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Q.1
Two spheres A and B of radius a and b respectively are at the same potential. The ratio the surface charge density of A and B is
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a) a/b
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b) b/a
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c)a2 / b2
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d)b2 / a2
Explanation
Potential on surface of sphere=σr / εo Both spheres at same potentialσaa / εo=σbb / εoσa / σb=b/aAnswer: (b)
Q.2
A uniform wire of length 5m is carrying a steady current. The electric field inside it is 0.2 V/m. The potential difference across the ends of wire is
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a) 1.0 volt
0%
b) 0.5 volt
0%
c)0.1 volt
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d)5 volt
Explanation
V=E.d=02×5=1V Answer:(a)
Q.3
A charge is divided into parts q1 and (q-q1). What is the ratio q/q1 so that the force between the two parts placed a given distance apart is maximum?
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a) 1:1
0%
b) 2:1
0%
c) 1:2
0%
d) 1:4
Explanation
F is maximum when (q-q1)=q1 or q/q1=2 : 1 Answer: (b)
Q.4
Let P(r)=Qr/ πR4 be the charge density distribution for a solid sphere of radius R and total charge Q. For a point 'p' inside the sphere at distance r1 from the centre of the sphere, the magnitudes of electric field is .. [ AIEEE 2009]
0%
a)
0%
b)
0%
c)
0%
d)zero
Explanation
let us consider a spherical shell of thickness dr and radius r. the volume of this shell=4πr2dr. The charge enclosed within shell.The charge enclosed by the sphere of radius r1∴ The electric field at point p inside the sphere at a distance r1 from the centre of the sphere is Answer: (b)
Q.5
A hollow metal sphere of radius 5cms is charged such that the potential on its surface is 10 Volts. The potential at the centre of the sphere is [ IIT 1983]
0%
a) zero
0%
b) 10 Volts
0%
c)same as at a point 5 cm away from the surface
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d)same as at a point 25 cm away from the surface
Explanation
Potential inside the hollow sphere is same at every point and is equal to potential on the surfaceAnswer: (b)
Q.6
Two point charges +q and -q are held fixed at ( -d,0) and (d,0) respectively of a x-y coordinate system. Then [ IIT 1995]
0%
a)the electric field E at all points on the x-axis has the same direction
0%
b) electric field at all points on y-axis is along x-axis
0%
c)Work has to be done in bringing a test charge from ∞ to the origin
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d)the dipole moment is 2qd along x-axis
Explanation
option (a) :electric field at any point on axial line is along the direction of dipole moment which is on negative x-axis Thus incorrectoption (b): If we take any point on Y axis then we find net electric field along +X-direction Thus option is correctoption c: Potential at equatorial ine of dipole is zero, so potential at (0,) is zero. Potential at ∞ is also zero thus no work is done Option (c) is incorrectOption d) The direction of dipole moment is from -ve to +ve. Therefore option is incorrect Answer:(b)
Q.7
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and capacitor are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal. The final energy of the configuration is [ IIT 1995]
0%
a) zero
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b) (3/2) CV2
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c) (25/6) CV2
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d) (9/2) CV2
Explanation
C and 2C are in parallel to each other ∴ Resultant capacitance=2C+C=3C Net potential=2V - V=V ∴ Final energy=½ CR VR Final energy=½3CV2=(3/2) CV2 Answer: (b)
Q.8
An electron of mass me initially at rest, moves through a certain distance in a uniform electric field in time tA proton of mass mp, also, initially at rest, takes time t2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio t2 / t1 is nearly equal to [ IIT 1997]
0%
a)1
0%
b) ( mp / me)1/2
0%
c)( me / mp)1/2
0%
d)1836
Explanation
A charge on proton and electron is same therefore electric force on both will be same but acceleration will be different due to different mass Displacement of electron of x in time t1 acceleration=Ee/ me x=ut + ½ at12 x=½ Ee/ me ×t12 Displacement of x of proton in time t2 acceleration=Ee/ mp x=ut + ½ at22 x=½ Ee/ mp ×t12 ∴ t2 / t1=( mp / me ) 1/2Answer: (b)
Q.9
