Description

This mock test of MCQs Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics for JEE helps you for every JEE entrance exam.
This contains 56 Multiple Choice Questions for JEE MCQs 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 MCQs 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 MCQs Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics exercise for a better result in the exam. You can find other MCQs 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.

QUESTION: 1

The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density ρ (r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure The electric field is only along the radial direction.

**Q. The electric field at r = R is**

Solution:

When the point of observation is on the surface of sphere then the whole charge inside the sphere (when distributed symmetrically about the centre) behaves as a point charge on the centre. Therefore until the charge distribution is symmetrical about the centre it does not matter what is the ratio a/R. The electric field remains constant and is equal to

QUESTION: 2

The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density ρ (r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure The electric field is only along the radial direction.

**Q. For a = 0, the value of d (maximum value of ρ as shown in the figure) is –**

Solution:

For a = 0, the graph is as shown. The equation for the graph line is

The charge in the dotted element shown in Fig (2) is dq = ρ × 4πr^{2}dr

QUESTION: 3

The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density ρ (r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure The electric field is only along the radial direction.

**Q. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies.**

Solution:

If the volume charge density is constant then E ∝ r.

QUESTION: 4

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 source (HV) is connected across the conducting plates such that the bottom plate is at +V_{0} and the top plate at –V_{0}. Due 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)

**Q. Which one of the following statements is correct?**

Solution:

After colliding the top plate, the ball will gain negative charge and get repelled by the top plate and bounce back to the bottom plate.

QUESTION: 5

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 source (HV) is connected across the conducting plates such that the bottom plate is at +V_{0} and the top plate at –V_{0}. Due 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)

**Q. The aver age current in the steady state registered by the ammeter in the circuit will be**

Solution:

From (i), (ii) and (iii)

QUESTION: 6

**STATEMENT-1 : **For practical purposes, the earth is used as a reference at zero potential in electrical circuits. and

**STATEMENT-2 :** The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by

Solution:

Both the statements are true and statement-2 is the correct explanation of statement-1

QUESTION: 7

On moving a charge of 20 coulomb by 2 cm, 2 J of work is done, then the potential difference between the points is

Solution:

QUESTION: 8

If there are n capacitors in parallel connected to V volt source, then the energy stored is equal to

Solution:

The equivalent capacitance of n identical capacitors of capacitance C is equal to nC. Energy stored in this capacitor

QUESTION: 9

A charged particle q is placed at the centre O of cube of length L (A B C D E F G H). Another same charge q is placed at a distance L from O. Then the electric flux through ABCD is

Solution:

Both the charges are identical and placed symmetrically about ABCD. The flux crossing ABCD due to each

charge is but in opposite directions. Therefore the resultant is zero.

QUESTION: 10

If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is

Solution:

For equilibrium of charge Q

QUESTION: 11

Capacitance (in F) of a spherical conductor with radius 1 m is

Solution:

For an isolated sphere, the capacitance is given by

QUESTION: 12

If the electric flux entering and leaving an enclosed surface respectively is φ_{1} and φ_{2} , the electric charge inside the surface will be

Solution:

The flux entering an en closed sur face is taken as negative and the flux leaving the surface is taken as positive, by convention. Therefore the net flux leaving the enclosed surface = φ_{2} -φ_{1}

∴ the charge enclosed in the surface by Gauss’s law is q = ∈_{0} (φ_{2} -φ_{1})

QUESTION: 13

A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

Solution:

The capacitance of a parallel plate capacitor in which a metal plate of thickness t is inserted is given by

QUESTION: 14

A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point P a distance R/2 from thecentre of the shell is

Solution:

Electric potential due to charge Q placed at the centre of the spherical shell at point P is

Electric potential due to charge q on the surface of the spherical shell at any point inside the shell is

∴ The net electric potential at point P is

QUESTION: 15

The work done in placing a charge of 8 × 10^{-18} coulomb on a condenser of capacity 100 micro-farad is

Solution:

The work done is stored as the potential energy. The potential energy stored in a capacitor is given by

QUESTION: 16

Three charges –q_{1} , +q_{2} and –q_{3} are place as shown in the figure. The x - component of the force on –q_{1} is proportional to

Solution:

Force on charge q_{1} due to q_{2} is

Force on charge q_{1} due to q_{3} is

The X - component of the force (Fx) on

QUESTION: 17

The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter the change in the resistance of the wire will be

Solution:

R_{f} = n^{2}R_{i}

Here n = 2 (length becomes twice)

∴ R_{f} = 4R_{i}

New reresistance = 400 of R_{i}

∴ Increase = 300%

QUESTION: 18

Two spherical conductors B and C having equal radii and carrying equal charges on them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that B but *uncharged* is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is

Solution:

x is distance between the spheres. After first operation charge on B is halved i.e Q/2 and charge on third sphere becomes Q/2. Now it is touched to C, charge then equally distributes them selves to make potential same, hence charge on C becomes

QUESTION: 19

A charge particle ‘q’ is shot towards another charged particle ‘Q’ which is fixed, with a speed ‘v’. It approaches ‘Q’ upto a closest distance r and then returns. If q were given a speed of ‘2v’ the closest distances of approach would be

Solution:

QUESTION: 20

Four charges equal to -Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is

Solution:

Net field at A should be zero

QUESTION: 21

A charged oil drop is suspended in a uniform field of 3×10^{4} v/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9×10^{–15} kg and g = 10 m/s^{2})

Solution:

At equilibrium, electric force on drop balances weight of drop.

