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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]****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

So, total capacitance is given by

For αd << 1 , we have

Now, using binomial expansion, we get

Ans:

Given that, C = 60 pF = 60 × 10

We know that, energy when capacitor is fully charged is given by

Energy when capacitor is disconnected and connected to uncharged capacitor is

Where C' = C+C = 2C and V' = V/2

Then,

So, energy lost is**Q 3. Effective capacitance of parallel combination two capacitors C _{1} and C_{2} is 10 μF. When these capacitors are individually connected to a voltage source of 1 V, the energy stored in the capacitor C_{2} is 4 times that of C_{1}. If these capacitors are connected in series, their effective capacitance will be [2020]** (3)

(1) 4.2 μF

(2) 3.2 μF

(3) 1.6 μF

(4) 8.4 μF

Ans:

Given that

C

Energy stored in capacitor is given by

We know that

⇒C

Solving Eqs. (1) and (2), we get

C

Effective capacitance of series combination is

(1) R

(2) (R

(3) R

(4) (R

Given that

We know that, electric filed is given by

So,

We know that, electrostatic potential is given by

So,

Ans:

Let small element dx at distance x has capacitance dC.

We know that, capacitance of parallel plat capacitor is given by

So,

Since α is very small so tan α ≈ α .

Therefore,

(1) Q

(2) Q

(3) Q

(4) Q

Ans:

In first circuit diode is reversed biased so it is open circuit.

so, Q

In second circuit diode is forward biased so it it short circuit.

So, current flow in the circuit at time t is given by

At t = RC, we get

Ans:

Given that

Electric filed due to infinite plane is given by

For first infinite plane, electric filed is given by

For second infinite plane, electric filed is given by

So, resultant electric field in region is**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]****Ans: **(1)**Solution:****Given that, **

Q_{A} = -4q, Q_{B} = 2q, Q_{C} = -2q

x component of electric field due to point charge A at centre O is given by

x component of electric field due to point charge B at centre O is given by

x component of electric field due to point charge C at centre O is given by

Total x component electric field due to point charges A, B, and C at centre O is given by

E_{T} = E_{A} + E_{B} + E_{C}**Q 9. In finding the electric field using Gauss law the formula is applicable. In the formula ε _{0 }is permittivity of free space, A is the area of Gaussian surface and q_{enc} is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?**(2)

(1) Only when the Gaussian surface is an equipotential surface.

(2) Only when the Gaussian surface is an equipotential surface and is constant on the surface.

(3) Only when = constant on the surface.

(4) For any choice of Gaussian surface. [2020]

Ans.

Gauss law is given by

Gauss law only holds when gaussian surface is an equipotential surface and electric field is constant throughout the surface.

Given that v

Coulomb force is given by F = qE

⇒ ma = qE

⇒ a = qE/m

Final velocity of charge particle in electric filed is

Only option (2) graph satisfy the condition.

(1) 21/34

(2) 18/34

(3) 17/54

(4) 18/54

Ans.

Given that

So, cavity in solid sphere has density

Electric filed of a solid sphere is given by

Electric filed at point A is only due to cavity given by

(1)

Electric filed at point B is due to solid sphere and cavity given by

(2)

From Eqs. (1) and (2), we get

Ans:

Given that

Since 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,

**Q 13. An electric field passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕ _{I}_{ }and ϕ_{II} respectively. The difference between (ϕ_{I}_{ }- ϕ_{II}) is (in Nm^{2} /C) ________. [2020]** -48

Ans:

Given that

Electric flux through closed area is given by

Electric flux through surface ABCD is

Electric flux through surface BCGF is

Here x = 3 so ϕ

therefore,

ϕ

**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√2Ans:** (2)

Let the total charge on the ring is Q then, electric field at the axis of a charged ring is given by

For maximum electric field,

(1) –Q/4

(2) +Q/2

(3) +Q/4

(4) –Q/2

Ans:

Charge q is placed at center so, it is equilibrium condition for +Q placed at x = 0 For equilibrium,

[2019]

Ans:

We know that electric field vector is

Let E_{1} and E_{2} are the electric field of q_{1} and q_{2}.

For charge q_{1} = √10 μC

Therefore, total electric field vector is**Q 17. Charge is distributed within a sphere of radius R with a volume charge density where A and a are constants. If Q is the total charge of this charge distribution, the radius R is [2019]Ans:** (2)

The distance from A at which both of them produce the same potential is [2019]

Ans:

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.

Therefore, distance

Ans:

Potential at origin is

Now, work done, W = qΔV

(1) 3F

(2) F/3

(3) 9F

(4) 27F

Ans:

Electric field on equatorial plane of dipole is

Dividing Eq. (2) by Eq. (1), we get

⇒ F' = 27F

**(1) -q/(1+√2)**

**(2) +q**

**(3) -2q**

**(4) -√2q/(√2+ 1)Ans:** (4)

Let a be the two equal sides of isosceles triangle.

Thus, net electrostatic energy is given as

Since, the net electrostatic energy is zero.

Therefore,

**Ans:** (4)

Dipole moment is P

= ql cos 30° (1)

P

= ql cos 30° (2)

Now, net dipole moment is

Since, direction of dipole moment is negative y axis.

Ans:

Consider a small element dx at a distance x.

Integrating both the sides, we get

**Q 23. A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K _{1}, K_{2}, K_{3}, K_{4} arranged as shown in the figure. The effective dielectric constant K will be [2019]**(N*)

Ans:

Answer does not match with any given options

The arrangement of capacitor shown in the figure

This Answer does not match with any given options.

(1) 4

(2) 14

(3) 12

(4) 36

Let K be the dielectric constant of material

Ans:

Let potential at distance r from then common center is

But surface change density for all are same.

Put these values in Eq. (1), we get

(1) 692 pJ

(2) 508 pJ

(3) 560 pJ

(4) 600 pJ

Ans:

Initial energy of capacitor

Since battery is disconnected so charge remains same.

Final energy of capacitor is

Therefore, work done is

W = U

= 600 – 92 = 508 pJ

(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.

Growth and decay of current in LR circuit is

Therefore, the given charge shows the potential of a uniformly charged spherical shell.

(1) + 18 μC

(2) + 12 μC

(3) −18 μC

(4) −12 μC

Ans:

6 μF and 4 μF are parallel to each other. Total charge is 30 μF.

Therefore, charge on right plate is +18 μC.

(1) −20 × 10

(2) −7 × 10

(3) −10 × 10

(4) −9 × 10

Ans:

Potential energy is

(1)

(2)

(3)

(4)

Ans:

Three capacitors are in parallel.

(1)

(2)

(3)

(4)

Ans:

At any instant time t total energy of charge distribution is constant.

Also the slope of V-R curve will go on decreasing.

Ans:

From given circuit

(1) 7.85 × 10

(2) 6.85 × 10

(3) 8.85 × 10

(4) 9.85 × 10

By Gauss law electric field is

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