A non conducting ring of radius 0.5 m carries a total charge of 1.1×10-10C distributed non-uniformly on its circumference producing an electric field E everywhere in space. The value of the integral ( l=0 being center of ring) in volts is [ IIT 1997]
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a) +2
0%
b) -1
0%
c)-2
0%
d)zero
Explanation
potential at centre Vo=kq/R and potential infinity=0Answer: (a)
Q.10
Two identical metal plates are given positive charge Q 1 and Q2 ( Q2< Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitor with capacitance C, the potential difference between them is [ IIT 1999]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let A be the area of plate and d be the distance between the plates Within capacitor electric field due to individual plate Answer:(d)
Q.11
For the circuit shown in figure. Which of the following statement is true[ IIT 1999]
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a) With S1 closed V1=15V, V2=20V
0%
b) With S3 closed V1=V2=25V
0%
c) With S1 and S2 closed V1=V2=0
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d) With S1 and S3 closed V1=30V, V2=20V
Explanation
With the closing of switch S3 and S1 the negative charge on C2 will attract the positive charge on C1. The negative charge on C1 will attract positive charge on C1. No transfer of charge will take place. Therefore p.d. across C1 and C2 will be 30V and 20 V Answer: (d)
Q.12
if a capacitor C, 3C, 5C ...∞ is one network and 2C, 4C, 6C ...∞ is another network are connected in series. If the effective capacitance of first network is C1 and that of network second be C2 then C1C2 /(C2 -C1) equal to
0%
a) C / log2
0%
b) log2/C
0%
c) ∞
0%
d) (log2)2 / C
Explanation
Answer: (a)
Q.13
Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in figure. The net electrostatic energy of the configuration ia zero if Q is equal to .. [ IIT 2000]
0%
a)-q / (1+√2)
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b) -2q / (2+√2)
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c)-2q
0%
d)+q
Explanation
we have Answer: (b)
Q.14
A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants k1, k2 and k3 as shown. 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 [ IIT 2000]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let C1 be the capacitance with K1C2 be the capacitance with K2C3 be the capacitance with K3For C1 Area=A/2 and distance=d/2 For C2 Area=A/2 and distance=d/2 For C2 Area=A and distance=d/2 From figure it is clear that C1 and C2 are parallel let it be C' and C' is in series with C3 But C"=εoKA/ d comparing value of K option (b) is correctAnswer: (b)
Q.15
Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is ..[ IIT 2001]
0%
a)zero
0%
b)q/2
0%
c)q
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d)2q
Explanation
Since capacitor B is not grounded thus there will not be any transfer of charge and charge on capacitor B will be zero Answer:(a)
Q.16
A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x=+1 cm and C be the point on the y-axis at y=+1cm. Then the potential at the point A, B, and C satisfy [ IIT 2001]
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a) VA < VB
0%
b) VA > VB
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c) VA < VC
0%
d) VA > VC
Explanation
Electric field is along positive x-direction and dv=-E dx. Thus as we go away on positive direction on x axis potential will reduce thus VA > VB Answer: (b)
Q.17
Two equal point charges are fixed at x=-a and x=+a on the x-axis. Another point charge Q is placed at the origin. The change in the elecrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to [ IIT 2002]
0%
a)x
0%
b) x2
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c)x3
0%
d)1/x
Explanation
Initial potential energy of Q Final potential energy of Q Change in potential energy of Q=Uf - Ui Thus change in potential energy ∝ x2Answer: (b)
Q.18
Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other VThe negative ends of the capacitor are connected together. When the positive ends are also connected, the decrease in the energy of the combined system is [ IIT 2002]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Before connecting Charge on first capacitor q1=CV1Charge on second capacitor=q2=CV2 Total Energy of capacitors before joining After joining total charge will be conserved and capacitance will be 2C final energy Change in energy Uf-Ui Answer: (c)
Q.19
A metallic shell has a point charge 'q' kept inside its cavity. Which one of the following diagram correctly represnets the electric lines of forces?