QUESTION: 22

Two point charges + 8q and – 2q are located at x = 0 and x = L respectively. The location of a point on the x axis at which the net electric field due to these two point charges is zero is

Solution:

QUESTION: 23

Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and -q. The potential difference between the centres of the two rings is

Solution:

QUESTION: 24

A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is ‘C’ then the resultant capacitance is

Solution:

As n plates are joined, it means (n – 1) capacitor joined in parallel.

∴ resultant capacitance = (n – 1) C

QUESTION: 25

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 the figure. The surface charge density σ of the sheet is proportional to

Solution:

QUESTION: 26

A fully char ged capacitor has a capacitan ce ‘C’. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raised by ‘ DT ’, the potential difference ‘V’ across the capacitance is

Solution:

Applying conservation of energy,

QUESTION: 27

An electric dipole is placed at an angle of 30° to a nonuniform electric field. The dipole will experience

Solution:

The electric field will be different at the location of the two charges. Therefore the two forces will be unequal.

This will result in a force as well as torque.

QUESTION: 28

Two insulating plates are both uniformly charged in such a way that the potential difference between them is V_{2} – V_{1} = 20 V. (i.e., plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e = 1.6 × 10^{–19} C, m_{e} = 9.11 × 10^{–31} kg)

Solution:

= 2.65 ×10^{6} m / s

QUESTION: 29

Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is

Solution:

After connection, V_{1} = V_{2}

The ratio of electric fields

Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.

QUESTION: 30

An electric charge 10^{–3} μ C is placed at the origin (0, 0) of X – Y co-ordinate system. Two points A and B are situated at and (2, 0) respectively. The potential difference between the points A and B will be

Solution:

The distance of point from the origin,

The distance of point B(2, 0) from the origin,

∴ Potential difference between the points A and B is zero.

QUESTION: 31

Charges are placed on the vertices of a square as shown. Let be the electric field and V the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then

Solution:

As shown in the figure, the resultant electric fields before and after interchanging the charges will have the same magnitude, but opposite directions.

Also, the potential will be same in both cases as it is a scalar quantity.

QUESTION: 32

The potential at a point x (measured in μ m) due to some charges situated on the x-axis is given by V(x) = 20/(x^{2} – 4) volt

The electric field E at x = 4 m m is given by

Solution:

Positive sign indicates tha -direction.

QUESTION: 33

A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

Solution:

The potential energy of a charged capacitor before removing the dielectric slat is

The potential energy of the capacitor when the dielectric slat is first removed and the reinserted in the gap between the plates is

There is no change in potential energy, therefore work done is zero.

QUESTION: 34

If gE and gM are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio

Solution:

Electronic charge does not depend on acceleration due to gravity as it is a universal constant.

So, electronic charge on earth = electronic charge on moon

∴ Required ratio = 1.

QUESTION: 35

A parallel plate capacitor with air between the plates has capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k_{1} = 3 and thickness d/3 while the other one has dielectric constant k_{2} = 6 and thickness 2d/3. Capacitance of the capacitor is now

Solution:

The given capacitance is equal to two capacitances connected in series where

The equivalent capacitance C_{eq} is

QUESTION: 36

A thin spherical sh ell of radus R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E(r) produced by the shell in the range 0 __<__ r < ∞, where r is the distance from the centre of the shell?

Solution:

The electric field inside a thin spherical shell of radius R has charge Q spread uniformly over its surface is zero.

Outside the shell the electric field is These

characteristics are represented by graph (a).

QUESTION: 37

Two points P and Q are maintained at the potentials of 10 V and – 4 V, respectively. The work done in moving 100 electrons from P to Q is :

Solution:

= (– 100 × 1.6 × 10^{–19}) (– 4 – 10)

= +2.24 × 10^{–16}J

QUESTION: 38

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then Q/q equals:

Solution:

Let F be the force between Q and Q. The force between q and Q should be attractive for net force on Q to be zero. Let F' be the force between Q and q . For equilibrium

QUESTION: 39

This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

**Statement-1 :** For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.

**Statement-2 : **The net work done by a conservative force on an object moving along a closed loop is zero.

Solution:

Statement 1 is true.

Statement 2 is true and is the correct explanation of (1)

QUESTION: 40

Let be the charge density distribution for asolid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance r_{1} from the centre of the sphere, the magnitude of electric field is :

Solution:

Let us consider a spherical shell of thickness dx and radius x. The volume of this spherical shell = 4πr^{2} dr .

