance becomes CThe ratio C2/C1 is
0%
a) 6/5
0%
b) 5/3
0%
c) 7/5
0%
d) 7/3
Explanation
Three capacitors Capacitor a and b are in series and capacitor c is parallel to series combination Similarly Cb = 4C1 Series combination effective capacitance Ce Ce is in Parallel with C effective is C2 C2=Ce + Cc Answer:(d)
Q.46
Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution, a spherical cavity of radius R2, centred at P with distance OP = a = R1 - R2 (see figure) is made. If the electric field inside the cavity at position r is E(r) , then the correct statement(s) is (are) … [ IIT Advance 2015]
0%
a) E is uniform, its magnitude is independent of R2 but its direction depends on r
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b) E is uniform, its magnitude depends on R2 and its direction depends on r
0%
c) E is uniform, its magnitude is independent of a but its direction depends on a
0%
d) E is uniform and both its magnitude and direction depend on a
Explanation
Option d is correct Answer:(d)
Q.47
PARAGRAPH Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage across (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at -VDue to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity) Q550A) Which one of the following statements is correct?
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a) The balls will bounce back to the bottom plate carrying the opposite charge they went up with
0%
b) The balls will stick to the top plate and remain there
0%
c) The balls will execute simple harmonic motion between the two plates
0%
d) The balls will bounce back to the bottom plate carrying the same charge they went up with
Explanation
Potential on surface of the ball is V Potential difference between two plates =2V0 Electric filed between the plates is E= 2V0/h Thus force on balls F=q×V If m is the mass of ball then acceleration of ball a = F/m Time taken by ball to travel distance h Now current due to one ball i=Q/t Answer:(a)
Q.48
A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct ?
0%
a) The circumference of the flat surface is an equipotential
0%
b) The electric flux passing through the curved surface of the hemisphere is
0%
c) Total flux through the curved and the flat surface is Q/ε0
0%
d) The component of the electric field normal to the flat surface is constant over the surface
Explanation
Option a correct as all points on the circumference are equidistance Electric flux passing through the curved surface = electric flux passing through flat surface. And are equal but it inters through curved surface and leave curved surface thus flux passing through surface =0 [Option c wrong] Electric filed passing through the flat surface= Consider a ring of radius r and thickness dr sin component of Electric field gets cancelled but "cos" component of electric filed get added. R2 + r2 = t2 rdr=tdt ∴ If r = 0 , t= R, r=R , t= R√2 Option b correct and option c and d wrong Answer:(a, b)
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