Question 1 :Define the term 'dielectric constant' of a medium in terms of capacitance of a capacitor.

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Part Answer1

Dielectric constant is the ratio of the capacitance of a capacitor with dielectric between the plates (Cm) to thecapacitance of the same capacitor with vacuum or air (Co) between the plates, i.e.,

K = CmCo

Question 2 :In a medium the force of attraction between two point electric charges, distance d apart is F. What distanceapart should these be kept in the same medium so that the force between them becomes (i) 3F (ii) F3 ?

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Part Answer1

For a given pair of point charges in a mediumF 1 d2(i) For the force to become 3F, the separation d must become d 3 .

Part Answer2

(ii) For the force to become F/3, the separation d must become 3 d.

Question 3 :Why does the electric field inside a dielectric decrease when it is placed in an external electric field?Suggested Answers

Part Answer1

An electric field Ep is induced inside the dielectric in a direction opposite to the direction of external electric field

E

O.

SCIENCE PART MOCK TEST - 1

TEST ID SCI001

Class - XII

ANSWER KEY

Question 1 :

Question 2 :

Question 3 :

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Part Answer2

Thus net field is

E

= E

O Ep

Question 4 :Describe schematically the equipotential surfaces corresponding to :(a) A constant electric field in the z-direction.

(b A field that uniformly increases in magnitude but remains in a constant (say, z) direction.(c) A single positive charge at the origin, and

(d) A uniform grid consisting of long equally spaced parallel charged wires in a plane.

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Part Answer1

(a) Planes parallel to x - y plane.

Part Answer2

(b) Planes parallel to x - y plane . Planes differing by a fixed potential get closer as field increases.

Part Answer3

(c) Concentric spheres with centre at the origin.

Part Answer4

(d) A periodically varying shape near the grid which gradually reaches the shape of planes parallel to the grid at fardistances.

Question 5 :An electrical technician requires a capacitance of 2F in his circuit across a potential difference of 1kV. Alarge number of 1F capacitors are available to him each of which can withstand a potential difference of not more than

400V. Suggest a possible arrangement that requires the minimum number of capacitors.Suggested Answers

Part Answer1

Total capacitance required = 2FPotential difference, V=1000VCapacity of capacitors available = 1FMaximum potential difference that a capacitor can withstand = 400VSuppose, there are 'm' rows and in each row there are 'n' capacitors joined in series.Potential across each row = 1000 voltPotential across each capacitor in a row = 1000n

Question 4 :

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Part Answer2

According to the condition 1000n = 400n = 2.5or n = 3

Part Answer3

Total capacitance in a row = = 13Capacitance of 'm' rows = m3According to requirement m3 = 2 m = 6

Part Answer4

Total no. of capacitors required = 36 = 18

Question 6 :An electric dipole with dipole moment 4109Cm is aligned at 30 with the direction of a uniform electricfield of magnitude 5104NC1. Calculate the magnitude of the torque acting on the dipole.

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Part Answer1

Here p = 4109Cm, = 30, E = 5104NC1

Torque = pE sin

Part Answer2

= 41095104sin30= 4510512 = 10

4Nm.

Question 7 :State Gauss's theorem in electrostatics. A charge of 17.7104 C is distributed uniformly over a largesheet of area 200 m2. Calculate the electric field intensity at a distance 20 cm from it in air.

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Question 6 :

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Part Answer1

Gauss' theorem: Total electric flux through a closed surface is 10 times the magnitude of net charge enclosed by the

surface.

E =q0

Part Answer2

b( ) q = 17.7104 C, A = 200m2

E = 20 =q

2A0 = qA

Part Answer3

E = 17.7104

22008.8541012

Part Answer4

= 5105NC1

Question 8 :Electric field intensity at point 'B' due to a point charge 'Q' kept at point 'A' is 24 NC-1 and the electricpotential at point 'B' due to same charge is 12 JC-1. Calculate the distance AB and also the magnitude of charge Q.

