Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Class 12 Physics 35 Years JEE Mains &Advance Past year Paper

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Q 1. A parallel plate capacitor has plates of area A separated by distance d between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1+αx ) where x is the distance measured from one of the plates. If (αd ) << 1, the total capacitance of the system is best given by the expression    [2020]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans: (1)
Solution:
Given that, parallel plate capacitor filled with dielectric constant k(x) which varies as, k(x) = K(1+αx)
where x is distance measured from one of the plates.
Now, capacitance of a small element dx is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev 
So, total capacitance is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
For αd << 1 , we have
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Now, using binomial expansion, we get
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 2. A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time and charge is redistributed between them is (in nJ) _________.     [2020]
Ans:
(6)
Solution:
Given that, C = 60 pF = 60 × 10−12 F and V = 20 V.
We know that, energy when capacitor is fully charged is given by

Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Energy when capacitor is disconnected and connected to uncharged capacitor is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Where C' = C+C = 2C and V' = V/2
Then,
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
So, energy lost is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 3. Effective capacitance of parallel combination two capacitors C1 and C2 is 10 μF. When these capacitors are individually connected to a voltage source of 1 V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be    [2020]
(1) 4.2 μF
(2) 3.2 μF
(3) 1.6 μF
(4) 8.4 μF
Ans:
(3)
Solution:
Given that
CP = C1 + C2 =  10μF   ...(1)
Energy stored in capacitor is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
We know that
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
⇒C2 = 4C1      (since V = 1V)   ...(2)
Solving Eqs. (1) and (2), we get
C1 = 2μF, C2 = 8μF
Effective capacitance of series combination is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 4.  Consider two charged metallic spheres S1 and S2 of radii R1 and R2 respectively. The electric fields E1 (on S1) and E2 (on S2) on their surface are such that E1/E2 = R1/R2. Then the ratio V1 ( on S1) /V2 ( on S2) of the electrostatic potentials on each sphere is    [2020]
(1) R1/ R2
(2) (R1/ R2)2
(3) R2/ R1
(4) (R1/ R2)3

Ans: (2)
Solution:
Given that
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
We know that, electric filed is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
So,
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
We know that, electrostatic potential is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
So,
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 5. A capacitor is made of two square plates each of side α making a very small angle α between them, as shown in figure. The capacitance will be close to    [2020]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRevPrevious year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(1)
Solution:
Let small element dx at distance x has capacitance dC.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRevWe know that, capacitance of parallel plat capacitor is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
So,
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Since α is very small so tan α ≈ α .
Therefore,
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 6. Two identical capacitors A and B, charged to the same potential 5 V are connected in two different circuits as shown below at time t = 0. If the charge on capacitors A and B at time t = CR is QA and QB respectively, then (Here e is the base of natural logarithm)    [2020]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev(1) Q= VC/e, Q= CV/2
(2) Q= VC, QB = CV
(3) QA = VC, QB =VC/e
(4) QA = CV/2, QB = VC/e
Ans: 
(3)
Solution:
In first circuit diode is reversed biased so it is open circuit.
so, QA = CV
In second circuit diode is forward biased so it it short circuit.
So, current flow in the circuit at time t is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
At t = RC, we get
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 7. Two infinite planes each with uniform surface charge density +σ are kept in such a way that the angle between them is 30º. The electric field in the region shown between them is given by    [2020]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRevPrevious year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(4)
Solution:
Given that

Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRevElectric filed due to infinite plane is given by

Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
For first infinite plane, electric filed is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
For second infinite plane, electric filed is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
So, resultant electric field in region is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 8. Three charged particles A, B and C with charges −4q, 2q and −2q are present on the circumference of a circle of radius d. The charged particles A, C and Centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is    [2020]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRevPrevious year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans: (1)
Solution:
Given that, 
QA = -4q, QB = 2q, QC = -2q
x component of electric field due to point charge A at centre O is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
x component of electric field due to point charge B at centre O is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
x component of electric field due to point charge C at centre O is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Total x component electric field due to point charges A, B, and C at centre O is given by
ET = EA + EB + EC
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 9. In finding the electric field using Gauss law the formula Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev is applicable. In the formula εis permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?
(1) Only when the Gaussian surface is an equipotential surface.
(2) Only when the Gaussian surface is an equipotential surface and Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev is constant on the surface.
(3) Only when Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev = constant on the surface.
(4) For any choice of Gaussian surface.     [2020]
Ans. 
(2)
Solution: 
Gauss law is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Gauss law only holds when gaussian surface is an equipotential surface and electric field is constant throughout the surface.