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
electric field inside the metaliic portion is zero hence option (a) and (d) are incorrect.Electric field lines are normal to a surface. Hence option (b) is incorrect.Only option (c) represents the correct aswer Answer:(c)
Q.20
Six charges of equal magnitude, 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what it would have been if only one +ve charge is placed at R [ IIT 2004]
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a) +, +, +, -, -, -
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b) -, +, +, +, -, -
0%
c) -, +, +, -, +, -
0%
d) +, -, +, -, +, -
Explanation
If opposite charges are kept on opposite verices electric field will be added . and if like charge are kept at vertices electric field will be cancelled. Thus option C is correct Answer: (c)
Q.21
A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be [ IIT 2004]
0%
a)due to q1 and q2 only
0%
b) due to q2 only
0%
c)zero
0%
d)due to all
Explanation
The flux through the Gaussian surface is due to the charges inside Gaussian surface. But electric field on the Gaussian surface will be due to the charges present on the Gaussian surface and outside it. It will be due to all the chargesAnswer: (d)
Q.22
Three infinitly long charge sheets are placed as shown in figure. the electric field at point P is [ IIT 2005]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Drection of electric field produced at point P due to all the paltes in along negative x axis The total electric field E=E1 + E2 + E3 Answer: (c)
Q.23
A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral [ IIT 2007]
0%
a)A potential difference appears between the two cylinders when a charge density is given to the inner cylinder
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b) A potential difference appears between the two cylinders when a charge density is given to the outer cylinder
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c)No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinder
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d)No potential difference appears between the two cylinders when same charge density is given to both the cylinders
Explanation
When a charge density is given to the inner cylinder, the potential developed at its surface is different from that on the outer cylinder. this is beacuse the potential decreases with distance from a charged conducting cylinder when the point of consideration is outside the cylinder. But when a charge densitty is given to the outer cylinder, it will chage its potential by the same amount as that of the inner cylinder. Therefore no potential difference will be produced between the cylinders in this case Answer:(a)
Q.24
Consider a neutral conducting spher. A positive point charge is placed outside the sphere. The net charge on the sphere is then [ IIT 2007]
0%
a) negative and distributed uniformly over the surface of the sphere
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b) negativ and appers only at the point on the sphere closer to the point charge
0%
c) negative and distributed non-uniformly over the entire surface of the sphere
0%
d) Zero
Explanation
When a positive point charge is placed outside a conducting sphere, a rearrangement of charges takes place on the surface. But the total charges on the sphere is zero as no charge has left or entered the sphere Answer: (d)
Q.25
A spherical portion has been removed from a solid surface having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the empited space is .. [ IIT 2007]
0%
a)zero everywhere
0%
b) non-zero and uniform
0%
c)non-uniform
0%
d)zero only at its center
Explanation
Answer: (b)
Q.26
A proton is released from rest at a distance of 10-4Å from nucleus of mercury atom (Z=80). The kinetic energy of the proton when it is far away from the nucleus is
0%
a) 12eV
0%
b) 12 KeV
0%
c) 1.2MeV
0%
d) 12 MEV
Explanation
Kinetic energy of proton at inifinity=Potentil energy of proton at 10-4Å Answer: (d)
Q.27
positive and negative point charges of equal magnitude are kept at (0, 0, a/2) and ( 0, 0, -a/2) respectively. the work done by the electric field when another positive point charge is moved from ( -a, 0, 0) to ( 0, a, 0) is [ IIT 2007]
0%
a) positive
0%
b) negative
0%
c)zero
0%
d)depends on the path connectingthe initial and final positions
Explanation
Magnitude of position for the point is same thus potential at both the point is same , hence zero workAnswer: (c)
Q.28
Consider a system of three charges q/3, q/3 and 2q/3 placed at points A, B and C resppectively, as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB=660°.. [ IIT 2008]
0%
a) The electric field at point O is q / 8πεoR2 directed along negative x-axis
0%
b) the potential energy of the system is zero
0%
c)the magnitude of the force between the charges at C and B is q2 / 54πεoR2
0%
d)the potential at point O is q/ 12πεoR
Explanation
option a : Electric field due to A and B at O is equal and opposite producing a resultant which is zero. The elctric field at O is due to charge at C=q / 6πεoR2 option is not correctOption b and d : Potential at O can not be zero as all the points are at equii distance from O and charges total is zon zero. Thus potential energy can not be zero. Option b and D are not correct option c: Magnitude of force between B and C is=Option is correct Answer:(c)