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Part Answer1

E = 24NC1, V = 12 JC1

E = kQr2

...(i)

V = kQr ...(ii)i.e. E = Vr

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Part Answer2

r = VE =1224 = 0.5m

Part Answer3

From equation (ii)Q = Vrk =

120.59109

C

Part Answer4

= 6.71010C

Question 9 :An electric dipole of length 4 cm, when placed with its axis making an angle of 60 with a uniform electricfield experiences a torque of 4 3 Nm. Calculate the (i) magnitude of the electric field (ii) potential energy of the dipole, if

the dipole has charges of 8 nC.Suggested Answers

Part Answer1

Given : 2l = 4102m; q = 8109 C; = 60 = 4 3 Nm

i( ) = pE

= q 2l( ) E sinE = q 2l( )sin

= 4 3 281094102 3

Part Answer2

E = 141011 = 0.251011N C

E = 2.51010N C

Part Answer3

ii( ) Here = 60p = q2l = 81094102p = 321011CmE = 2.51010N C or V m

PE = pE cos= 3210112.51010cos60

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Part Answer4

P.E. = 3210112.51010 12

P.E. = 4 J

Question 10 :Deduce an expression for the capacitance of a parallel plate capacitor with air as the medium between theplates.

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Part Answer1

Consider a capacitor with surface charge density on its plates. Suppose area of each plate is 'A' and separationbetween the plates is 'd'.

Part Answer2

We know Q = CV

C = QVHere Q = A 1( )

V = EodV = o d Eo =

o

2( )

Part Answer3

From equations (1) and (2), we getC = Ad o

Part Answer4

C = o Ad

Question 11 :Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational forcebetween two stationary point masses, both have inverse-square dependence on the distance between the charges/masses.(a) Compare the strength of these forces by determining the ratio of their magnitude (i) for an electron and a proton and

(ii) for two protons.(b) Estimate the accelerations for electron and proton due to the electrical force of their mutual attraction when they are 1

= 1010m apart?mp = 1.671027kg, me = 9.111031kg

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Question 10 :

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Part Answer1

(a)(i) Consider an electron and a proton at a distance r apart.

Electrostatic force, Fe = ke2

r2and Gravitational force

FG =Gmpmer2

FeFG = ke

2Gmpme

Part Answer2

=9109 1.61019

2

6.6710119.1110311.671027

= 2.31039

Part Answer3

(ii) For two protons at a distance r apart

FeFG

= ke

2

G mp2

=9109 1.61019

2

6.671011 1.6710272

= 1.31036

Part Answer4

(b) F = ke2

r2

=9109 1.61019

2

10102 = 2.310

8N

As F=ma

Part Answer5

ae = Fme =2.3108

9.111031= 2.51022m s2

ap = Fmp =2.3108

1.671027= 1.41019m s2

Question 12 :What is electric flux? Write its S.I. units. Using Gauss theorem, deduce an expression for the electric fieldat a point due to a uniformly charged infinite plane sheet.

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Part Answer1

The total number of electric lines of force passing through the area normally is called its electric flux.

Part Answer2

Its SI unit is Vm or Nm2/C.

Part Answer3

Consider an infinite plane sheet of charge having surface charge density . Let P be a point at a distance r from it. Letus consider a Gaussian surface in the form of a cylinder of length 2r and area of cross-section A arranged such that itslateral surface is parallel to the field lines and its ends are normal to the field lines as shown.

Part Answer4

For lateral surface

E

.d A

= 0 E A

For end faces of the cylinder

E

.d A

= EAand q = A

Part Answer5

According to Gauss' theorem

E

.d A

= q0On including both circular faces E. 2A( ) = A0

Part Answer6

E = 20

Question 13 :(a) Why two electric lines of force do not intersect each other?(b) Draw one equipotential surface (i) in a uniform electric field and (ii) for a point charge Q < 0( ).

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Question 13 :

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(a) Two electric lines of force do not intersect each other because then there will be two tangents at the point givingtwo directions of electric field, which is not possible.

Part Answer2

Part Answer3

(ii)

Question 14 :Two point charges 5 108 C and 2 108 C are separated by a distance of 20 cm in air asshown in the figure.

(i) Find at what distance from point A would the electric potential be zero.

(ii) Also calculate the electrostatic potential energy of the system.Suggested Answers

Part Answer1

Given : qA = 5108C, qB = 2108 C, r = 20cm

(i) Suppose electr