Q 10. A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x travelled by it is correctly given by (graphs are schematic and not drawn to scale).
(1)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(2)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(3)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(4)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans. (2)
Given that   vi = 0
Coulomb force is given by  F = qE
⇒ ma = qE
⇒ a = qE/m
Final velocity of charge particle in electric filed is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev 
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Only option (2) graph satisfy the condition.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Q 11. Consider a sphere of radius R which carries a uniform charge density ρ. If a sphere of radius R/2 is carved out of it, as shown, the ratio Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev of magnitude of electric field Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev respectively, at points A  and B due to the remaining portion is    [2020]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev(1) 21/34
(2) 18/34
(3) 17/54
(4) 18/54
Ans.
  (2)
Given that
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
So, cavity in solid sphere has density  Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Electric filed of a solid sphere is given by  
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Electric filed at point A is only due to cavity given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev    (1)
Electric filed at point B is due to solid sphere and cavity given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev    (2)
From Eqs. (1) and (2), we get
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 12. An electric dipole of moment Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev cm is at the origin (0, 0, 0). The electric field due to this dipole at  Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev (note that  Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev is parallel to    [2020]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(3)
Solution:
Given that
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Since Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev are perpendicular to each other so, point lies on the equatorial plane.
Therefore, electric field at the point will be antiparallel to the dipole moment
So,  Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev


Q 13.  An electric field Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI and ϕII respectively. The difference between  (ϕI  - ϕII) is (in Nm2 /C) ________.    [2020]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRevAns:
-48
Solution:
Given that
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev 
Electric flux through closed area is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Electric flux through surface ABCD is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Electric flux through surface BCGF is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Here x = 3 so ϕII  = 16 x 3 = 48 Nm2 /C
therefore,
ϕI  - ϕII = 0 - 48 = -48Nm2/C

Q 14. For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its center. Then value of h is    [2019]
(1) R/√5
(2) R/√2
(3) R
(4) R√2
Ans:
(2)
Solution:
Let the total charge on the ring is Q then, electric field at the axis of a charged ring is given by
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
For maximum electric field, Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 15. Three charges +Q, q, +Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by +Q, placed at x = 0, is zero, then value of q is    [2019]
(1) –Q/4
(2) +Q/2
(3) +Q/4
(4) –Q/2
Ans:
(1)
Solution:
Charge q is placed at center so, it is equilibrium condition for +Q placed at x = 0 For equilibrium,
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 16. Two point-charges Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev and q2 (−25μC) are placed on the x-axis at x = 1 m and x = 4m, respectively. The electric field (in V/m) at a point y = 3 m on y-axis is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev     [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(1)
Solution:
We know that electric field vector is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Let E1 and E2 are the electric field of q1 and q2.
For charge q1 = √10 μC
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Therefore, total electric field vector is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 17. Charge is distributed within a sphere of radius R with a volume charge density Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev where A and a are constants. If Q is the total charge of this charge distribution, the radius R is    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(2)
Solution:
Let Q be charge is distribution within a sphere of radius R
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 18. Two electric dipoles- A, B with respective dipole moments Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev are placed on the x-axis with a separation R, as shown in the figure.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
The distance from A at which both of them produce the same potential is    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans: 
(4)
Solution:
On the x axis in left on A potential due to A and B. But potential due to A is higher than that of B, between sign of potential due to A and B is opposite. So, potential can be same only in right of B on x-axis.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Therefore, distance
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 19. Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, −2) and (0, −2). The work required to put a fifth charge Q at the origin of the coordinate system will be    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(2)
Solution:
Potential at origin is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Now, work done, W = qΔV
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 20. Charges –q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the mid-point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P′ such that OP' = (y/3), the force on Q will be close to  Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(1) 3F
(2) F/3
(3) 9F
(4) 27F
Ans: 
(4)
Solution:
Electric field on equatorial plane of dipole is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Dividing Eq. (2) by Eq. (1), we get
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
⇒ F' = 27F