The charge enclosed within shell

The charge enclosed in a sphere of radius r_{1} is

∴ The electric field at point p inside the sphere at a distance r_{1} from the centre of the sphere is

QUESTION: 41

A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field the centre is

Solution:

Let us consider a differential element dl. charge on this element.

Electric field at O due to dq is

Th e compon ent dE cos θ will be coun ter bala n ced by another element on left portion. Hence resultant field at O is the resultant of the component dE sinθ only.

The direction of E is towards negative y-axis.

QUESTION: 42

Let there be a spherically symmetric charge distribution with charge density varying as upto r = R, and ρ(r) = 0 for r > R , where r is the distance from the origin. The electric field at a distance r(r < R) from the origin is given by

Solution:

Let us con sider a spher ical shell of r adius x and thickness dx.

Charge on this shell

∴ Total charge in the spherical region from centre to r (r < R ) is

QUESTION: 43

Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d << l) apart because of their mutual repulsion.

The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity v. Then as a function of distance x between them,

Solution:

At any instant

T cosθ = mg ....(i)

T sin θ = F_{e} ....(ii)

For small θ, sin θ ≈ tan θ ∴ q^{2} ∝ x^{3}

QUESTION: 44

The electrostatic potential inside a charged spherical ball is given by φ = ar^{2} + b where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is :

Solution:

Electric field

By Gauss's theorem

From (i) and (ii),

QUESTION: 45

In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as function of distance from the centre, The graph which would correspond to the above will be:

Solution:

QUESTION: 46

This questions has statement-1 and statement-2. Of the four choices given after the statements, choose the one that best describe the two statements.

An insulating solid sphere of radius R has a uniformly positive charge density r. As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinite is zero.

**Statement -1 :** When a charge q is take from the centre of the surface of the sphere its potential energy changes by

**Statement -2 :** The electric field at a distance r (r <R) from the centre of the sphere is

Solution:

The electric field inside a uniformly charged sphere is

The electric potential inside a uniformly charged sphere

∴ Potential difference between centre and surface

QUESTION: 47

Two capacitors C_{1} and C_{2} are charged to 120 V and 200 V respectively. It is found that connecting them together the potential on each one can be made zero. Then

Solution:

For potential to be made zero, after connection

QUESTION: 48

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 is placedat the origin. If charge q_{0} is given a small displacement (y<<a) along the y-axis, the net force acting on the particle is proportional to

Solution:

QUESTION: 49

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 distance L from the end A is

Solution:

Electric potential is given by,

QUESTION: 50

Assume that an electric field exists in space. Then the potential difference V_{A} - V_{O}, where V_{O} is the potential at the origin and V_{A} the potential at x = 2 m is:

Solution:

Poten tial difference between any two poin ts in an electric field is given by,

QUESTION: 51

A parallel plate capacitor is made of two circular plates separated by a distance 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 10^{4} Vm the charge density of the positive plate will be close to:

Solution:

Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,

Then, charge density

σ = Kε_{0}E

= 2.2 × 8.85 × 10^{–12} × 3 × 10^{4} ≈ 6 × 10^{–7} C/m^{2}

QUESTION: 52

In the given circuit, charge Q_{2} on the 2µF capacitor changes as C is varied from 1µF to 3µF. Q2 as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale)

Solution:

Therefore graph d correctly dipicts.

*Multiple options can be correct

QUESTION: 53

A uniformly charged solid sphere of radius R has potential V_{0} (measured with respect to ∞) on its surface. For this sphere the equipotential surfaces with potentials and have radius R_{1}, R_{2}, R_{3} and R_{4 }respectively. Then

Solution:

At the centre of sphare r = 0. Here

R_{4} = 4R

Also, R_{1} = 0 and R_{2} < (R_{4} – R_{3})

QUESTION: 54

A long cylindrical shell carries positive surface charge s in the upper half and negative surface charge - σ in the lower half. The electric field lines around the cylinder will look like figure given in : (figures are schematic and not drawn to scale)

Solution:

Field lines originate perpendicular from positive charge and terminate perpendicular at negative charge. Further this system can be treated as an electric dipole.

QUESTION: 55

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distance 30 m from it, would equal :

Solution:

∴ Voltage across 9μF is also 2V

∴ Charge on 9μF capacitor = 9 × 2 = 18μC

∴ Total charge on 4 μF and 9μF = 42μc

QUESTION: 56

The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), have volume charge density

where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is :

Solution:

Applying Gauss's law

For E to be independent of 'r'

- MCQs Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics
Test | 56 questions | 110 min

- Multiple Correct MCQ Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics
Test | 30 questions | 60 min

- Single Correct MCQ Of Electrostatics, Past Year Questions JEE Advance, Class 12, Physics
Test | 34 questions | 70 min

- MCQs Of Current Electricity, Past Year Questions JEE Advance, Class 12, Physics
Test | 55 questions | 110 min

- MCQ(Practice)- Electrostatics, Class 12, Physics
Test | 50 questions | 90 min