Multiple Correct MCQ Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics


30 Questions MCQ Test Class 12 Physics 35 Years JEE Mains &Advance Past year Paper | Multiple Correct MCQ Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics


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This mock test of Multiple Correct MCQ Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Multiple Correct MCQ Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics (mcq) to study with solutions a complete question bank. The solved questions answers in this Multiple Correct MCQ Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Multiple Correct MCQ Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics exercise for a better result in the exam. You can find other Multiple Correct MCQ Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics extra questions, long questions & short questions for JEE on EduRev as well by searching above.
*Multiple options can be correct
QUESTION: 1

Two equal negative charges –q are fixed at points (0, – a) and (0, a) on y – axis. A positive charge Q is released from rest at the point (2a, 0) on the x - axis. The charge Q will

Solution:

Let us consider the positive charge Q at any instant of time t at a distance x from the origin. It is under the influence of two forces   On resolving these two forces we find that F sin θ cancels out. The resultant force is

FR = 2F cos θ


Since FR is not proportional to x, the motion is NOT simple harmonic. The charge Q will accelerate till the origin and gain velocity. At the origin the net force is zero but due to momentum it will cross the origin and more towards left. As it comes on negative x-axis, the force is again towards the origin.   

*Multiple options can be correct
QUESTION: 2

A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q0, V0, E0 and U0 respectively. A dielectric slab is now introduced to fill the space between the plates with battery still in connection.
The corresponding quantities now given by Q, V, E and U are related to th e previous one as

Solution:


*Multiple options can be correct
QUESTION: 3

A charge q is placed at the centre of the line  joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to :

Solution:

q has to be negative for equilibrium.


Considering equilibrium of 1

*Multiple options can be correct
QUESTION: 4

A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles :

Solution:

Charge on plate is q


Charge on plate is q


*Multiple options can be correct
QUESTION: 5

A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the same two surfaces is :

Solution:

The potential inside the shell will be the sa me everywhere as on its surface. As we add – 3Q charge on the surface, the potential on the surface changes by the same amount as that inside. Therefore the potential difference remains the same.

*Multiple options can be correct
QUESTION: 6

Seven capacitors each of capacitan ce 2μF are to be connected in a configuration to obtain an effective capacitance of   Which of the combination (s) shown in figure will achieve the desired result ?

Solution:

The equivalent capacitance


*Multiple options can be correct
QUESTION: 7

A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then

Solution:

Q will remain same as no charge is leaving or entering the plates during the process of slab insertion

Now, Q = C ' V ' = C ' E 'd

Work done is the change in energy stored

*Multiple options can be correct
QUESTION: 8

Two identical thin rings, each of radius R metres, are coaxially placed a distance R metres apart. If Q1 coulomb, and Q2 coulomb, are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is

Solution:

The work done in moving a charge from A to B

W = (T.P.E.)A – (T.P.E.)B where T.P.E. = Total Potential Energy




*Multiple options can be correct
QUESTION: 9

The magnitude of electric field  the annular region of a charged cylindrical capacitor.

Solution:

Let λ be the charge per unit length. Let us consider a Gaussian surface (dotted cylinder).
Applying Gauss's law


For the flat portions of Gaussian surface, the angle between electric field and surface is 90°.
Hence flux through flat portions is zero.  

NOTE : By symmetry, the electric field on the curved surface is same throughout.

The angle between   (for curved surface)

*Multiple options can be correct
QUESTION: 10

A metallic solid sphere is placed in a uniform electric fied. The lines of force follow the path(s) shown in Figure as

Solution:

The electric lines of force cannot enter the metallic sphere as electric field inside the solid metallic sphere is zero. Also, the origination and termination of the electric lines of force from the metallic surface is normally (directed towards the centre).

*Multiple options can be correct
QUESTION: 11

A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 3d. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to 3d,

Solution:

In region I and III, there will be electric field   directedfrom positive to negative. In region II, due to orientation of dipoles, there is an electric field  present in opposite
direction of   But since  is also present, the net electric field is  in the direction of    as shown in the diagram. (∵ E0 > Ek)


NOTE : When one moves opposite to the direction of electric field, the potential always increases. The stronger the electric field, the more is the potential increase. Since in region II, the electric field is less as compared to I and III therefore the increase in potential will be less but there has to be increase in potential in all the regions from x = 0 to x = 3d. Also where

E is uniform, dV/dx = constt.