Q 21. Three charges Q, +q and +q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

(1) -q/(1+√2)

(2) +q

(3) -2q

(4) -√2q/(√2+ 1)
Ans:
(4)
Solution:
Let a be the two equal sides of isosceles triangle.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Thus, net electrostatic energy is given as
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Since, the net electrostatic energy is zero.
Therefore,
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 22. Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure:     [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(4)
Solution:
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Dipole moment is P1 = ql cos θ
= ql cos 30°    (1)
P2 = ql cos θ
= ql cos 30°    (2)
Now, net dipole moment is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Since, direction of dipole moment is negative y axis.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 22. A parallel plate capacitor is made of two square plates of side ‘a’, separated by a distance d (d ≪ a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. The value of capacitance of this arrangement is    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(2)
Solution:
Consider a small element dx at a distance x.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Integrating both the sides, we get
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 23. A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans: 
(N*)
Answer does not match with any given options
Solution:
The arrangement of capacitor shown in the figure
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
This Answer does not match with any given options.

Q 24. A parallel plate capacitor is of area 6 cm2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K1 = 10, K2 = 12 and K3 = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(1) 4
(2) 14
(3) 12
(4) 36

Ans: (3)
Solution:
Let K be the dielectric constant of material
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 25. A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(4)
Solution:
Let potential at distance r from then common center is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
But surface change density for all are same.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Put these values in Eq. (1), we get
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 26. A parallel-plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is     [2019]
(1) 692 pJ
(2) 508 pJ
(3) 560 pJ
(4) 600 pJ
Ans:
(2)
Solution:
Initial energy of capacitor
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Since battery is disconnected so charge remains same.
Final energy of capacitor is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Therefore, work done is
W = Ui – U
= 600 – 92 = 508 pJ

Q 27. The given graph shows variation (with distance r from centre) of    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(1) electric field of a uniformly charged sphere.
(2) potential of a uniformly charged spherical shell.
(3) potential of a uniformly charged sphere.
(4) electric field of a uniformly charged spherical shell.

Ans: (2)
Solution:
Growth and decay of current in LR circuit is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Therefore, the given charge shows the potential of a uniformly charged spherical shell.

Q 28. In the figure shown below, the charge on the left plate of the 10 μF capacitor is −30 μC. The charge on the right plate of the 6 μF capacitor is    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(1) + 18 μC
(2) + 12 μC
(3) −18 μC
(4) −12 μC
Ans:
(1)
Solution:
6 μF and 4 μF are parallel to each other. Total charge is 30 μF.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Therefore, charge on right plate is +18 μC.

Q 29. An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10−29 C m. What is the potential energy of the electric dipole?    [2019]
(1) −20 × 10−18 J
(2) −7 × 10−27 J
(3) −10 × 10−29 J
(4) −9 × 10−20 J
Ans:
(2)
Solution:
Potential energy is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 30. Seven capacitors, each of capacitance 2 μF, are to be connected in a configuration to obtain an effective capacitance of Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev. Which of the combinations, shown in figures below, will achieve the desired value?     [2019]
(1)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(2)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(3)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(4)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(2)
Solution: Effective capacitance is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Three capacitors are in parallel.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 30. There is a uniform, spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that best represents the speed V(R(t)) of the distribution as a function of its instantaneous radius R(t) is    [2019]
(1)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(2)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(3)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(4)
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(3)
Solution:
At any instant time t total energy of charge distribution is constant.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Also the slope of V-R curve will go on decreasing.
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 31. In the circuit shown, find C if the effective capacitance of the whole circuit is to be 0.5 μF. All values in the circuit are in μF.   [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Ans:
(1)
Solution:
From given circuit
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

Q 32. A parallel-plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is    [2019]
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
(1)  7.85 × 10−10 C
(2) 6.85 × 10−10 C
(3) 8.85 × 10−10 C
(4) 9.85 × 10−10 C

Ans: (3)
Solution:
By Gauss law electric field is
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev
Previous year questions (2016-20) - Electrostatics (Part - 1) Notes | EduRev

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