Q.29
A parallel plate capacitor C with plates of unit area and separation d is filled with liquid of dielectric constant K=The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed v, the time constant as a function of time t is [ IIT 2008]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let the level of liquid at instant of time 't' be x. Then v=-dx/dt dx=-vdt Thus we can consider two cpacitor with distance (d-x) and x are connected in series . equivalent capacitance Time constant τ=R Ceq Answer: (a)
Q.30
Theree concentric metallic spherical shell of radii R, 2R, 3R are given charges Q1, Q2 and Q3 respectively. It is found that the surface charge densities of outer surfaces of the shells are equal. Then the ratio of the charges given to the shell Q1: Q2 : Q3 is [IIT 2009]
0%
a)1:2:3
0%
b) 1:3:5
0%
c)1:4:9
0%
d)1:8:18
Explanation
Due to induction and added charge :Sphere with radius 2R will have total charge Q1+ Q2 sphere of radius 3R is Q1 + Q2 + Q1Charge on sphere with radius R=Q1 since surface charge density of all the sphere is same Q1 : Q2:Q3=1:3:5Answer: (b)
Q.31
A disc of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at ( -a/2, 0, 0). A rod of length a carrying a uniformly distributioed charge 8C is placed on the x-axis from x=a/4 to x=5a/Two point charges -7C and 3C are palaced at (a/4, -a/4, 0) and (-3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x=± a/2, y=±a/2, x=± a/The electric flux throgh this cubical surface is [ IIT 2009]
0%
a) -2C/εo
0%
b) 2C/εo
0%
c) 10C/εo
0%
d)12C/εo
Explanation
From the figure it is cklaear that Half of the disc is eneclosed in cube thus charge eneclosed=3Csides of cubes are a/2Total lengeth of rod is a and charge is 8C and rod is palced at x=a/4 thus cube enclosed a/4 length=2C Charge 7C is at (a/4, -a/4, 0) which will be enclosed by cube Charge 3C is at (-3a/4, 3a/4, 0) is out of the cube Thus total charge enclosed=3C+2C-7C=-2C. Therefore the elctric flux through the cube is φ=-2C/εoAnswer: (a)
Q.32
Eight drops of mercury of equal radii and possessing equal charges combined to form a big drop. Then the capacitance of bigger drop compared to each individual drop is [ MNR 1987]
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a) 8 times
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b) 4 times
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c)2 times
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d)32 times
Explanation
Capacitance of spherical drop ∝ radiusvolume of big drop=8× volume of small drop Thus , radius of dig drop=2 × radius of small drop Capacitance of spherical drop ∝ radiusThus cpacitance of big drop=2× cpacitance of small drop Answer:(c)
Q.33
Two condensers of capacity 0.3µF and 0.6µF respectively are connected in series. The combination is connected across a potential of 6V. The ratio of energies stored by the condenser wil be [ MPPMT 1990]
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a) 1/2
0%
b) 2
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c) 1/4
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d) 4
Explanation
capcitors are connected in series thus charge on both capacitor is same Thus U1=q2 / 2C1 U2=q2 / 2C2 ∴ U1 / U2=C2 / C1 ∴ U1 / U2=2 Answer: (b)
Q.34
A parallel plate capacitor has a capacity C. The separation between plates is doubled and a dielectric medium is inserted between plates. The new capacitance is 3C. The dielectric constant of medium is
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a)1.5
0%
b) 3.0
0%
c)6.0
0%
d)12.0
Explanation
C'=KCAnswer: (c)
Q.35
If two conducting spheres are separately charged and then brought in contact
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a) the total energy of the two sphere is conserved
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b) the total charge on the spheres is conservered
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c)both the total energy and charge are conservered
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d)the final potential is always the mean of the original potential of the two spheres
Explanation
Answer: (b)
Q.36
A potential difference V is applied across two capacitors of capacitance C1 and C2 connected in series. Then the potential difference across C1 will be
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a) VC2 / C1
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b) V( C1 + C2) /C1
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c)VC2 / ( C1 + C2)
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d)VC1 / ( C1 + C2)
Explanation
equivalante capacitance of combination=(C1C2) / ( C1 + C2) Thus charge on the combination q=(VC1C2) / ( C1 + C2) ∴ potential difference across C1=VC2 / ( C1 + C2) Answer:(c)
Q.37
Three capacitors of capacitance 3µF, 9µF and 18µF are connected in series and another time in parallel. The ratio of equivalent capacitance in the two case ( Cs / Cp) will be [ CPMT 1990]
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a) 1:15
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b) 15:1
0%
c) 1:1
0%
d) 1:3
Explanation
Cs=2µF Cp=30µFs / Cp)=1:15Answer: (a)
Q.38
The electric field between the two spheres of a charged spherical condenser: [ MPPMT 1994]
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a)is zero
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b) is constant
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c)increases with distance from the centre
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d)decreses with distance from the centre
Explanation
Answer: (d)
Q.39
The capacity of a parallel plate condenser is 5µF. When a glass plate is placed between the plates of the condenser, its potential difference reduces to 1/8 of the original value. The value of the dielectric constant of glass is [ MPPMT 1985]
0%
a) 1.6
0%
b) 8
0%
c)5
0%
d)40
Explanation
K=Vo / V=8Answer: (b)
Q.40
A glass slab is put within the plates of a charged parallel plate condenser. Which of the following quanities does not change? [ MPPMT 1998]
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a) energy of the condenser
0%
b) capacity
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c)intensity of electric field
0%
d)charge
Explanation
Answer:(d)
Q.41
A 4 µF conductor is charged to 400V and then its plates are joined through a resistance of 1kΩ. The heat produced in the resistance is [ CBSE 1994]
0%
a) 0.61 J
0%
b) 1.28J
0%
c) 0.64J
0%
d) 0.32J
Explanation
Energy stored in cpaitor=energy lost in resistance energy=½ CV2 energy=½ 4×10-6 ×(400)2 energy=0.32J Answer: (d)
Q.42
An infinite number of identical capaciotrs each of capaciatance 1µF are connected as shown in figure. Then the equivalent capacitance between A and B is [ AP 1990]
0%
a)1 µF
0%
b) 2µF
0%
c)1/2µF
0%
d)∞
Explanation
Rows contains capaciors 1, 1/2, 1/4, 1/8, 1/16,... µFabove value capacitors are connected in parallel thus C=1+ (1/2)+(1/4)+(1/8)+(1/16) +...It is gemometric progression∴ C=first term/ (1 -difference) Here a=1 and r=1/2C=1/(1-1/2)=2µFAnswer: (b)
Q.43
A capacitor of capacitance 2µF is charged to a potential difference of 200 V. After disconnecting from battery, it is connected in parallel with an another uncharged capacitor. The common potential is 40V. The capacitance of the second capacitor is [ CPMT 1991]
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a) 2µF
0%
b) 4µF
0%
c)8µF
0%
d)16µF
Explanation
If V is common potential, From formula Answer: (c)
Q.44
Two point charges at a certain distance experience a force of 5N. Each charge is doubled in magnitude and distance between the two is halved. The interacting force would be
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a) 16 N
0%
b) 80 N
0%
c)5 N
0%
d)20 N
Explanation
Answer:(b)
Q.45
Four metallic plates each with a surface area of one side A, are palced at a distance d from each other. The plates are connected as shown in the figure. Then the capacitance of the system between a and b is :
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a)3εoA / d
0%
b) 2εoA / d
0%
c)2εoA / 3d
0%
d)3εoA / 2d
Explanation
Effective capacity between a and b is C/2+C=3C/2=3εoA/2d Answer:(d)
Q.46
Two parallel plates, separated by a distance of 5mm, are kept at a potential difference of 50V. A particle of mass 10-15kg and charge 10-11 C enters in it with a velocity 107 m/s. The acceleration of the particle will be [ MPPMT 1997]
0%
a) 108 m/s2
0%
b) 5 ×105 m/s2
0%
c) 105 m/s2
0%
d) 2 × 103 m/s2
Explanation
E=V/d=50/5×10-3=104 V/m a=F/m=Eq/m=104 ×10-11 / 10-15 a=108 m/s2 Answer: (a)
Q.47
Pulling the plates of charged capacitor apart
0%
a)increases the capacitance
0%
b) increases the potential difference
0%
c)does not affect potential difference
0%
d)decreases the potential difference
Explanation
Answer: (b)
Q.48
N identical spherical drop charged to same potential V are combined to form the big drop will be [ JIPMER 1998]
0%
a) V
0%
b) V×N
0%
c)V/N
0%
d)V ×N2/3
Explanation
Radius of big sphere R=N1/3 r Thus potential of big sphere=kNq/ ( N1/3r Potential=N2/3VAnswer: (d)
Q.49
The plates of a parallel plate capacitor are charged up to 100V. A 2mm thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6 mm. The dielectric constant of the plates is [ MPPMT 1991]
0%
a) 5
0%
b) 1.25
0%
c)4
0%
d)2.5
Explanation
V=q/Clet charge on the capacitor after insertion of plate be q1 and capacitance be C1∴ q/C=q1 / C1let thickness of plate be t and dielectric constant be k then effective increase in the distance t/k let d' be the new distance between the plate then resultant effective distance=d'- t+ t/k Answer:(a)
Q.50
Four equal capacitors, each with a capacitance C are connected to a battery of e.m.f 10V as shown in figure. The midpont of the capacitor system is connected to earth. Then the potentials of B and D are respectively
0%
a) +10V, zero voly
0%
b) +5V , -5V
0%
c) -5V, +5V
0%
d) zero volt, 10 volt
Explanation
Total capacitance=C/4 Charge flowing through capacitors=CV=(C/4) × 10=10C/4=2.5C Potential at F is zero Potential across each capacitor between B and F is=v=q/C V=2.5C / C=2.5 V. Thus potential across two capacitor=2.5+2.5=+5V Similarly Potential across each capacitor Between F and D=-2.5 Thus potential across two capacitor=-2.5-2.5=+5V Answer: (b)
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