*Multiple options can be correct
QUESTION: 12

A charge +q is fixed at each of the points x = x0,  x = 3x0, x = 5x0,..... x = ∞ on the x axis, and a charge –q is fixed at each of the points x = 2x0, x = 4x0, x = 6x0,.... x = ∞. Here x0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/(4πε0r).
Then, the potential at the origin due to the above system of charges is

Solution:

Potential at origin will be given by

*Multiple options can be correct
QUESTION: 13

A positively charged thin metal ring of radius R is fixed in the xy plane with its centre at the origin O. A negatively charged particle P is released from rest at the point (0, 0, z0) where z0 > 0. Then the motion of P is

Solution:

Let Q be the charge on the ring, the negative charge – q is released from point P (0, 0, Z0). The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be

Therefore, force on charge P will be towards centre as shown, and its magnitude is

Similarly, when it crosses the origin, the force is again towards centre O.
Thus the motion of the particle is periodic for all values of Z0 lying between 0 and ∞.

Secondly if 

i.e. the restoring force Fe∝ – Z0. Hence the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position)

*Multiple options can be correct
QUESTION: 14

A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centre

Solution:

KEY CONCEPT :  The expressions of the electric field inside the sphere   outside the sphere

Hence, E increases for r < R and decreases for R < r < ∞.

*Multiple options can be correct
QUESTION: 15

An ellipsoidal cavity is carved within a perfect conductor. A positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surface as shown in the figure. Then

Solution:

When two points are connected with a conducting path in electrostatic condition, then the potential of the two points is equal. Thus potential at A = Potential at B (c) is the correct option.
Option (d) is a result of Gauss's law

Total electric flux through cavity

Option (a) and (b) are dependent on the curvature which is different at points A and B.

*Multiple options can be correct
QUESTION: 16

A spherical symmetric charge system is centered at origin. Given, Electric potential

Solution:

(a) The  whole charge Q will be enclosed in a sphere of diameter 2R0. (b) Electric field E = zero inside the sphere.
Hence electric field is discontinued at r = R0.
(c) Changes in V and E are continuously present for r > R0. Option (c) is incorrect.
(d) For r < R 0, t he potential V is constant and the electric intensity is zero.
Obviously, the electrostatic energy is zero for r < R0.

*Multiple options can be correct
QUESTION: 17

Under the influence of the Coulomb field of charge + Q, a charge –q is moving around it in an elliptical orbit. Find out the correct statement(s).

Solution:

The situation is shown in the figure which is similar to a planet revolving around sun. The distance of – q from + Q is changing, therefore, force between the charges will change.   

The speed of the charge – q will be greater when the charge is nearer to + Q as compared to when it is far.
Therefore,  the angular velocity of charge – q is also variable. The direction of the velocity changes continously, therefore, linear momentum is also variable.
The angular momentum of (– q) about Q is constant because the torque about + Q is zero.

*Multiple options can be correct
QUESTION: 18

A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that

Solution:

The electric field lines are orginating from Q1 and terminating on Q2. Therefore Q1 is positive and Q2 is negative.
As the number of lines associated with Q1 is greater than that associated with Q2, therefore |Q1| > |Q2|.
Option (a) is correct.
At a finite distance on the left of Q1, the electric field intensity cannot be zero because the electric field created by Q1 will be greater than Q2. This is because the magnitude of Q1 is greater and the distance smaller 

At a finite distance to the right of Q2, the electric field is zero. Here, the electric field created by Q2 at a particular point will cancel out the electric field created by Q1.

*Multiple options can be correct
QUESTION: 19

A spherical metal shell A of radius RA and a solid metal sphere B of radius RB(< RA) are kept far apart and each is given charge ‘+Q’. Now they are connected by a thin metal wire. Then

Solution:

Electric field inside a spherical metallic shell with charge on the surface is always zero. Therefore option [a] is correct.
When the shells are connected with a thin metal wire then electric potentials will be equal, say V.

As RA > RB therefore QA > AB. option [b] is also correct.


Option (d) is also correct

*Multiple options can be correct
QUESTION: 20

Which of the following statement(s) is/are correct?

Solution:

(a) is not correct because it is valid only when E ∝ r–2
(b) is not correct
(c) is correct as between two point charges we will get a point where the electric field due to the two point charges cancel out each other.
(d) is correct when the work done is without accelerating the charge.

*Multiple options can be correct
QUESTION: 21

A cubical region of side a has its centre at the origin.
It encloses three fixed point charges, -q at (0, -a/4, 0), +3q at (0, 0, 0) and -q at (0, +a/4, 0). Choose the correct options(s)

Solution:

The electric flux passing through 

  is same due to symmetry.

The net electric flux through the cubical region is

*Multiple options can be correct
QUESTION: 22

Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the figure. Given that   which of the following statement(s) is (are) correct?

Solution:

∴  EO = EA + ED + (EF + EC) cos 60º + (EB + EC) cos 60º

The electric potential at O is

PR is perpendicular bisector (the equatorial line) for the electric dipoles AB, FE and BC. Therefore the electric potential will be zero at any point on PR.
At any point ST, the electric field will be directed from S to T.
The potential decreases along the electric field line.

*Multiple options can be correct
QUESTION: 23

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 2R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio 

Solution:

Electric field E1 due to smaller sphere at P is

Electric field E2 due to bigger sphere at P is

Option (d) is correct.

*Multiple options can be correct
QUESTION: 24

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 C2. After some time, S2 is released and then S3 is pressed. After some time

 

Solution:

Step 1 : When S1 is pressed. The capacitor C1 gets charged such that its upper plate acquires a positive charge + 2 CV0 and lower plate – 2 CV0.
Step 2 : When S2 is pressed (S1 open). As C1 = C2 the charge gets distributed equal. The upper plates of C1 and C2 now take charge + CV0 each and lower plate – CV0 each.

(b) and (d) are correct option.

*Multiple options can be correct
QUESTION: 25

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

 

Solution:

Let us consider a point P on the overlapping region. The electric field intensity at P due to positively charged sphere 

The electric field intensity at P due to negatively charged
sphere   The total electric field,

Therefore the electric field is same in magnitude and direction option (c) and (d) are correct.

*Multiple options can be correct
QUESTION: 26

Let E1 (r), E2(r) and E3(r) be the respective electric field 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

Solution:


∴ Q = 2λr0                  ...(1)




*Multiple options can be correct
QUESTION: 27

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 C1. When the capacitor is charged, the plate area covered by the dielectric gets charge Q1 and the rest of the area gets charge Q2. The electric field in the dielectric is E, and that in the other portion is E2. Choose the correct option/ options, ignoring edge effects.

Solution:

This is a combination of two capacitors in parallel.
Therefore C = C1 + C2     ∴ C2 = C – C1


∴ (d) is a correct option.

     ∴ (c) is incorrect

Also V = E × d

   ∴ (a) is a correct option

*Multiple options can be correct
QUESTION: 28

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)

Solution:

Force on change q when it is given a small displacement x is Fnet = F1 – F2


 and is directed towards the mean position therefore the charge +q will execute SHM.

In case of charge (–q) F2 > F1 therefore the charge –q continues to move in the direction of its displacement.

[C] is the correct option.

*Multiple options can be correct
QUESTION: 29

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  then the correct statement(s) is (are)

Solution:

Assume the cavity to contain similar charge distribution of positive and negative charge as the rest of sphere.
Electric field at M due to uniformly distributed charge of the whole sphere of radius R1

Electric field at M due the negative charge distribution in the cavity

∴ The total electric field at M is

(d) is the correct option

*Multiple options can be correct
QUESTION: 30

A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When two dielectrics of different relative primitivities (ε1 = 2 and ε2 = 4) are introduced between the two plates as shown in the figure, the capacitance becomes C2. The ratio

Solution: